Chapter 9: Differential Equations

The given differential equation is $$\frac{d^4 y}{dx^4} + \sin(y''') = 0$$.

Order: The highest order derivative occurring in the equation is $$\frac{d^4 y}{dx^4}$$, a derivative of order $$4$$. Hence the order is $$4$$.

Degree: The equation contains the term $$\sin(y''')$$, in which the derivative $$y''' = \frac{d^3 y}{dx^3}$$ appears inside a sine function. Hence the equation is not a polynomial equation in its derivatives, and so its degree is not defined.

The given differential equation is $$y' + 5y = 0$$.

Order: The highest order derivative occurring in the equation is $$y'$$, a derivative of order $$1$$. Hence the order is $$1$$.

Degree: The equation is a polynomial equation in the derivative $$y'$$, and the highest power of $$y'$$ is $$1$$. Hence the degree is $$1$$.

The given differential equation is $$\left(\frac{ds}{dt}\right)^4 + 3s \frac{d^2 s}{dt^2} = 0$$.

Order: The highest order derivative occurring in the equation is $$\frac{d^2 s}{dt^2}$$, a derivative of order $$2$$. Hence the order is $$2$$.

Degree: The equation is a polynomial equation in the derivatives $$\frac{ds}{dt}$$ and $$\frac{d^2 s}{dt^2}$$. The power of the highest order derivative $$\frac{d^2 s}{dt^2}$$ is $$1$$. Hence the degree is $$1$$.

The given differential equation is $$\left(\frac{d^2 y}{dx^2}\right)^2 + \cos\left(\frac{dy}{dx}\right) = 0$$.

Order: The highest order derivative occurring in the equation is $$\frac{d^2 y}{dx^2}$$, a derivative of order $$2$$. Hence the order is $$2$$.

Degree: The equation contains the term $$\cos\left(\frac{dy}{dx}\right)$$, in which the derivative $$\frac{dy}{dx}$$ appears inside a cosine function. Hence the equation is not a polynomial equation in its derivatives, and so its degree is not defined.

The given differential equation can be written as $$\frac{d^2 y}{dx^2} - \cos 3x - \sin 3x = 0$$.

Order: The highest order derivative occurring in the equation is $$\frac{d^2 y}{dx^2}$$, a derivative of order $$2$$. Hence the order is $$2$$.

Degree: The equation is a polynomial equation in the derivative $$\frac{d^2 y}{dx^2}$$ (the terms $$\cos 3x$$ and $$\sin 3x$$ involve only the independent variable $$x$$, not any derivative). The power of $$\frac{d^2 y}{dx^2}$$ is $$1$$. Hence the degree is $$1$$.

The given differential equation is

$$ (y''')^2 + (y'')^3 + (y')^4 + y^5 = 0 $$

Step 1 – Check whether the degree is defined
A degree is defined only when the equation can be written as a polynomial in the dependent variable $$y$$ and in all its derivatives. Here every term – $$ (y''')^2,\,(y'')^3,\,(y')^4,\,y^5 $$ – is an integral (whole-number) power of $$y$$ or of one of its derivatives. Hence the equation is a polynomial in the derivatives and its degree is therefore defined.

Step 2 – Order of the differential equation
The highest derivative present is the third derivative $$y'''$$. Therefore

$$ \text{order} = 3 $$

Step 3 – Degree of the differential equation
Because the equation is already a pure polynomial, we simply inspect the exponent of the highest-order derivative $$y'''$$. It appears as $$ (y''')^2 $$. Thus

$$ \text{degree} = 2 $$

Conclusion
The differential equation has order 3 and degree 2.

Step 1 – Write the equation clearly
$$y''' + 2y'' + y' = 0$$

Step 2 – Find the highest order derivative
The derivatives present are $$y'$$, $$y''$$ and $$y'''$$. The highest of these is the third derivative $$y'''$$.
Therefore, order = 3.

Step 3 – Check whether the degree is defined
A degree is defined only when the differential equation can be expressed as a polynomial in the derivatives of $$y$$, with no fractional or negative powers of the derivatives.
The given equation is already a polynomial in the derivatives and every derivative appears to the first power only. Hence the equation satisfies the condition for the degree to be defined.

Step 4 – Determine the degree
Since each derivative is raised to the power 1 and the highest power involved is 1, the degree = 1.

Conclusion
Order = 3, Degree = 1.

The given differential equation is

$$\frac{dy}{dx}+y=e^x$$

Step 1 – Order
The highest derivative appearing is the first derivative $$\frac{dy}{dx}$$. Hence the order is 1.

Step 2 – Degree
Write the equation in polynomial form in the derivatives:

$$\frac{dy}{dx}+y-e^x=0$$

The highest (and only) power of the highest-order derivative $$\frac{dy}{dx}$$ is 1. Therefore the degree is 1.

Thus, the differential equation has order 1 and degree 1.

The given differential equation is

$$y'' + (y')^2 + 2y = 0$$

• Order
  The highest derivative that appears is the second derivative $$y''$$. Therefore, the order of the differential equation is 2.
• Degree
  The equation is already a polynomial in $$y'', y'$$ and $$y$$. The highest power (exponent) of the highest–order derivative $$y''$$ is 1. Hence, the degree of the differential equation is 1.

Given differential equation

$$y'' + 2y' + \sin y = 0$$

1. Order
   The highest order derivative present is $$y''$$ (second derivative).
   Therefore, the order is $$2$$.
2. Degree
   First write the equation free from radicals or fractional powers of the derivatives. The given form already satisfies this condition: the derivatives $$y''$$ and $$y'$$ appear only to the first power.
   Since the equation is a polynomial in the derivatives (there are no non-polynomial functions of $$y'$$ or $$y''$$), the highest power of the highest order derivative, $$y''$$, is $$1$$.
   Hence the degree is $$1$$.

Thus, the differential equation has order $$2$$ and degree $$1$$.

The given differential equation is $$\left(\frac{d^2 y}{dx^2}\right)^3 + \left(\frac{dy}{dx}\right)^2 + \sin\left(\frac{dy}{dx}\right) + 1 = 0$$.

For the degree of a differential equation to be defined, the equation must be expressible as a polynomial in each of its derivatives after removing any radicals or fractional powers.

The terms $$\left(\frac{d^2 y}{dx^2}\right)^3$$ and $$\left(\frac{dy}{dx}\right)^2$$ are indeed polynomial in the derivatives, but the term $$\sin\left(\frac{dy}{dx}\right)$$ is a transcendental (non-polynomial) function of $$\frac{dy}{dx}$$. Hence the whole equation is not a polynomial in its derivatives.

Because this condition fails, the degree of the differential equation is not defined.

Therefore, the correct choice is (D).

First write the given differential equation clearly:

$$2x^2 \frac{d^2 y}{dx^2} - 3 \frac{dy}{dx} + y = 0$$

Recall the definition:

• The order of a differential equation is the highest order of the derivative of the unknown function that appears in the equation.

Identify all derivatives present in the equation:

• $$\frac{d^2 y}{dx^2}$$  — this is a second-order derivative.
• $$\frac{dy}{dx}$$  — this is a first-order derivative.
• The function $$y$$ itself (which can be considered the 0th-order derivative).

The highest order derivative that appears is $$\frac{d^2 y}{dx^2}$$, i.e. order 2.

Therefore, the order of the differential equation is $$2$$, which matches option (A).

Given $$y = e^{-3x}$$.

First derivative

Using the chain rule,

$$\frac{dy}{dx} = \frac{d}{dx}\bigl(e^{-3x}\bigr) = e^{-3x}\cdot(-3) = -3e^{-3x}.$$

Second derivative

$$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\bigl(-3e^{-3x}\bigr) = -3\cdot e^{-3x}\cdot(-3) = 9e^{-3x}.$$

Verification in the differential equation

Substitute $$\frac{d^{2}y}{dx^{2}},\;\frac{dy}{dx}$$ and $$y$$ in $$\frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} - 6y = 0$$:

$$\text{L.H.S.}=9e^{-3x}+(-3e^{-3x})-6e^{-3x}=(9-3-6)e^{-3x}=0\cdot e^{-3x}=0.$$

The left-hand side is zero, so the equation is satisfied.

Hence $$y = e^{-3x}$$ is a solution of the given differential equation.

We have to show that the function $$y = a\cos x + b\sin x$$, where $$a, b \in \mathbf{R}$$, satisfies the differential equation $$\dfrac{d^2 y}{dx^2} + y = 0$$.

Step 1 – First derivative
Differentiate $$y$$ with respect to $$x$$:

$$\frac{dy}{dx}=\frac{d}{dx}\,(a\cos x+b\sin x)=a\,( -\sin x)+b\,(\cos x)=-a\sin x+b\cos x.$$

Step 2 – Second derivative
Differentiate $$\dfrac{dy}{dx}$$ once more:

$$\frac{d^2y}{dx^2}=\frac{d}{dx}\,(-a\sin x+b\cos x)=-a\cos x-b\sin x.$$

Step 3 – Substitute into the differential equation

Left-hand side (LHS) of the given equation:

$$\frac{d^2y}{dx^2}+y=(-a\cos x-b\sin x)+(a\cos x+b\sin x).$$

Step 4 – Simplify

Combine like terms:

$$(-a\cos x+a\cos x)+(-b\sin x+b\sin x)=0+0=0.$$

Thus $$\dfrac{d^2y}{dx^2}+y=0,$$ so the given function indeed satisfies the differential equation.

We have to check whether the function $$y = e^x + 1$$ satisfies the differential equation $$y'' - y' = 0$$.

Step 1: First derivative

$$y' = \dfrac{dy}{dx} = \dfrac{d}{dx}(e^x) + \dfrac{d}{dx}(1) = e^x + 0 = e^x.$$

Step 2: Second derivative

$$y'' = \dfrac{d^2y}{dx^2} = \dfrac{d}{dx}(e^x) = e^x.$$

Step 3: Substitute in the differential equation

$$y'' - y' = e^x - e^x = 0.$$

The left–hand side equals the right–hand side (which is $$0$$), hence the given function satisfies the differential equation.

Therefore, $$y = e^x + 1$$ is a solution of $$y'' - y' = 0$$.

We are asked to check whether the function $$y = x^2 + 2x + C$$ satisfies the differential equation $$y' - 2x - 2 = 0$$, where $$C$$ is an arbitrary constant.

Step 1 : Find the derivative of $$y$$

Differentiate term by term with respect to $$x$$:

$$y' = \frac{d}{dx}(x^2) + \frac{d}{dx}(2x) + \frac{d}{dx}(C) = 2x + 2 + 0 = 2x + 2.$$

Step 2 : Substitute $$y'$$ into the differential equation

L.H.S. of the given differential equation is

$$y' - 2x - 2 = (2x + 2) - 2x - 2.$$

Simplifying, we get

$$y' - 2x - 2 = 2x + 2 - 2x - 2 = 0.$$

Step 3 : Conclusion

The left-hand side equals the right-hand side (which is $$0$$) for every real $$x$$. Therefore, the function $$y = x^2 + 2x + C$$ is indeed a solution of the differential equation $$y' - 2x - 2 = 0$$.

Given function
$$y = \cos x + C$$ where $$C$$ is an arbitrary constant.

Step 1 : Differentiate $$y$$
$$\frac{dy}{dx} = y' = \frac{d}{dx}(\cos x) + \frac{d}{dx}(C) = -\sin x + 0 = -\sin x.$$

Step 2 : Substitute in the differential equation
The differential equation is $$y' + \sin x = 0.$$ Replacing $$y'$$ by $$-\sin x$$, we get
$$(-\sin x) + \sin x = 0.$$

Step 3 : Verify
The left-hand side is $$0$$, which equals the right-hand side. Hence the equation is satisfied for every value of $$x$$ and for every constant $$C$$.

Conclusion
Therefore, $$y = \cos x + C$$ is indeed a solution of the differential equation $$y' + \sin x = 0$$.

Step 1 : Differentiate y with respect to x

The given function is $$y=(1+x^2)^{1/2}$$.
Using the chain rule,

$$\frac{dy}{dx}=\frac12\,(1+x^2)^{-1/2}\;\times\;2x$$

Simplifying,

$$y'=\frac{x}{\sqrt{1+x^2}}$$

Step 2 : Compute the right–hand side of the given DE

The differential equation to be verified is $$y'=\frac{xy}{1+x^2}$$.
Substitute the given expression for y:

$$\frac{xy}{1+x^2}=\frac{x\,\sqrt{1+x^2}}{1+x^2}$$

Cancel the common factor $$\sqrt{1+x^2}$$ in numerator and denominator:

Since $$1+x^2=(\sqrt{1+x^2})^2$$, we have

$$\frac{x\,\sqrt{1+x^2}}{1+x^2}=\frac{x\,\sqrt{1+x^2}}{(\sqrt{1+x^2})^2}=\frac{x}{\sqrt{1+x^2}}$$

Step 3 : Compare LHS and RHS

We obtained
Left–hand side (the derivative): $$y'=\dfrac{x}{\sqrt{1+x^2}}$$
Right–hand side (from the DE): $$\dfrac{xy}{1+x^2}=\dfrac{x}{\sqrt{1+x^2}}$$

The two expressions are identical, hence the given function satisfies the differential equation.

Conclusion: The function $$y=\sqrt{1+x^2}$$ is indeed a solution of the differential equation $$y'=\frac{xy}{1+x^2}$$.

Given function: $$y = \mathrm{A}x$$, where $$\mathrm{A}$$ is a constant.

Step 1 — Differentiate

$$y' = \dfrac{dy}{dx} = \mathrm{A}$$

Step 2 — Substitute in the differential equation

Left–hand side (LHS): $$xy' = x(\mathrm{A}) = \mathrm{A}x$$

Right–hand side (RHS): $$y = \mathrm{A}x$$

Step 3 — Compare LHS and RHS

LHS = RHS for every $$x \neq 0$$.

Therefore, $$y = \mathrm{A}x$$ is indeed a solution of the differential equation $$xy' = y$$.

Let the proposed solution be $$y = x \sin x$$.

1. Compute $$y'$$.

By the product rule,

$$y' = \dfrac{d}{dx}(x \sin x) = \sin x + x \cos x.$$

2. Left-hand side of the differential equation.

$$\text{LHS} = x y' = x(\sin x + x \cos x) = x \sin x + x^2 \cos x.$$

3. Right-hand side of the differential equation.

Compute the expression under the square root:

$$x^2 - y^2 \;=\; x^2 - (x \sin x)^2 \;=\; x^2\bigl(1 - \sin^2 x\bigr) \;=\; x^2 \cos^2 x.$$

The stated domain — namely $$x \neq 0$$ together with the condition $$x \gt y$$ or $$x \lt -y$$ — guarantees $$|x| \gt |y|$$, so $$x^2 - y^2 \geq 0$$ and the square root is well defined. Using $$\sqrt{a^2} = |a|$$,

$$\sqrt{x^2 - y^2} \;=\; \sqrt{x^2 \cos^2 x} \;=\; |x \cos x|.$$

Therefore

$$\text{RHS} \;=\; y + x\sqrt{x^2 - y^2} \;=\; x \sin x + x\,|x \cos x|.$$

4. Comparison of LHS and RHS.

For $$x \neq 0$$, the equality $$\text{LHS} = \text{RHS}$$ reduces to

$$x^2 \cos x \;=\; x\,|x \cos x|,$$

which is equivalent to $$x \cos x = |x \cos x|$$, i.e. $$x \cos x \geq 0$$.

5. Conclusion.

Hence the function $$y = x \sin x$$ satisfies the differential equation $$x y' = y + x\sqrt{x^2 - y^2}$$ on the part of the stated domain where $$x \cos x \geq 0$$ (for example, on $$0 \leq x \leq \tfrac{\pi}{2}$$, where both $$x \geq 0$$ and $$\cos x \geq 0$$). On this subset the verification is complete.

We have to show that the implicit relation

$$xy = \log y + C$$

indeed satisfies the differential equation

$$y' = \frac{y^{2}}{1-xy}\;(xy \neq 1).$$

Take $$y=y(x)$$ and differentiate the given relation with respect to $$x$$.

Step 1 : Differentiate both sides.

Left side: $$\frac{d}{dx}(xy)=x\,y'+y$$ (product rule).

Right side: $$\frac{d}{dx}(\log y + C)=\frac{1}{y}\,y'+0.$$

Hence

$$x\,y' + y = \frac{1}{y}\,y'.$$

Step 2 : Collect the $$y'$$ terms.

$$x\,y' - \frac{1}{y}\,y' = -y.$$

Factor $$y'$$:

$$y'\left(x-\frac{1}{y}\right)=-y.$$

Step 3 : Isolate $$y'.$$

$$y' = \frac{-y}{\displaystyle x-\frac{1}{y}}.$$

Rewrite the denominator over a common denominator $$y$$:

$$x-\frac{1}{y}=\frac{xy-1}{y}.$$

Therefore

$$y' = \frac{-y}{(xy-1)/y}=\frac{-y^{2}}{xy-1}.$$

Multiply numerator and denominator by $$-1$$:

$$y' = \frac{y^{2}}{1-xy}.$$

This is exactly the right–hand side of the required differential equation, provided $$xy\neq1$$ so that the denominator is non-zero.

Hence the given function satisfies the differential equation.

Let the given relation be written as
$$x = y - \cos y\;.$$

Differentiate both sides with respect to $$x$$ (implicit differentiation):

Left-hand side: $$\dfrac{d}{dx}(x)=1.$$
Right-hand side: $$\dfrac{d}{dx}\,[y-\cos y]=(1+\sin y)\,y'.$$ Hence

$$1=(1+\sin y)\,y' \;\;\Longrightarrow\;\; y'=\dfrac{1}{1+\sin y}.$$

Now evaluate the left side of the differential equation $$(y\sin y+\cos y+x)\,y'$$ using the obtained $$y'$$.

First simplify the bracket using the original relation $$x=y-\cos y$$:

$$y\sin y+\cos y+x=y\sin y+\cos y+(y-\cos y)=y\sin y+y.$$

Therefore

$$\bigl(y\sin y+\cos y+x\bigr)\,y'=\bigl(y\sin y+y\bigr)\,\dfrac{1}{1+\sin y} = y\,(\sin y+1)\,\dfrac{1}{1+\sin y}=y.$$

The right side of the given differential equation is also $$y$$, so the equality

$$(y\sin y+\cos y+x)\,y' = y$$

holds identically. Hence the function defined by $$x=y-\cos y$$ satisfies the differential equation.

Verified.

We are given the relation

$$x + y = \tan^{-1} y$$

and we have to show that it satisfies the differential equation

$$y^2\,y' + y^2 + 1 = 0.$$

Step 1 – Differentiate the given relation.

Differentiate both sides with respect to $$x$$. Remember that $$y = y(x)$$, so every appearance of $$y$$ must be treated using the chain rule.

Derivative of the left side:

$$\frac{d}{dx}(x) + \frac{d}{dx}(y) = 1 + y'.$$

Derivative of the right side:

$$\frac{d}{dx}\bigl(\tan^{-1} y\bigr) = \frac{1}{1 + y^2}\,y'.$$

Hence

$$1 + y' = \frac{y'}{1 + y^2}.$$

Step 2 – Eliminate the denominator.

Multiply both sides by $$1 + y^2$$:

$$(1 + y^2)(1 + y') = y'.$$

Step 3 – Expand and collect terms.

$$1 + y^2 + y' + y^2y' = y'.$$

Subtract $$y'$$ from both sides:

$$1 + y^2 + y^2y' = 0.$$

Step 4 – Write in the required form.

Re-ordering gives

$$y^2y' + y^2 + 1 = 0,$$

which is exactly the differential equation to be verified.

Therefore, the function defined implicitly by $$x + y = \tan^{-1} y$$ is a solution of $$y^2y' + y^2 + 1 = 0.$$

Given differential equation: $$x + y \dfrac{dy}{dx} = 0$$ with the condition $$y \neq 0$$.

Proposed solution: $$y = \sqrt{a^2 - x^2} ,\; x \in (-a,\,a).$$

Step 1: Differentiate the proposed solution.

Square both sides to simplify differentiation:

$$y^2 = a^2 - x^2.$$

Differentiating both sides with respect to $$x$$:

$$2y \dfrac{dy}{dx} = -2x.$$ Dividing by $$2y$$ (allowed because $$y \neq 0$$):

$$\dfrac{dy}{dx} = -\dfrac{x}{y}.$$

Step 2: Substitute $$\dfrac{dy}{dx}$$ and $$y$$ in the differential equation.

Left-hand side (LHS): $$x + y \dfrac{dy}{dx} = x + y\left(-\dfrac{x}{y}\right).$$

Simplify: $$x + y\left(-\dfrac{x}{y}\right) = x - x = 0.$$

Step 3: Compare with the right-hand side (RHS).

RHS of the differential equation is $$0.$$ Since LHS 3D RHS 3D $$0$$ for every $$x \in (-a,\,a)$$ (with $$y \neq 0$$), the given function satisfies the differential equation.

Conclusion: $$y = \sqrt{a^2 - x^2}$$ is indeed a solution of $$x + y \dfrac{dy}{dx} = 0$$ on the stated domain.

The order of a differential equation is the highest order of the derivative that appears in it.

For an ordinary differential equation of order $$n$$, its general solution must be sufficiently broad to contain $$n$$ linearly independent particular solutions; consequently, it involves exactly $$n$$ independent (arbitrary) constants. These constants are needed so that we can fit the solution to any prescribed set of $$n$$ initial (or boundary) conditions.

Therefore, for a differential equation of fourth order (i.e. $$n = 4$$), the general solution will contain:

$$4$$ independent arbitrary constants.

Thus the correct choice is (D).

Step 1 : Recall the definitions
For a differential equation of order $$n$$:

• The general solution contains exactly $$n$$ arbitrary constants.
• A particular solution is obtained by assigning specific numerical values to those constants; hence, no arbitrary constant is left.

Step 2 : Apply to the given order
Here $$n = 3$$ (third-order differential equation).

• General solution  ⟶  3 arbitrary constants.
• Particular solution  ⟶  all 3 constants fixed, so 0 arbitrary constants remain.

Step 3 : Choose the correct option
The number of arbitrary constants in the particular solution is $$0$$; therefore, option (D) is correct.

We are given the first-order differential equation

$$\frac{dy}{dx}=\frac{x+1}{2-y},\qquad (y\neq 2)$$

1. Separate the variables

Multiply both sides by $$2-y$$ and by $$dx$$ to collect the variables:

$$(2-y)\,dy=(x+1)\,dx$$

2. Integrate both sides

$$\int(2-y)\,dy = \int(x+1)\,dx$$

Evaluate the integrals term by term.

• Left side: $$\int(2-y)\,dy = \int2\,dy-\int y\,dy = 2y-\dfrac{y^{2}}{2}+C_{1}$$
• Right side: $$\int(x+1)\,dx = \int x\,dx+\int1\,dx = \dfrac{x^{2}}{2}+x+C_{2}$$

3. Combine the constants

Equate the two antiderivatives and absorb $$C_{1}$$ and $$C_{2}$$ into a single constant $$C$$:

$$2y-\dfrac{y^{2}}{2}=\dfrac{x^{2}}{2}+x+C$$

4. Simplify

Multiply by $$2$$ to remove the fraction:

$$4y-y^{2}=x^{2}+2x+C'$$

Rearrange all the terms on one side (renaming the constant again as $$C$$):

$$x^{2}+2x+y^{2}-4y=C$$

5. State the general solution

The required one-parameter family of solutions (valid for $$y\neq2$$) is

$$x^{2}+2x+y^{2}-4y=C,$$

where $$C$$ is an arbitrary real constant.

The given differential equation is

$$\frac{dy}{dx}=\frac{1+y^{2}}{1+x^{2}}.$$

This equation is separable, so we rewrite it by putting every expression containing $$y$$ on the left and everything containing $$x$$ on the right:

$$\frac{dy}{1+y^{2}}=\frac{dx}{1+x^{2}}.$$

Now integrate both sides with respect to their own variables:

$$\int\frac{dy}{1+y^{2}}=\int\frac{dx}{1+x^{2}}.$$

Using the standard integral $$\displaystyle\int\frac{du}{1+u^{2}}=\tan^{-1}u+C$$, we obtain

$$\tan^{-1}y = \tan^{-1}x + C,$$

where $$C$$ is the (real) constant of integration.

This implicit relation is already an acceptable general solution. If we wish, we can solve explicitly for $$y$$. Applying the tangent function to both sides,

$$y = \tan\bigl( \tan^{-1}x + C \bigr).$$

Thus the required general solution may be written in either of the following equivalent forms:

• $$\tan^{-1}y - \tan^{-1}x = C,$$
• $$y = \tan\bigl( C + \tan^{-1}x \bigr).$$

Given differential equation
$$\frac{dy}{dx} = -4x y^{2}$$

Step 1  (Separate the variables)
Bring all terms in $$y$$ to one side and all terms in $$x$$ to the other side:

$$\frac{dy}{y^{2}} = -4x\,dx$$

Step 2  (Integrate both sides)

Left side:
$$\int y^{-2}\,dy = -y^{-1} + C_{1}$$ because $$\int y^{n}\,dy = \frac{y^{n+1}}{n+1}$$ for $$n \neq -1$$ and here $$n = -2$$.

Right side:
$$\int -4x\,dx = -4 \cdot \frac{x^{2}}{2} + C_{2} = -2x^{2} + C_{2}$$.

Combine the two (absorb the constants into a single constant $$C$$):

$$-\frac{1}{y} = -2x^{2} + C$$

Step 3  (Simplify the constant)
Multiply by $$-1$$ to make the coefficient of $$\dfrac{1}{y}$$ positive:

$$\frac{1}{y} = 2x^{2} - C$$

Let $$C' = -C$$ (still an arbitrary constant). Then

$$\frac{1}{y} = 2x^{2} + C'$$

Step 4  (Use the initial condition)
We are told $$y = 1$$ when $$x = 0$$.

Substitute these values into the equation $$\dfrac{1}{y} = 2x^{2} + C'$$:

$$\frac{1}{1} = 2(0)^{2} + C' \;\Rightarrow\; 1 = 0 + C' \;\Rightarrow\; C' = 1$$

Step 5  (Write the particular solution)

Insert $$C' = 1$$ back into $$\dfrac{1}{y} = 2x^{2} + C'$$:

$$\frac{1}{y} = 2x^{2} + 1$$

Finally, solve for $$y$$:

$$y = \frac{1}{1 + 2x^{2}}$$

Therefore, the required particular solution is $$y = \dfrac{1}{1 + 2x^{2}}$$.

Given differential equation:

$$x\,dy = (2x^2 + 1)\,dx \quad (x \neq 0).$$

First isolate $$dy$$:

$$dy = \frac{2x^2 + 1}{x}\,dx.$$

Simplify the right-hand side:

$$dy = \left( \frac{2x^2}{x} + \frac{1}{x} \right)dx = \left(2x + \frac{1}{x}\right)dx.$$

Integrate both sides:

$$\int dy = \int \left(2x + \frac{1}{x}\right)dx.$$

Calculate the integrals term by term:

• $$\int 2x\,dx = x^2 + C_1$$
• $$\int \frac{1}{x}\,dx = \ln|x| + C_2$$

Combining the two antiderivatives into a single constant $$C = C_1 + C_2$$ gives

$$y = x^2 + \ln|x| + C.$$

The curve passes through $$(1,1)$$, so substitute $$x = 1$$ and $$y = 1$$:

$$1 = 1^2 + \ln|1| + C \;\;\Rightarrow\;\; 1 = 1 + 0 + C \;\;\Rightarrow\;\; C = 0.$$

Hence the required equation of the curve is

$$y = x^2 + \ln|x|.$$

Let the required curve be represented by $$y = y(x)$$. The slope of the tangent (that is, the derivative) is given as

$$\frac{dy}{dx} = \frac{2x}{y^{2}}.$$

This is a first-order differential equation that is separable. We rearrange the variables so that all terms containing $$y$$ are on one side and those containing $$x$$ on the other:

$$y^{2}\,dy = 2x\,dx.$$

Integrate both sides:

$$\int y^{2}\,dy = \int 2x\,dx.$$

Compute the antiderivatives:

$$\frac{y^{3}}{3} = x^{2} + C,$$

where $$C$$ is the constant of integration.

To find $$C$$ we use the fact that the curve passes through $$(-2,\,3)$$. Substitute $$x = -2$$ and $$y = 3$$ into the last equation:

$$\frac{(3)^{3}}{3} = (-2)^{2} + C \;\Longrightarrow\; \frac{27}{3} = 4 + C \;\Longrightarrow\; 9 = 4 + C.$$

Therefore, $$C = 5$$.

Substituting $$C = 5$$ back, we obtain

$$\frac{y^{3}}{3} = x^{2} + 5.$$

Multiplying by 3 to clear the denominator gives the required Cartesian equation of the curve:

$$y^{3} = 3x^{2} + 15.$$

Given that the principal increases continuously at a constant rate of 5 % per year, the amount $$P(t)$$ at time $$t$$ years satisfies the differential equation

$$\dfrac{\mathrm dP}{\mathrm dt}=0.05P$$

This is a first-order linear differential equation whose general solution is obtained by separating the variables:

$$\dfrac{1}{P}\,\mathrm dP=0.05\,\mathrm dt\quad\Rightarrow\quad \int\dfrac{1}{P}\,\mathrm dP=\int0.05\,\mathrm dt$$

$$\ln|P|=0.05t+C$$

Writing the constant of integration as $$\ln C_1$$ gives

$$P=C_1e^{0.05t}$$

At $$t=0$$, the principal is Rs 1000, so

$$1000=C_1e^{0.05\times0}=C_1\;\;\Rightarrow\;\;C_1=1000$$

Hence the amount after $$t$$ years is

$$P(t)=1000e^{0.05t}$$

The amount doubles when $$P(t)=2000$$:

$$2000=1000e^{0.05t}\;\;\Rightarrow\;\;2=e^{0.05t}$$

Taking the natural logarithm,

$$\ln2=0.05t\;\;\Rightarrow\;\;t=\dfrac{\ln2}{0.05}$$

Since $$\ln2\approx0.6931$$,

$$t\approx\dfrac{0.6931}{0.05}\approx13.86\text{ years}$$

Therefore, the principal will double in about 13.9 years (approximately 14 years).

We have to solve the first-order differential equation

$$\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}$$

Re-write it in differential form:

$$dy=\frac{1-\cos x}{1+\cos x}\,dx$$

Integrate both sides:

$$\int dy=\int\frac{1-\cos x}{1+\cos x}\,dx$$

Step 1   Simplify the integrand

Use the half-angle identities

• $$1-\cos x = 2\sin^2\dfrac{x}{2}$$
• $$1+\cos x = 2\cos^2\dfrac{x}{2}$$

Therefore

$$\frac{1-\cos x}{1+\cos x}=\frac{2\sin^2\dfrac{x}{2}}{2\cos^2\dfrac{x}{2}}=\tan^2\dfrac{x}{2}$$

Thus the equation becomes

$$\int dy=\int \tan^2\dfrac{x}{2}\,dx$$

Step 2   Integrate

Let $$\theta=\dfrac{x}{2}\;\Longrightarrow\;dx=2\,d\theta$$.

Then

$$\int \tan^2\dfrac{x}{2}\,dx=\int \tan^2\theta\,(2\,d\theta)=2\int\tan^2\theta\,d\theta$$

Recall $$\tan^2\theta=\sec^2\theta-1$$, so

$$2\int\tan^2\theta\,d\theta=2\int(\sec^2\theta-1)\,d\theta =2\bigl(\tan\theta-\theta\bigr)+C_1$$

Undo the substitution $$\theta=\dfrac{x}{2}$$:

$$2\bigl(\tan\theta-\theta\bigr)=2\left(\tan\dfrac{x}{2}-\dfrac{x}{2}\right)$$

Step 3   Write the general solution

Combining the two integrals gives

$$y=2\tan\dfrac{x}{2}-x+C$$

where $$C$$ is an arbitrary constant.

Hence, the general solution of the differential equation is

$$y=2\tan\dfrac{x}{2}-x+C.$$

We are given the first-order differential equation

$$\frac{dy}{dx} = \sqrt{4 - y^{2}}, \qquad -2 \lt y \lt 2.$$

This equation is separable; all terms involving $$y$$ can be moved to one side and $$dx$$ to the other.

1. Separate the variables.

$$\frac{dy}{\sqrt{4 - y^{2}}} = dx.$$

2. Integrate both sides.

$$\int \frac{dy}{\sqrt{4 - y^{2}}} = \int dx.$$

Left integral. Write the denominator as $$\sqrt{2^{2} - y^{2}}$$ and apply the standard formula

$$\int \frac{du}{\sqrt{a^{2} - u^{2}}} = \sin^{-1}\!\left(\frac{u}{a}\right) + C.$$

With $$u = y$$ and $$a = 2$$:

$$\int \frac{dy}{\sqrt{4 - y^{2}}} = \sin^{-1}\!\left(\frac{y}{2}\right) + C_{1}.$$

Right integral. $$\int dx = x + C_{2}.$$

Combining the two constants ($$C = C_{2} - C_{1}$$):

$$\sin^{-1}\!\left(\frac{y}{2}\right) = x + C.$$

3. Solve for $$y$$.

Taking sine on both sides,

$$\frac{y}{2} = \sin(x + C) \;\;\Longrightarrow\;\; y = 2\sin(x + C).$$

Since $$|\sin(x + C)| \le 1$$, the function automatically satisfies $$-2 \le y \le 2$$, which is consistent with the prescribed domain $$-2 \lt y \lt 2$$.

4. General solution.

$$y = 2\sin(x + C), \qquad C \in \mathbb{R}.$$

The given differential equation is

$$\frac{dy}{dx} + y = 1 \quad (y \neq 1).$$

This is a first-order linear ordinary differential equation of the standard form

$$\frac{dy}{dx} + P(x)\,y = Q(x)$$

with $$P(x) = 1$$ and $$Q(x) = 1$$.

Step 1: Integrating factor (I.F.).

$$\text{I.F.} = e^{\int P(x)\,dx} = e^{\int 1\,dx} = e^{x}.$$

Step 2: Multiply the equation by the I.F.

$$e^{x}\,\frac{dy}{dx} + e^{x}\,y = e^{x}.$$

Step 3: Recognise the left side as a perfect derivative.

Since $$\dfrac{d}{dx}(e^{x}y) = e^{x}\,\dfrac{dy}{dx} + e^{x}\,y$$, the equation becomes

$$\frac{d}{dx}\bigl(e^{x}y\bigr) = e^{x}.$$

Step 4: Integrate both sides with respect to $$x$$.

$$e^{x}y = e^{x} + C,$$

where $$C$$ is the constant of integration.

Step 5: Solve for $$y$$.

Dividing by $$e^{x}$$ (which is never zero):

$$y = 1 + C\,e^{-x}.$$

The condition $$y \neq 1$$ prohibits $$C = 0$$, but the form is otherwise unrestricted. Thus

$$y = 1 + C\,e^{-x}, \; C \in \mathbb{R}\setminus\{0\}.$$

This is the required general solution.

Step 1 : Identify useful derivatives
For both variables we know

• $$\dfrac{d}{dx}(\tan x)=\sec^2 x\,,$$ so $$\sec^2 x\,dx=d(\tan x)$$
• $$\dfrac{d}{dy}(\tan y)=\sec^2 y\,,$$ so $$\sec^2 y\,dy=d(\tan y)$$

Step 2 : Rewrite the given differential equation
The equation is

$$\sec^2 x\,\tan y\,dx+\sec^2 y\,\tan x\,dy=0$$

Substituting the differentials obtained above, let
$$u=\tan x,\;du=\sec^2 x\,dx\qquad\text{and}\qquad v=\tan y,\;dv=\sec^2 y\,dy.$$

Then the equation becomes

$$v\,du+u\,dv=0.$$

Step 3 : Recognise an exact differential
Using the product rule for differentials,

$$d(uv)=u\,dv+v\,du.$$

Comparing with the previous step we see that the left–hand side is exactly $$d(uv)$$. Hence

$$d(uv)=0.$$

Step 4 : Integrate
Integrating directly,

$$uv=C,$$ where $$C$$ is the constant of integration.

Step 5 : Replace u and v
Recall $$u=\tan x$$ and $$v=\tan y$$, so the general solution is

$$\tan x\,\tan y=C.$$

We have

$$ (e^{x}+e^{-x})\,dy-(e^{x}-e^{-x})\,dx=0. $$

Re-arrange to make $$dy$$ the subject:

$$ (e^{x}+e^{-x})\,dy=(e^{x}-e^{-x})\,dx. $$

Divide by $$(e^{x}+e^{-x})\,dx$$ (which is never zero) to obtain a first-order separable form:

$$ \frac{dy}{dx}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}. $$

Simplify the right–hand side by recognising hyperbolic functions:

$$ e^{x}-e^{-x}=2\sinh x, \quad e^{x}+e^{-x}=2\cosh x. $$

Hence

$$ \frac{dy}{dx}=\frac{2\sinh x}{2\cosh x}=\tanh x. $$

Integrate both sides with respect to $$x$$:

$$ \int dy=\int \tanh x\,dx. $$

Recall that $$\tanh x=\dfrac{\sinh x}{\cosh x}$$. Put $$u=\cosh x\ \Rightarrow\ du=\sinh x\,dx$$; then

$$ \int \tanh x\,dx=\int \frac{\sinh x}{\cosh x}\,dx=\int \frac{du}{u}=\ln|u|+C=\ln|\cosh x|+C. $$

Therefore

$$ y=\ln|\cosh x|+C. $$

Re-writing (absorbing signs into the constant if desired) gives the general solution

$$ y-\ln|\cosh x|=C'. $$

Given differential equation:

$$\frac{dy}{dx} = (1 + x^2)(1 + y^2)$$

Step 1 : Separate the variables

Bring every term containing $$y$$ to the left and every term containing $$x$$ to the right:

$$\frac{1}{1 + y^2}\,dy = (1 + x^2)\,dx$$

Step 2 : Integrate both sides

$$\int \frac{1}{1 + y^2}\,dy = \int (1 + x^2)\,dx$$

Left-hand integral (standard result):

$$\int \frac{1}{1 + y^2}\,dy = \tan^{-1} y + C_1$$

Right-hand integral:

$$\int (1 + x^2)\,dx = \int 1\,dx + \int x^2\,dx = x + \frac{x^3}{3} + C_2$$

Step 3 : Combine the constants

Let $$C = C_2 - C_1$$ (still an arbitrary constant). Therefore

$$\tan^{-1} y = x + \frac{x^3}{3} + C$$

Step 4 : Write the solution explicitly for $$y$$ (optional)

Taking tangent on both sides,

$$y = \tan\left( x + \frac{x^3}{3} + C \right)$$

Thus, the required general solution is either of the equivalent forms shown above.

We are given the differential equation

$$y \log y\,dx - x\,dy = 0$$

Rewrite it so that all terms containing $$y$$ are on one side and those containing $$x$$ are on the other.

$$y \log y\,dx = x\,dy \;\;\Longrightarrow\;\; \frac{dy}{dx} = \frac{y\log y}{x}$$

This form is separable. Bring every factor that involves $$y$$ to the left and every factor that involves $$x$$ to the right:

$$\frac{dy}{y\log y} = \frac{dx}{x}$$

Integrate both sides.

• For the left integral, set $$t = \log y \;\Rightarrow\; dt = \dfrac{dy}{y}$$. Then

$$\int\frac{dy}{y\log y} = \int\frac{dt}{t} = \log|t| + C_1 = \log|\log y| + C_1$$

• The right integral is the familiar logarithm:

$$\int\frac{dx}{x} = \log|x| + C_2$$

Combine the two integration constants into a single constant $$C$$ (because $$C = C_2 - C_1$$ is still arbitrary):

$$\log|\log y| = \log|x| + C$$

Use the property $$\log a - \log b = \log\dfrac{a}{b}$$ to rewrite and then exponentiate:

$$\log\Bigl|\frac{\log y}{x}\Bigr| = C \;\;\Longrightarrow\;\; \Bigl|\frac{\log y}{x}\Bigr| = e^{C}$$

Let $$k = e^{C} > 0$$ (absorbing the absolute–value sign and the possibility of a negative ratio into the arbitrary constant). Hence

$$\log y = kx$$

Finally, exponentiate once more:

$$y = e^{kx}$$

Putting $$A = e^{k} (>0)$$, the solution can be written even more simply as

$$y = A^{x}$$

where $$A$$ is an arbitrary positive constant. This is the required general solution.

Given differential equation:

$$x^{5}\,\frac{dy}{dx} = -y^{5}$$

Rewrite $$\displaystyle \frac{dy}{dx} = -\frac{y^{5}}{x^{5}}$$.

Separate the variables:

$$\frac{dy}{y^{5}} = -\frac{dx}{x^{5}}$$

Integrate both sides:

$$\int y^{-5}\,dy = -\int x^{-5}\,dx$$

Using $$\int t^{n}\,dt = \frac{t^{n+1}}{n+1} \;(n\neq -1)$$ we get

$$\frac{y^{-4}}{-4} = -\frac{x^{-4}}{-4} + C$$

Simplify:

$$-\frac{1}{4}y^{-4} = \frac{1}{4}x^{-4} + C$$

Multiply by 4:

$$-y^{-4} = x^{-4} + C_1$$

Move the term on the left or change the constant sign (since an arbitrary constant remains arbitrary):

$$x^{-4} + y^{-4} = C$$

Finally, writing with positive exponents in the denominators, the general solution is

$$\boxed{\displaystyle \frac{1}{x^{4}} + \frac{1}{y^{4}} = C}\;.$$

We have the first-order differential equation

$$\frac{dy}{dx}=\sin^{-1}x.$$

Write the differentials:

$$dy=\sin^{-1}x\,dx.$$

Integrate both sides with respect to $$x$$:

$$\int dy = \int \sin^{-1}x\,dx.$$

This gives

$$y = \int \sin^{-1}x\,dx + C,$$

where $$C$$ is the constant of integration. Hence we only need to evaluate the integral

$$I = \int \sin^{-1}x\,dx.$$

Step 1: Integrate by parts

Choose

• $$u = \sin^{-1}x \;\;\Rightarrow\;\; du = \frac{1}{\sqrt{1-x^{2}}}\,dx,$$
• $$dv = dx \;\;\Rightarrow\;\; v = x.$$

Then by the integration–by–parts formula $$\int u\,dv = uv - \int v\,du$$, we obtain

$$I = x\sin^{-1}x - \int \frac{x}{\sqrt{1-x^{2}}}\,dx.$$

Step 2: Evaluate the remaining integral

Consider

$$J = \int \frac{x}{\sqrt{1-x^{2}}}\,dx.$$

Put $$t = 1 - x^{2}\;\;\Rightarrow\;\; dt = -2x\,dx\;\;\Rightarrow\;\; x\,dx = -\tfrac12 dt.$$ Then

$$J = \int \frac{x}{\sqrt{1-x^{2}}}\,dx = -\tfrac12 \int \frac{dt}{\sqrt{t}}.$$

Since $$\int t^{-1/2}\,dt = 2\sqrt{t}$$, we get

$$J = -\tfrac12 \bigl(2\sqrt{t}\bigr) = -\sqrt{t} = -\sqrt{1-x^{2}}.$$

Step 3: Put it all together

Substituting $$J$$ back into $$I$$:

$$I = x\sin^{-1}x - (-\sqrt{1-x^{2}}) = x\sin^{-1}x + \sqrt{1-x^{2}}.$$

Step 4: General solution

Therefore

$$y = x\sin^{-1}x + \sqrt{1-x^{2}} + C.$$

This is the required general solution of the given differential equation.

We start with the given differential equation

$$e^x \tan y\,dx+(1-e^x)\sec^2 y\,dy=0$$

Shift one term to the other side so that the two differentials appear on opposite sides:

$$e^x \tan y\,dx=-(1-e^x)\sec^2 y\,dy$$

Divide by the non–zero factors $$\tan y\,(1-e^x)$$ to separate the variables:

$$\frac{e^x}{1-e^x}\,dx=-\frac{\sec^2 y}{\tan y}\,dy$$

Observe that each fraction is an exact derivative:

• Put $$t=1-e^x\;\Rightarrow\;dt=-e^x\,dx$$, hence $$\dfrac{e^x}{1-e^x}\,dx=-\dfrac{dt}{t}=-d\bigl(\ln|t|\bigr)$$.
• Since $$\dfrac{d}{dy}\bigl(\ln|\tan y|\bigr)=\dfrac{1}{\tan y}\sec^2 y$$, we have $$\dfrac{\sec^2 y}{\tan y}\,dy=d\bigl(\ln|\tan y|\bigr)$$.

Using these, rewrite the separated equation as

$$-d(\ln|1-e^x|)=-d(\ln|\tan y|)$$

Integrate both sides:

$$-\ln|1-e^x|=-\ln|\tan y|+C$$

(where $$C$$ is the constant of integration). Multiply by $$-1$$ and absorb the sign into the constant to get

$$\ln|1-e^x|=\ln|\tan y|+C'$$

Exponentiating gives

$$|1-e^x|=k\,|\tan y|\quad\text{with}\;k=e^{C'}>0$$

Since the constant $$k$$ can take any non-zero value, we drop the absolute signs and rename the constant as $$C$$ (positive, negative or zero):

$$1-e^x=C\,\tan y$$

Thus the required general solution is

$$1-e^x=C\tan y$$

Given differential equation:

$$(x^{3} + x^{2} + x + 1)\,\frac{dy}{dx} = 2x^{2} + x.$$

Rewrite as $$\dfrac{dy}{dx} = f(x)$$:

$$\frac{dy}{dx} = \frac{2x^{2} + x}{x^{3} + x^{2} + x + 1}.$$

Factorise the denominator:

$$x^{3} + x^{2} + x + 1 = x^{2}(x + 1) + (x + 1) = (x + 1)(x^{2} + 1).$$

Hence

$$\frac{dy}{dx} = \frac{2x^{2} + x}{(x + 1)(x^{2} + 1)}.$$

Integrating both sides with respect to $$x$$:

$$y = \int \frac{2x^{2} + x}{(x + 1)(x^{2} + 1)}\,dx + C.$$

Partial-fraction decomposition.

Write

$$\frac{2x^{2} + x}{(x + 1)(x^{2} + 1)} = \frac{A}{x + 1} + \frac{Bx + D}{x^{2} + 1}.$$

Multiplying through by the denominator,

$$2x^{2} + x = A(x^{2} + 1) + (Bx + D)(x + 1).$$

Comparing coefficients,

$$A + B = 2, \quad B + D = 1, \quad A + D = 0.$$

Solving: $$D = -A$$, $$B = A + 1$$, and substituting in $$A + B = 2$$ gives $$A = \tfrac12$$, $$B = \tfrac32$$, $$D = -\tfrac12$$.

Thus

$$\frac{2x^{2} + x}{(x + 1)(x^{2} + 1)} = \frac{1/2}{x + 1} + \frac{(3/2)x - 1/2}{x^{2} + 1}.$$

Integrate term by term.

$$y = \frac12\int \frac{dx}{x + 1} + \frac32\int \frac{x\,dx}{x^{2} + 1} - \frac12\int \frac{dx}{x^{2} + 1} + C.$$

Evaluating each piece (using $$u = x^{2} + 1$$ in the middle integral):

• $$\dfrac12\int \dfrac{dx}{x + 1} = \dfrac12\ln|x + 1|.$$
• $$\dfrac32\int \dfrac{x\,dx}{x^{2} + 1} = \dfrac34\ln(x^{2} + 1).$$
• $$-\dfrac12\int \dfrac{dx}{x^{2} + 1} = -\dfrac12\tan^{-1}x.$$

Therefore

$$y = \frac12\ln|x + 1| + \frac34\ln(x^{2} + 1) - \frac12\tan^{-1}x + C.$$

Apply the condition $$y = 1$$ at $$x = 0$$:

$$1 = \frac12\ln 1 + \frac34\ln 1 - \frac12\tan^{-1}0 + C = C.$$

Hence $$C = 1$$ and the required particular solution is

$$y = \frac12\ln|x + 1| + \frac34\ln(x^{2} + 1) - \frac12\tan^{-1}x + 1.$$

We are given the differential equation

$$x\,(x^{2}-1)\,\frac{dy}{dx}=1\;.$$

First isolate $$\dfrac{dy}{dx}$$:

$$\frac{dy}{dx}=\frac{1}{x\,(x^{2}-1)}=\frac{1}{x\,(x-1)(x+1)}\;.$$

Write the right–hand side in partial fractions.
Assume

$$\frac{1}{x\,(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}\;.$$

Multiply by $$x(x-1)(x+1)$$ to get

$$1=A(x-1)(x+1)+B\,x(x+1)+C\,x(x-1)\;.$$

Expand and collect like terms:

$$A(x^{2}-1)+B(x^{2}+x)+C(x^{2}-x)=(A+B+C)x^{2}+(B-C)x-A\;.$$

Equating coefficients with $$1=0\,x^{2}+0\,x+1$$ gives

• $$(A+B+C)=0$$
• $$(B-C)=0$$
• $$-A=1\;\Rightarrow\;A=-1$$

From $$B-C=0$$ we get $$B=C$$, and using $$A+B+C=0$$:

$$-1+2B=0\;\Rightarrow\;B=C=\frac12\;.$$

Hence

$$\frac{1}{x(x^{2}-1)}=-\frac1x+\frac{1}{2(x-1)}+\frac{1}{2(x+1)}\;.$$

Rewrite the differential equation as

$$dy=\Biggl(-\frac1x+\frac{1}{2(x-1)}+\frac{1}{2(x+1)}\Biggr)dx\;,$$

and integrate term-wise:

$$\begin{aligned}y&=\int\Biggl(-\frac1x+\frac{1}{2(x-1)}+\frac{1}{2(x+1)}\Biggr)dx\\[4pt]&=-\ln|x|+\frac12\ln|x-1|+\frac12\ln|x+1|+C\;.(*)\end{aligned}$$

The condition $$y=0$$ when $$x=2$$ fixes $$C$$:

At $$x=2$$, $$y=0$$, so

$$0=-\ln2+\tfrac12\ln1+\tfrac12\ln3+C=-\ln2+\tfrac12\ln3+C\;.$$

Thus

$$C=\ln2-\frac12\ln3=\ln\left(\frac{2}{\sqrt3}\right).$$

Substitute $$C$$ back in $$(*)$$ to obtain the required particular solution:

$$\boxed{\;y=-\ln|x|+\tfrac12\ln|x-1|+\tfrac12\ln|x+1|+\ln\left(\dfrac{2}{\sqrt3}\right)\;}$$

This can also be expressed in a single logarithm as

$$y=\ln\Biggl(\dfrac{2\sqrt{x^{2}-1}}{\sqrt3\,x}\Biggr).$$

We have the differential equation

$$\cos\left(\frac{dy}{dx}\right)=a\qquad (a\in\mathbf R).$$

For the cosine of a real number to equal a, we must have $$|a|\le 1.$$ (If $$|a|>1$$, the equation has no real solution.)

Assuming $$|a|\le 1,$$ apply the inverse cosine to both sides:

$$\frac{dy}{dx}=\cos^{-1}(a).$$

The right-hand side is a constant (denote it by $$k$$):

$$k=\cos^{-1}(a).$$

Integrate with respect to $$x$$:

$$\int dy = \int k\,dx \;\;\Longrightarrow\;\; y = kx + C,$$

where $$C$$ is the constant of integration.

Use the initial condition $$y=1$$ when $$x=0$$:

$$1 = k\cdot 0 + C \;\;\Longrightarrow\;\; C = 1.$$

Substitute $$C$$ back:

$$y = kx + 1.$$

Finally replace $$k$$ by its value $$\cos^{-1}(a)$$:

$$y = \bigl(\cos^{-1}(a)\bigr)x + 1, \qquad |a|\le 1.$$

(Any other branch of the inverse cosine, such as $$2\pi n \pm \cos^{-1}(a),\;n\in\mathbf Z,$$ would also give a constant slope and hence a family of straight-line solutions. The line above corresponds to the principal value.)

The given differential equation is $$\frac{dy}{dx}=y\tan x$$.

This equation is separable. Move all terms in $$y$$ to the left and all terms in $$x$$ to the right:

$$\frac{1}{y}\,dy = \tan x\,dx.$$

Integrate both sides:

$$\int \frac{1}{y}\,dy = \int \tan x\,dx.$$

The left integral gives $$\ln|y|+C_1$$, and the right integral uses the fact that $$\int \tan x\,dx = -\ln|\cos x|+C_2$$. Combining the two arbitrary constants into one constant $$C$$, we write

$$\ln|y| = -\ln|\cos x| + C.$$

Simplify the logarithms. Using $$-\ln|\cos x| = \ln|\sec x|$$,

$$\ln|y| = \ln|\sec x| + C.$$

Exponentiate to solve for $$y$$:

$$|y| = e^{C}\,|\sec x|.$$

Let $$e^{C}=K$$, where $$K>0$$. Dropping the absolute value by allowing $$K$$ to take any non-zero value (positive or negative), we have

$$y = K\sec x.$$

Use the initial condition $$y=1$$ when $$x=0$$ to find $$K$$. Since $$\sec 0 = 1$$,

$$1 = K \cdot 1 \Longrightarrow K = 1.$$

Therefore the required particular solution is

$$y = \sec x.$$

We are given the differential equation

$$y' = \dfrac{dy}{dx} = e^x \sin x$$

To find the required curve, integrate both sides with respect to $$x$$:

$$\int y'\,dx = \int e^x \sin x\,dx$$

$$y = \int e^x \sin x\,dx + C$$

where $$C$$ is the constant of integration.

Step 1: Evaluate the integral $$\int e^x \sin x\,dx$$.

Let $$I = \int e^x \sin x\,dx$$.

Use integration by parts: choose $$u = \sin x$$ and $$dv = e^x dx$$, so $$du = \cos x\,dx$$ and $$v = e^x$$.

Then

$$I = e^x \sin x - \int e^x \cos x\,dx$$

Denote $$J = \int e^x \cos x\,dx$$. Again integrate by parts with $$u = \cos x$$ and $$dv = e^x dx$$:

$$J = e^x \cos x - \int (-e^x \sin x)\,dx = e^x \cos x + \int e^x \sin x\,dx = e^x \cos x + I$$

Substitute this expression for $$J$$ back into the first equation:

$$I = e^x \sin x - (e^x \cos x + I)$$

$$I = e^x \sin x - e^x \cos x - I$$

$$2I = e^x(\sin x - \cos x)$$

$$I = \dfrac{e^x}{2}(\sin x - \cos x) + C_1$$

The constant $$C_1$$ can be absorbed into the main constant of the solution, so we write

$$\int e^x \sin x\,dx = \dfrac{e^x}{2}(\sin x - \cos x) + C$$

Step 2: Write the general solution.

Hence

$$y = \dfrac{e^x}{2}(\sin x - \cos x) + C$$

Step 3: Use the initial condition $$(0,0)$$.

At $$x = 0$$ we have $$y = 0$$:

$$0 = \dfrac{e^0}{2}(\sin 0 - \cos 0) + C = \dfrac{1}{2}(0 - 1) + C = -\dfrac{1}{2} + C$$

Thus $$C = \dfrac{1}{2}$$.

Step 4: Write the particular solution.

$$y = \dfrac{e^x}{2}(\sin x - \cos x) + \dfrac{1}{2}$$

or equivalently

$$y = \dfrac{1}{2}\bigl[e^x(\sin x - \cos x) + 1\bigr].$$

Given differential equation:

$$xy\,\frac{dy}{dx} = (x + 2)(y + 2).$$

1. Bring the derivative to one side.

$$\frac{dy}{dx} = \frac{(x + 2)(y + 2)}{xy} = \frac{x + 2}{x} \cdot \frac{y + 2}{y}.$$

2. Separate the variables.

$$\frac{y}{y + 2}\,dy = \frac{x + 2}{x}\,dx.$$

3. Integrate both sides.

Rewrite the left-hand integrand:

$$\frac{y}{y + 2} = \frac{y + 2 - 2}{y + 2} = 1 - \frac{2}{y + 2}.$$

Hence

$$\int \left(1 - \frac{2}{y + 2}\right) dy = \int \left(1 + \frac{2}{x}\right) dx.$$

Compute the integrals:

• Left: $$y - 2\ln|y + 2| + C_{1}.$$
• Right: $$x + 2\ln|x| + C_{2}.$$

Let $$C = C_{2} - C_{1}$$ (a single arbitrary constant).

4. General integral form.

$$y - 2\ln|y + 2| = x + 2\ln|x| + C.$$

5. Use the given point $$(1, -1)$$ to find $$C$$.

Substitute $$x = 1$$, $$y = -1$$:

• Left side: $$-1 - 2\ln|{-1} + 2| = -1 - 2\ln 1 = -1.$$
• Right side: $$1 + 2\ln|1| + C = 1 + C.$$

Equating: $$-1 = 1 + C \;\Rightarrow\; C = -2$$.

6. Required solution curve.

$$y - 2\ln|y + 2| = x + 2\ln|x| - 2.$$

This implicit relation is the solution curve passing through $$(1, -1)$$, valid for $$x \neq 0$$ and $$y \neq -2$$.

Let the required curve be represented by $$y=f(x)$$. At any point $$(x,y)$$ on this curve, the slope of the tangent is $$\dfrac{dy}{dx}$$.

The given condition states that the product of this slope and the $$y$$-coordinate equals the $$x$$-coordinate:

$$y\,\dfrac{dy}{dx}=x$$

This is a first–order differential equation. We solve it by the method of separation of variables.

Rewrite the equation so that all terms in $$y$$ appear on one side and all terms in $$x$$ on the other:

$$y\,\dfrac{dy}{dx}=x\;\;\Longrightarrow\;\;y\,dy=x\,dx$$

Integrate both sides:

$$\int y\,dy=\int x\,dx$$

$$\frac{y^{2}}{2}=\frac{x^{2}}{2}+C$$

Multiply every term by 2 for convenience:

$$y^{2}=x^{2}+C_1$$

(Here $$C_1=2C$$ is an arbitrary constant.)

To determine $$C_1$$, use the fact that the curve passes through $$(0,-2)$$.

Substitute $$x=0$$ and $$y=-2$$ in $$y^{2}=x^{2}+C_1$$:

$$(-2)^{2}=0^{2}+C_1\;\;\Longrightarrow\;\;4=C_1$$

Therefore $$C_1=4$$, and the required equation of the curve is

$$y^{2}-x^{2}=4$$

This is the equation of a rectangular hyperbola whose transverse axis is the $$y$$-axis.

Let $$(x, y)$$ be an arbitrary point on the required curve.

1. Express the given geometric condition as a differential equation

The slope of the tangent at $$(x, y)$$ is $$\dfrac{dy}{dx}$$.

The slope of the line segment joining $$(x, y)$$ to the fixed point $$(-4, -3)$$ is

$$\dfrac{y - (-3)}{x - (-4)} = \dfrac{y + 3}{x + 4}.$$

According to the statement, the tangent’s slope is twice this slope, so

\[\frac{dy}{dx} = 2\,\frac{y + 3}{x + 4}. \tag{1}\]

2. Separate the variables

Rewrite (1):

\[\frac{dy}{y + 3} = 2\,\frac{dx}{x + 4}. \tag{2}\]

3. Integrate both sides

Integrating (2),

$$\int \frac{1}{y + 3}\,dy = 2 \int \frac{1}{x + 4}\,dx.$$

This gives

\[\ln|y + 3| = 2\ln|x + 4| + C, \tag{3}\]

where $$C$$ is the constant of integration.

4. Simplify the integrated result

Apply properties of logarithms to (3):

$$\ln|y + 3| = \ln|x + 4|^{2} + C.$$

Exponentiating both sides,

$$|y + 3| = e^{C}(x + 4)^{2}.$$

Absorbing $$e^{C}$$ and the possible sign into a single non-zero constant $$k$$, we write

\[y + 3 = k(x + 4)^{2}. \tag{4}\]

5. Determine $$k$$ from the given point

The curve passes through $$(-2, 1)$$. Substitute $$x = -2,\; y = 1$$ in (4):

$$1 + 3 = k(-2 + 4)^{2} \;\Rightarrow\; 4 = k(2)^{2} = 4k \;\Rightarrow\; k = 1.$$

6. Write the particular solution

With $$k = 1$$, equation (4) becomes

$$y + 3 = (x + 4)^{2}.$$

Hence

$$y = (x + 4)^{2} - 3.$$

Expanding if desired:

$$y = x^{2} + 8x + 13.$$

Thus, the required curve is the parabola $$y = x^{2} + 8x + 13$$.

The volume $$V$$ of a sphere of radius $$r$$ is given by

$$V = \dfrac{4}{3}\pi r^{3}.$$

Because the balloon is being inflated at a constant rate,

$$\frac{\mathrm dV}{\mathrm dt}=k,$$

where $$k$$ is a constant. Integrating with respect to time $$t$$ gives

$$V = kt + C,$$

where $$C$$ is the constant of integration.

At the initial instant $$t = 0$$ the radius is $$3$$ units, so

$$V(0)=\dfrac{4}{3}\pi(3)^{3}=36\pi.$$

Thus $$C = 36\pi,$$ and hence

$$V = kt + 36\pi.\qquad(1)$$

After $$3\,\text{s}$$ the radius is $$6$$ units, so

$$V(3)=\dfrac{4}{3}\pi(6)^{3}=288\pi.$$

Substituting $$t = 3$$ and $$V = 288\pi$$ in (1):

$$288\pi = k(3) + 36\pi \;\;\Longrightarrow\;\; k = 84\pi.$$

Therefore (1) becomes

$$V = 84\pi t + 36\pi.\qquad(2)$$

Replace $$V$$ by $$\dfrac{4}{3}\pi r^{3}$$ in (2):

$$\dfrac{4}{3}\pi r^{3} = 84\pi t + 36\pi.$$

Divide every term by $$\pi$$ and multiply by $$\dfrac{3}{4}$$:

$$r^{3} = \dfrac{3}{4}\bigl(84t + 36\bigr).$$

Compute the numerical factors:

$$r^{3} = 63t + 27.$$

Taking the cube root yields the required radius after $$t$$ seconds:

$$r(t)=\bigl(63t + 27\bigr)^{1/3}\;\text{units}.$$

(You may also write $$r(t)=3\,(1 + \tfrac{7}{3}t)^{1/3}$$.)

Let $$P(t)$$ (in rupees) be the principal after $$t$$ years.

The statement "principal increases continuously at the rate of $$r\%$$ per year" means

$$\frac{dP}{dt} = \frac{r}{100}\,P.$$

This is a separable first-order differential equation. Separating variables and integrating:

$$\int \frac{dP}{P} = \int \frac{r}{100}\,dt \;\Longrightarrow\; \ln P = \frac{r}{100}\,t + C.$$

Exponentiating,

$$P = C_{1}\,e^{\frac{r}{100}t}, \qquad C_{1} = e^{C}.$$

Initial condition: at $$t = 0$$, $$P = 100$$. So $$C_{1} = 100$$ and

$$P(t) = 100\,e^{\frac{r}{100}t}.$$

Since the amount doubles in 10 years, $$P(10) = 200$$:

$$200 = 100\,e^{\frac{10r}{100}} \;\Longrightarrow\; 2 = e^{\frac{r}{10}}.$$

Taking the natural logarithm,

$$\ln 2 = \frac{r}{10}.$$

Given $$\ln 2 = 0.6931$$,

$$r = 10 \times \ln 2 = 10 \times 0.6931 = 6.931.$$

Thus, the required rate is approximately $$6.931\%$$ per year.

Let the amount at time t years be $$P(t)\,(\text{in rupees})$$. Since the principal grows continuously at 5 % per year, the rate of increase of $$P$$ is proportional to $$P$$ itself:

$$\frac{dP}{dt}=0.05\,P$$

Step 1: Separate the variables

$$\frac{1}{P}\,dP = 0.05\,dt$$

Step 2: Integrate both sides

$$\int \frac{1}{P}\,dP = \int 0.05\,dt$$

$$\ln|P| = 0.05t + C$$

Step 3: Exponentiate to solve for $$P$$

$$P = A\,e^{0.05t}$$ where $$A = e^{C}$$ is the constant of integration.

Step 4: Use the initial condition

At $$t = 0$$, a deposit of Rs 1000 is made, so $$P(0)=1000$$.

$$1000 = A\,e^{0.05\times 0}=A\times 1 \;\Rightarrow\; A = 1000$$

Hence the model becomes

$$P(t)=1000\,e^{0.05t}$$

Step 5: Find the amount after 10 years

Substitute $$t = 10$$:

$$P(10)=1000\,e^{0.05\times 10}=1000\,e^{0.5}$$

Given $$e^{0.5}=1.648$$, we get

$$P(10)=1000\times 1.648 = 1648$$

Result: After 10 years the deposit will be worth Rs 1648.

Let $$N(t)$$ denote the number of bacteria (count) after $$t$$ hours.
Because the rate of growth is proportional to the number present,

$$\frac{\mathrm dN}{\mathrm dt}=kN \; ,\; k>0$$

Separating the variables and integrating,

$$\int \frac{\mathrm dN}{N}=\int k\,\mathrm dt \;\;\Longrightarrow\;\; \ln N=kt+C$$

Writing the constant of integration as $$\ln C_1$$, the general solution is

$$N=C_1e^{kt}$$

Step 1 : Use the initial condition

At $$t=0, \; N=100000$$, therefore

$$100000=C_1e^{k\,(0)} \;\;\Longrightarrow\;\; C_1=100000$$

Hence,

$$N(t)=100000\,e^{kt}$$

Step 2 : Determine $$k$$ from the 2-hour information

After $$2$$ h the count has increased by $$10\%$$, i.e. $$N(2)=110000$$. Put $$t=2$$:

$$110000=100000\,e^{2k} \;\;\Longrightarrow\;\; e^{2k}=\frac{110000}{100000}=1.1$$

Taking natural logarithms,

$$2k=\ln 1.1 \;\;\Longrightarrow\;\; k=\tfrac{1}{2}\,\ln 1.1$$

Step 3 : Find the time when $$N=200000$$

Let $$t=t_0$$ be the required time (in hours) to reach $$200000$$ bacteria.

$$200000=100000\,e^{k t_0}\;\;\Longrightarrow\;\; e^{k t_0}=2$$

Taking natural logarithms again,

$$k t_0=\ln 2 \;\;\Longrightarrow\;\; t_0=\frac{\ln 2}{k}$$

Substitute $$k=\tfrac{1}{2}\,\ln 1.1$$:

$$t_0=\frac{\ln 2}{\tfrac{1}{2}\,\ln 1.1}=\frac{2\,\ln 2}{\ln 1.1}$$

Numerical value

$$t_0\;\approx\;\frac{2\,(0.693147)}{0.095310}\;=\;14.5\text{ hours (approximately)}$$

Thus, starting from a count of $$1{,}00{,}000$$, it will take about $$14.5$$ hours for the culture to reach $$2{,}00{,}000$$ bacteria.

The given differential equation is $$\dfrac{dy}{dx}=e^{x+y}$$.

Write the right side as a product:

$$\dfrac{dy}{dx}=e^{x}\,e^{y}.$$

Separate the variables (all $$y$$ terms on one side, all $$x$$ terms on the other):

$$\dfrac{dy}{e^{y}} = e^{x}\,dx \;\;\Longrightarrow\;\; e^{-y}\,dy = e^{x}\,dx.$$

Integrate both sides:

$$\int e^{-y}\,dy = \int e^{x}\,dx.$$

Compute the integrals:

$$-\,e^{-y}=e^{x}+C_{1},$$

where $$C_{1}$$ is the constant of integration.

Multiply by $$-1$$ and rename the constant (say, $$C=-C_{1}$$):

$$e^{-y}= -e^{x}+C\;\;\Longrightarrow\;\; e^{x}+e^{-y}=C.$$

Hence the general solution is $$e^{x}+e^{-y}=C.$$ This matches option (A).

Step 1 : Rewrite the differential equation

The given equation is

$$ (x-y)\frac{dy}{dx}=x+2y. $$

Re-arranging,

$$ \frac{dy}{dx}=\frac{x+2y}{x-y}. $$

Step 2 : Show that it is homogeneous

In the right–hand side numerator $$x+2y$$ each term has degree 1, and in the denominator $$x-y$$ each term also has degree 1. Hence the ratio is a homogeneous function of degree 0, so the differential equation is homogeneous.

Step 3 : Substitute $$y=vx$$

Let $$y=vx \;\;(v\;\text{depends on }x).$$ Then

$$ \frac{dy}{dx}=v+x\,\frac{dv}{dx}. $$

Substituting this and $$y=vx$$ into the differential equation gives

$$ v+x\,\frac{dv}{dx}=\frac{x+2vx}{x-vx}=\frac{1+2v}{1-v}. $$

Step 4 : Obtain a separable equation in $$v$$ and $$x$$

Bring the terms together:

$$ x\,\frac{dv}{dx}=\frac{1+2v}{1-v}-v. $$

Simplify the right–hand side:

$$ \frac{1+2v}{1-v}-v=\frac{1+2v-v(1-v)}{1-v}=\frac{1+2v-v+v^{2}}{1-v}=\frac{v^{2}+v+1}{1-v}. $$

Hence

$$ x\,\frac{dv}{dx}=\frac{v^{2}+v+1}{1-v}. $$

Step 5 : Separate the variables

$$ \frac{1-v}{v^{2}+v+1}\,dv = \frac{dx}{x}. $$

Step 6 : Integrate both sides

We need

$$ I=\int\frac{1-v}{v^{2}+v+1}\,dv. $$

Write

$$ 1-v = -\tfrac12(2v+1)+\tfrac32, $$

so that

$$ I=-\tfrac12\int\frac{2v+1}{v^{2}+v+1}\,dv+\tfrac32\int\frac{dv}{v^{2}+v+1}. $$

The first integral is a logarithm:

$$ -\tfrac12\ln|v^{2}+v+1|. $$

For the second, complete the square:

$$ v^{2}+v+1=(v+\tfrac12)^{2}+\tfrac34, $$

giving

$$ \tfrac32\int\frac{dv}{(v+\tfrac12)^{2}+(\tfrac{\sqrt3}{2})^{2}}=\tfrac32\cdot\frac{2}{\sqrt3}\tan^{-1}\!\left(\frac{2v+1}{\sqrt3}\right)=\sqrt3\,\tan^{-1}\!\left(\frac{2v+1}{\sqrt3}\right). $$

Thus

$$ \int\frac{1-v}{v^{2}+v+1}\,dv = -\tfrac12\ln|v^{2}+v+1| + \sqrt3\,\tan^{-1}\!\left(\frac{2v+1}{\sqrt3}\right)+C. $$

Equating to $$\int\tfrac{dx}{x}=\ln|x|+C,$$ we have

$$ -\tfrac12\ln|v^{2}+v+1| + \sqrt3\,\tan^{-1}\!\left(\frac{2v+1}{\sqrt3}\right)=\ln|x|+C. $$

Step 7 : Replace $$v$$ by $$y/x$$ and simplify

Because $$v=\dfrac{y}{x},$$

$$ v^{2}+v+1=\frac{y^{2}+xy+x^{2}}{x^{2}}, $$

and

$$ \ln|v^{2}+v+1|=\ln|y^{2}+xy+x^{2}|-2\ln|x|. $$

Substituting in the integrated result cancels the $$\ln|x|$$ terms, leaving

$$ \sqrt3\,\tan^{-1}\!\left(\frac{2y+x}{\sqrt3\,x}\right)-\tfrac12\ln|y^{2}+xy+x^{2}|=C. $$

Step 8 : The general solution

The required implicit solution is therefore

$$ \sqrt3\,\tan^{-1}\!\left(\frac{2y+x}{\sqrt3\,x}\right)-\frac12\ln|y^{2}+xy+x^{2}|=C, $$

where $$C$$ is an arbitrary constant.

Step 1 : Rewrite the given differential equation

The equation is

$$x\cos\bigl(\tfrac{y}{x}\bigr)\,\frac{dy}{dx}=y\cos\bigl(\tfrac{y}{x}\bigr)+x.$$

Divide both sides by $$x\cos\bigl(\tfrac{y}{x}\bigr)\;(\cos\bigl(\tfrac{y}{x}\bigr)\neq0)$$:

$$\frac{dy}{dx}=\frac{y}{x}+\sec\bigl(\tfrac{y}{x}\bigr).$$

Step 2 : Show that the equation is homogeneous

A first-order ODE is homogeneous if it can be written as $$\frac{dy}{dx}=F\!\bigl(\tfrac{y}{x}\bigr).$$

Here the right–hand side depends on $$y$$ and $$x$$ only through the ratio $$\tfrac{y}{x}$$ (we have the term $$\tfrac{y}{x}$$ itself and $$\sec\bigl(\tfrac{y}{x}\bigr)$$). Hence the differential equation is homogeneous.

Step 3 : Use the substitution $$y=vx$$

Put $$y=vx$$, where $$v=v(x)$$. Then

$$\frac{dy}{dx}=v+x\frac{dv}{dx}.$$

Substitute this and $$\tfrac{y}{x}=v$$ into the rewritten ODE:

$$v+x\frac{dv}{dx}=v+\sec(v).$$

Cancel $$v$$ from both sides:

$$x\frac{dv}{dx}=\sec(v).$$

Step 4 : Separate the variables and integrate

$$\frac{dv}{dx}=\frac{\sec(v)}{x} \;\Longrightarrow\;\cos(v)\,dv=\frac{dx}{x}.$$

Integrate both sides:

$$\int\cos(v)\,dv=\int\frac{dx}{x} \;\Longrightarrow\; \sin v=\ln|x|+C,$$

where $$C$$ is the constant of integration.

Step 5 : Restore $$v=\tfrac{y}{x}$$

Because $$v=\tfrac{y}{x}$$, the integral result becomes

$$\sin\Bigl(\frac{y}{x}\Bigr)=\ln|x|+C.$$

This implicit relation is the required solution.

Consider

$$2y\,e^{x/y}\,dx+\bigl(y-2x\,e^{x/y}\bigr)dy=0\qquad(1)$$

Write

$$M(x,y)=2y\,e^{x/y},\;\;N(x,y)=y-2x\,e^{x/y}.$$

1. Homogeneity of the differential equation

Replace $x\to t x,\;y\to t y$:

$$M(tx,ty)=2(ty)\,e^{(tx)/(ty)}=2t y\,e^{x/y}=t\,M(x,y),$$ $$N(tx,ty)=(ty)-2(tx)e^{(tx)/(ty)}=t\bigl(y-2x\,e^{x/y}\bigr)=t\,N(x,y).$$

Both $M$ and $N$ are homogeneous of the same degree 1, hence (1) is a homogeneous differential equation.

2. Reduction to a separable form

Divide (1) by $dx$ and rearrange:

$$\left(y-2x\,e^{x/y}\right)\frac{dy}{dx}+2y\,e^{x/y}=0\;\;\Rightarrow\;\; \frac{dy}{dx}=-\frac{2y\,e^{x/y}}{y-2x\,e^{x/y}}\qquad(2)$$

Because $e^{x/y}$ involves $x/y$, put

$$y=v x\;\;\;(v\;\hbox{is a function of }x).$$

Then

$$\frac{dy}{dx}=v+x\frac{dv}{dx},\qquad e^{x/y}=e^{1/v}.$$

Substitute these into (2):

$$v+x\frac{dv}{dx}=-\frac{2v x\,e^{1/v}}{x\bigl(v-2e^{1/v}\bigr)} \;\;\Longrightarrow\;\;x\frac{dv}{dx}=-\frac{v^{2}}{v-2e^{1/v}}.$$

Separate the variables:

$$\frac{v-2e^{1/v}}{v^{2}}\,dv=-\frac{dx}{x}.\qquad(3)$$

3. Integration

Expand the left integrand:

$$\frac{v-2e^{1/v}}{v^{2}}=\frac1v-\frac{2e^{1/v}}{v^{2}}.$$

Integral on the left:

$$\int\frac{1}{v}\,dv-2\int\frac{e^{1/v}}{v^{2}}\,dv =\ln|v|-2\bigl(-e^{1/v}\bigr)=\ln|v|+2e^{1/v}.$$

Integral on the right is $-\ln|x|$. Hence (3) gives

$$\ln|v|+2e^{1/v}=-\ln|x|+C,$$

or

$$\ln|xv|+2e^{1/v}=C.\qquad(4)$$

4. Restore $v$ in terms of $x$ and $y$

Because $v=y/x$, we have $xv=y$ and $1/v=x/y$. Equation (4) becomes

$$\boxed{\ln|y|+2e^{x/y}=C}.\qquad(5)$$

5. Particular solution

The condition $x=0$ when $y=1$ gives, from (5),

$$\ln 1+2e^{0}=0+2=C\;\;\Longrightarrow\;\;C=2.$$

Insert $C=2$ in (5):

$$\boxed{\;2e^{x/y}+\ln y=2\;}.$$

This is the required particular solution (with $y>0$ because $y=1$ initially).

The differential equation that expresses the given condition is

$$\frac{dy}{dx}=\frac{x^2+y^2}{2xy}.$$

Multiply by $$2xy$$ to clear the denominator:

$$2xy\,\frac{dy}{dx}=x^2+y^2\quad\Longrightarrow\quad 2\frac{y}{x}\,\frac{dy}{dx}=1+\left(\frac{y}{x}\right)^2.$$

This form shows that the equation is homogeneous. Put

$$v=\frac{y}{x}\;\;(\text{so }y=vx)\;\;\Rightarrow\;\;\frac{dy}{dx}=v+x\,\frac{dv}{dx}.$$

Substitute in the differential equation:

$$2\frac{y}{x}\left(v+x\frac{dv}{dx}\right)=1+v^2$$

$$\Rightarrow\;2v\Bigl(v+x\frac{dv}{dx}\Bigr)=1+v^2$$

$$\Rightarrow\;2v^2+2xv\frac{dv}{dx}=1+v^2.$$

Rearrange:

$$2xv\frac{dv}{dx}=1+v^2-2v^2=1-v^2.$$

Separate the variables:

$$\frac{2v}{1-v^2}\,dv=\frac{dx}{x}.$$

Integrate both sides:

$$\int\frac{2v}{1-v^2}\,dv=\int\frac{dx}{x}+C.$$

Using the substitution $$w=1-v^2\;(dw=-2v\,dv)$$, the left integral becomes

$$\int\!-\frac{dw}{w}=-\ln|w|.$$

Hence

$$-\ln|1-v^2|=\ln|x|+C\_1.$$

Move the minus sign to the right and absorb it in the constant:

$$\ln|1-v^2|=-\ln|x|+C\_2\quad\Longrightarrow\quad\ln\bigl|x(1-v^2)\bigr|=C\_2.$$

Exponentiating,

$$x(1-v^2)=C,$$

where $$C$$ is an arbitrary non-zero constant. Replace $$v$$ by $$\dfrac{y}{x}$$:

$$x\Bigl(1-\Bigl(\frac{y}{x}\Bigr)^2\Bigr)=C\;\;\Longrightarrow\;\;\frac{x^2-y^2}{x}=C.$$

Multiplying by $$x$$ and renaming the constant $$C$$ as $$c$$, the required one-parameter family of curves is

$$x^2-y^2=cx.$$

Thus, every integral curve whose tangent satisfies $$\dfrac{dy}{dx}=\dfrac{x^2+y^2}{2xy}$$ is given by $$x^2-y^2=cx,$$ and conversely each member of this family satisfies the differential equation.

Given differential equation
$$ (x^2 + xy)\,dy = (x^2 + y^2)\,dx $$

1. Showing that it is homogeneous

• Write it in the form $$M(x,y)\,dx + N(x,y)\,dy = 0$$:
  $$ (x^2 + y^2)\,dx - (x^2 + xy)\,dy = 0 $$
  Here $$M(x,y)=x^2+y^2\;\text{and}\;N(x,y)=-(x^2+xy).$$
• Every term of $$M$$ and of $$N$$ is of degree 2 (e.g. $$x^2,\;y^2,\;xy$$ all have degree 2).

Since both $$M$$ and $$N$$ are homogeneous functions of the same degree (2), the given equation is a homogeneous differential equation.

2. Reducing it to a separable form

Divide by $$dx$$ (assuming $$dx\neq0$$):
$$ \frac{dy}{dx}=\frac{x^2+y^2}{x^2+xy}. $$

For a homogeneous first-order ODE we set $$y=vx$$ (so $$v=\dfrac{y}{x}$$). Then

$$ \frac{dy}{dx}=v+x\frac{dv}{dx}. $$

Substitute this and $$y=vx$$ into the right-hand side:

$$ v+x\frac{dv}{dx}=\frac{x^2+v^{2}x^{2}}{x^2+v x^{2}}=\frac{1+v^{2}}{1+v}. $$

Hence

$$ x\frac{dv}{dx}=\frac{1+v^{2}}{1+v}-v. $$

Simplify the right-hand side:

$$ \frac{1+v^{2}}{1+v}-v=\frac{1+v^{2}-v(1+v)}{1+v}=\frac{1-v}{1+v}. $$

Thus

$$ x\frac{dv}{dx}=\frac{1-v}{1+v}\quad\Longrightarrow\quad\frac{1+v}{1-v}\,dv=\frac{dx}{x}. $$

3. Integrating

Write the left integrand as partial fractions:

$$ \frac{1+v}{1-v}=-1+\frac{2}{1-v}. $$

Now integrate both sides:

$$ \int\left(-1+\frac{2}{1-v}\right)dv=\int\frac{dx}{x}. $$

Left side:

$$ -\int dv-2\int\frac{1}{1-v}dv=-v-2\ln|1-v|. $$

Right side: $$ \ln|x|+C. $$

Therefore

$$ -v-2\ln|1-v|=\ln|x|+C. $$

4. Re-substitution

Recall $$v=\dfrac{y}{x}$$. Replace v and, if desired, absorb the sign into the constant:

$$ \frac{y}{x}+2\ln\left|1-\frac{y}{x}\right|+\ln|x|=C. $$

This implicit relation is the required solution of the given homogeneous differential equation.

Step 1: Check for homogeneity.

Write the equation in the form $$\dfrac{dy}{dx} = f(x, y)$$, where

$$f(x, y) = \frac{x + y}{x} = 1 + \frac{y}{x}.$$

Replacing $$x \to tx$$, $$y \to ty$$ (with $$t > 0$$):

$$f(tx, ty) = 1 + \frac{ty}{tx} = 1 + \frac{y}{x} = f(x, y).$$

So $$f$$ is a homogeneous function of degree $$0$$, and the equation is homogeneous.

Step 2: Standard substitution.

Set $$y = vx$$, where $$v = v(x)$$. Then

$$\frac{dy}{dx} = v + x\,\frac{dv}{dx}.$$

Step 3: Substitute into the equation.

$$v + x\,\frac{dv}{dx} = 1 + v \;\Longrightarrow\; x\,\frac{dv}{dx} = 1 \;\Longrightarrow\; \frac{dv}{dx} = \frac{1}{x}.$$

Step 4: Integrate.

$$\int dv = \int \frac{dx}{x} \;\Longrightarrow\; v = \ln|x| + C,$$

where $$C$$ is a constant of integration.

Step 5: Back-substitute $$v = y/x$$.

$$\frac{y}{x} = \ln|x| + C \;\Longrightarrow\; y = x\,\ln|x| + C\,x.$$

This is the general solution of the given homogeneous differential equation.

Step 1 : Write the equation in the form $$M\,dx+N\,dy=0$$

$$(x-y)\,dy-(x+y)\,dx=0\;\;\Longrightarrow\;\;-\,(x+y)\,dx+(x-y)\,dy=0$$
Hence $$M=-(x+y)\;\text{and}\;N=(x-y).$$

Step 2 : Show that it is homogeneous

Both $$M$$ and $$N$$ contain only first–degree terms in $$x$$ and $$y$$, so each is a homogeneous function of degree 1.
Therefore the differential equation is homogeneous.

Step 3 : Convert to the standard form

Divide by $$dx$$ (and move the minus sign):
$$\frac{dy}{dx}=\frac{x+y}{x-y}.$$

Step 4 : Use the substitution $$y=vx\;\;(v=v(x))$$

Then $$\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}.$$

Substitute in the differential equation:
$$v+x\dfrac{dv}{dx}=\frac{x+vx}{x-vx}=\frac{1+v}{1-v}.$$

Step 5 : Separate the variables

$$x\dfrac{dv}{dx}=\frac{1+v}{1-v}-v=\frac{1+v^2}{1-v}.$$
$$\Longrightarrow\;\frac{1-v}{1+v^2}\,dv=\frac{dx}{x}.$$

Step 6 : Integrate

Left side:
$$\int\frac{1-v}{1+v^2}\,dv=\int\frac{1}{1+v^2}\,dv-\int\frac{v}{1+v^2}\,dv$$
$$=\tan^{-1}v-\tfrac12\ln(1+v^2).$$

Right side:
$$\int\frac{dx}{x}=\ln|x|+C.$$

Thus
$$\tan^{-1}v-\tfrac12\ln(1+v^2)=\ln|x|+C.$$

Step 7 : Return to the variables $$x,\,y$$

Because $$v=\dfrac{y}{x}$$, we have
$$\tan^{-1}\left(\frac{y}{x}\right)-\tfrac12\ln\left(1+\left(\frac{y}{x}\right)^2\right)=\ln|x|+C.$$

Write the logarithm more neatly:

$$\ln\left(1+\left(\frac{y}{x}\right)^2\right)=\ln\left(\frac{x^2+y^2}{x^2}\right)=\ln(x^2+y^2)-2\ln|x|.$$

Substituting, the $$\ln|x|$$ terms cancel, leaving

$$\boxed{\,\tan^{-1}\left(\dfrac{y}{x}\right)-\dfrac12\ln(x^2+y^2)=C\,}$$

This is the required solution of the given homogeneous differential equation.

Step 1: Check for homogeneity.

Write the equation in the form $$M(x, y)\,dx + N(x, y)\,dy = 0$$ with

• $$M(x, y) = x^{2} - y^{2},$$
• $$N(x, y) = 2xy.$$

Every term in $$M$$ and $$N$$ has degree $$2$$, so both are homogeneous of the same degree. Hence the equation is homogeneous.

Step 2: Convert to $$\dfrac{dy}{dx}$$ form.

$$(x^{2} - y^{2}) + 2xy\,\frac{dy}{dx} = 0 \;\Longrightarrow\; \frac{dy}{dx} = \frac{y^{2} - x^{2}}{2xy}.$$

Step 3: Homogeneous substitution.

Put $$y = vx$$, so $$v = \dfrac{y}{x}$$ and $$\dfrac{dy}{dx} = v + x\,\dfrac{dv}{dx}$$. Substituting:

$$v + x\,\frac{dv}{dx} = \frac{v^{2} - 1}{2v}.$$

Hence

$$x\,\frac{dv}{dx} = \frac{v^{2} - 1}{2v} - v = \frac{v^{2} - 1 - 2v^{2}}{2v} = -\frac{v^{2} + 1}{2v}.$$

Step 4: Separate the variables.

$$\frac{2v}{v^{2} + 1}\,dv = -\frac{dx}{x}.$$

Step 5: Integrate both sides.

Let $$w = v^{2} + 1$$, so $$dw = 2v\,dv$$. Then

$$\int \frac{2v\,dv}{v^{2} + 1} = \int \frac{dw}{w} = \ln|w| = \ln(v^{2} + 1).$$

Thus

$$\ln(v^{2} + 1) = -\ln|x| + C_{0}.$$

Step 6: Combine logarithms and back-substitute $$v = y/x$$.

$$\ln\bigl((v^{2} + 1)\,|x|\bigr) = C_{0}.$$

Exponentiating: $$(v^{2} + 1)\,x = C$$ (absorbing the sign and the constant into the new arbitrary constant $$C$$).

Replacing $$v^{2} = y^{2}/x^{2}$$:

$$x\,\frac{x^{2} + y^{2}}{x^{2}} = C \;\Longrightarrow\; \frac{x^{2} + y^{2}}{x} = C \;\Longrightarrow\; x^{2} + y^{2} = C\,x.$$

Step 7: General solution.

$$x^{2} + y^{2} = C\,x,$$

where $$C$$ is an arbitrary constant.

Given differential equation
$$x^{2}\frac{dy}{dx}=x^{2}-2y^{2}+xy$$

1. Checking homogeneity
First write the right–hand side as a sum of monomials and note their degrees (sum of powers of $$x$$ and $$y$$):

• $$x^{2}$$  →  degree $$=2$$
• $$-2y^{2}$$  →  degree $$=2$$
• $$+xy$$  →  degree $$=2$$

All terms on the right have the same degree, namely $$2$$. Hence the function
$$F(x,y)=x^{2}-2y^{2}+xy$$ is homogeneous of degree 2.

Divide the whole differential equation by $$x^{2}$$ to obtain a form in which the right–hand side depends only on $$y/x$$:

$$\frac{dy}{dx}=1-2\left(\frac{y}{x}\right)^{2}+\frac{y}{x}.$$

Because the right side is a function of the single ratio $$y/x$$, the differential equation is homogeneous.

2. Solving the homogeneous equation

Put $$y=vx$$, where $$v=v(x)$$. Then

$$\frac{dy}{dx}=v+x\frac{dv}{dx}.$$ Substitute these in the divided equation:

$$v+x\frac{dv}{dx}=1-2v^{2}+v.$$

Cancel the common term $$v$$ on both sides:

$$x\frac{dv}{dx}=1-2v^{2}.$$ This is separable:

$$\frac{dv}{1-2v^{2}}=\frac{dx}{x}.$$

Integration
Rewrite the denominator as $$1-(\sqrt2 v)^{2}$$ and integrate:

Let $$w=\sqrt{2}\,v\;(\Rightarrow v=\tfrac{w}{\sqrt2},\;dv=\tfrac{dw}{\sqrt2}).$$ Then

$$\int\frac{dv}{1-2v^{2}}=\int\frac{dw/\sqrt2}{1-w^{2}}=\frac1{\sqrt2}\int\frac{dw}{1-w^{2}}.$$

But $$\int\frac{dw}{1-w^{2}}=\frac12\ln\left|\frac{1+w}{1-w}\right|+C.$$ Hence

$$\frac1{\sqrt2}\left[\frac12\ln\left|\frac{1+w}{1-w}\right|\right]=\ln|x|+C.$$ That is,

$$\frac1{2\sqrt2}\ln\left|\frac{1+w}{1-w}\right|=\ln|x|+C.$$

Multiply by $$2\sqrt2$$ and write the constant again as $$C$$:

$$\ln\left|\frac{1+w}{1-w}\right|=2\sqrt2\,\ln|x|+C.$$

Exponentiating,

$$\left|\frac{1+w}{1-w}\right|=C\,x^{2\sqrt2},\qquad C>0.$$

Return to the original variables: $$w=\sqrt2 v=\sqrt2\,\frac{y}{x}.$$ Thus

$$\frac{1+\sqrt2\,y/x}{1-\sqrt2\,y/x}=C\,x^{2\sqrt2}.$$

Equivalently, after clearing the denominator, one may write the implicit solution as

$$\boxed{\displaystyle\frac{x+\sqrt2\,y}{x-\sqrt2\,y}=C\,x^{2\sqrt2}}.$$ Here $$C$$ is an arbitrary, non-zero constant (absorbing the sign of the absolute value).

This implicit relation gives the required solution of the given homogeneous differential equation.

Step 1 – Write the DE in dy/dx form
The given differential equation is $$x\,dy - y\,dx = \sqrt{x^{2}+y^{2}}\,dx$$.
Divide by $$dx$$ (assume $$dx \neq 0$$):
$$x\,\dfrac{dy}{dx} - y = \sqrt{x^{2}+y^{2}}$$
Hence
$$\dfrac{dy}{dx} = \dfrac{y+\sqrt{x^{2}+y^{2}}}{x}.$$

Step 2 – Show that the DE is homogeneous
Put $$v = \dfrac{y}{x} \;(\text{so that } y = vx)$$.
Then $$\sqrt{x^{2}+y^{2}} = x\sqrt{1+\bigl(\dfrac{y}{x}\bigr)^{2}} = x\sqrt{1+v^{2}}$$.
Therefore
$$\dfrac{dy}{dx} = \dfrac{y}{x} + \sqrt{1+\bigl(\dfrac{y}{x}\bigr)^{2}} = v + \sqrt{1+v^{2}},$$
which depends only on $$v = y/x$$. Hence the differential equation is homogeneous.

Step 3 – Substitute $$y = vx$$ and reduce to variables separable
Because $$y = vx$$, we have $$\dfrac{dy}{dx} = v + x\,\dfrac{dv}{dx}$$.
Equate this with the right‐hand side obtained above:
$$v + x\,\dfrac{dv}{dx} = v + \sqrt{1+v^{2}}.$$
Cancel $$v$$:
$$x\,\dfrac{dv}{dx} = \sqrt{1+v^{2}}.$$
Hence
$$\dfrac{dv}{\sqrt{1+v^{2}}} = \dfrac{dx}{x}.$$

Step 4 – Integrate
The left integral is standard:
$$\int \dfrac{dv}{\sqrt{1+v^{2}}} = \operatorname{arsinh}(v) = \ln\bigl|v + \sqrt{1+v^{2}}\bigr|.$$
The right integral is $$\int \dfrac{dx}{x} = \ln|x| + C$$, where $$C$$ is an arbitrary constant.
Thus
$$\ln\bigl|v + \sqrt{1+v^{2}}\bigr| = \ln|x| + C.$$

Step 5 – Remove the logarithms and back‐substitute $$v = y/x$$
Exponentiating, write $$k = e^{C} \;(k \neq 0)$$:
$$v + \sqrt{1+v^{2}} = kx.$$
Replace $$v$$ by $$y/x$$:
$$\dfrac{y}{x} + \sqrt{1+\bigl(\dfrac{y}{x}\bigr)^{2}} = kx.$$
Multiply by $$x$$:
$$y + x\sqrt{1+\dfrac{y^{2}}{x^{2}}} = kx^{2}.$$
Since $$x\sqrt{1+\dfrac{y^{2}}{x^{2}}} = \sqrt{x^{2}+y^{2}},$$ we finally get
$$\sqrt{x^{2}+y^{2}} + y = kx^{2}.$$

Step 6 – Write the general solution
Renaming the arbitrary constant $$k$$ as $$C$$ (allowed because $$C$$ can be any non-zero real number), the required solution is
$$\boxed{\sqrt{x^{2}+y^{2}} + y = Cx^{2}}.$$

Step 1 · Write the given differential equation

$$\{x\cos(\tfrac yx)+y\sin(\tfrac yx)\}\,y\,dx=\{y\sin(\tfrac yx)-x\cos(\tfrac yx)\}\,x\,dy$$

Step 2 · Show that it is homogeneous

Divide both sides by $$x y \;dx\;(x,y\neq0)$$:

$$\frac{dy}{dx}=\frac{y\bigl[x\cos(\tfrac yx)+y\sin(\tfrac yx)\bigr]}{x\bigl[y\sin(\tfrac yx)-x\cos(\tfrac yx)\bigr]}$$

Put $$v=\dfrac yx\;(\Rightarrow y=vx)$$. The right–hand side depends only on $$v$$, therefore the equation is homogeneous.

Step 3 · Transform with $$y=vx$$

Since $$y=vx$$, $$\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$$. Substitute into the equation:

$$v+x\frac{dv}{dx}=\frac{v\,[\cos v+v\sin v]}{v\sin v-\cos v}$$

Step 4 · Isolate $$\dfrac{dv}{dx}$$

$$x\frac{dv}{dx}=\frac{v[\cos v+v\sin v]-v(v\sin v-\cos v)}{v\sin v-\cos v}=\frac{2v\cos v}{v\sin v-\cos v}$$

Step 5 · Separate the variables

$$\frac{v\sin v-\cos v}{2v\cos v}\,dv=\frac{dx}{x}$$

Split the fraction:

$$\left(\frac12\tan v-\frac1{2v}\right)dv=\frac{dx}{x}$$

Step 6 · Integrate

Left:

$$\int\left(\frac12\tan v-\frac1{2v}\right)dv=-\tfrac12\ln|\cos v|-\tfrac12\ln|v|=-\tfrac12\ln|v\cos v|$$

Right:

$$\int\frac{dx}{x}=\ln|x|+C$$

Hence

$$-\tfrac12\ln|v\cos v|=\ln|x|+C\quad\Longrightarrow\quad\ln|v\cos v|=-2\ln|x|+C_1$$

Step 7 · Remove the logarithm and back-substitute

$$v\cos v=\frac{C}{x^2}\qquad(C\text{ is an arbitrary constant})$$

Because $$v=\dfrac yx$$, multiply by $$x^2$$:

$$x y\cos\left(\frac yx\right)=C$$

This implicit relation is the required solution.

Consider the differential equation

$$x\,\frac{dy}{dx} - y + x\,\sin\!\left(\frac{y}{x}\right) = 0.$$

1. Show that the equation is homogeneous.

A first-order differential equation is homogeneous if it can be written as $$\dfrac{dy}{dx} = F\!\left(\dfrac{y}{x}\right)$$.

Dividing the given equation by $$x$$ (with $$x \neq 0$$):

$$\frac{dy}{dx} - \frac{y}{x} + \sin\!\left(\frac{y}{x}\right) = 0 \;\Longrightarrow\; \frac{dy}{dx} = \frac{y}{x} - \sin\!\left(\frac{y}{x}\right).$$

The right-hand side depends only on $$y/x$$, so the equation is homogeneous.

2. Solve using the substitution $$y = vx$$.

With $$y = vx$$ and $$\dfrac{dy}{dx} = v + x\,\dfrac{dv}{dx}$$, the equation becomes

$$x\bigl(v + x\,\frac{dv}{dx}\bigr) - vx + x\sin v = 0.$$

The terms $$xv$$ and $$-vx$$ cancel, leaving

$$x^{2}\,\frac{dv}{dx} + x\sin v = 0.$$

Dividing by $$x^{2}$$ and separating variables:

$$\frac{dv}{\sin v} = -\frac{dx}{x}, \quad\text{i.e.}\quad \csc v\,dv = -\frac{dx}{x}.$$

3. Integrate.

Using the standard result $$\displaystyle\int \csc v\,dv = \ln\!\left|\tan\tfrac{v}{2}\right| + C$$:

$$\ln\!\left|\tan\tfrac{v}{2}\right| = -\ln|x| + C_{0}.$$

Combining logarithms:

$$\ln\!\left|x\,\tan\tfrac{v}{2}\right| = C_{0} \;\Longrightarrow\; x\,\tan\tfrac{v}{2} = C,$$

where $$C$$ is an arbitrary non-zero constant.

4. Restore $$v = y/x$$.

$$x\,\tan\!\left(\frac{y}{2x}\right) = C.$$

This implicit relation is the required general solution.

Given differential equation

$$y\,dx+x\log\left(\dfrac{y}{x}\right)dy-2x\,dy=0$$

1. Homogeneity check

• The function multiplying $$dx$$ is $$y$$, which is of degree 1 in $$x,y$$.
• The function multiplying $$dy$$ is $$x\bigl(\log(\tfrac{y}{x})-2\bigr)$$. Here $$x$$ is of degree 1 and $$\log(\tfrac{y}{x})-2$$ is of degree 0 (it contains only the ratio $$y/x$$). Hence the whole product has degree 1.

Because every term has the same degree (1), the differential equation is homogeneous.

2. Reduction to one variable

Rewrite the equation so that $$dy/dx$$ appears explicitly:

$$y\,dx+\bigl[x\log\!\left(\tfrac{y}{x}\right)-2x\bigr]dy=0$$

Divide by $$dx$$ (assuming $$dx\neq0$$):

$$y+x\left[\log\!\left(\tfrac{y}{x}\right)-2\right]\dfrac{dy}{dx}=0$$

$$\Rightarrow\;\dfrac{dy}{dx}=-\dfrac{y}{x\bigl[\log(\tfrac{y}{x})-2\bigr]}$$

Because the right–hand side is a function of $$y/x$$ alone, we use the homogeneous substitution

$$y=vx\quad\bigl(\,v=v(x)\bigr).$$

Then

$$\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}.$$

Insert this in the differential equation:

$$v+x\dfrac{dv}{dx}=-\dfrac{v}{\log v-2}.$$

Isolate $$x\,dv/dx$$:

$$x\dfrac{dv}{dx}=-\dfrac{v}{\log v-2}-v=-\dfrac{v}{\log v-2}-v\cdot\dfrac{\log v-2}{\log v-2}=-\dfrac{v\bigl[\log v-1\bigr]}{\log v-2}.$$

Hence

$$\dfrac{dv}{dx}=-\dfrac{v\bigl[\log v-1\bigr]}{x\bigl[\log v-2\bigr]}.$$

3. Separation of variables and integration

Separate the variables $$v$$ and $$x$$:

$$\dfrac{\log v-2}{v(\log v-1)}\,dv=-\dfrac{dx}{x}.$$

Integrate both sides:

$$\int\dfrac{\log v-2}{v(\log v-1)}\,dv=-\int\dfrac{dx}{x}+C.$$

To integrate the left side put $$u=\log v-1\;(\Rightarrow\;du=\tfrac{1}{v}dv)$$. Then

$$\int\dfrac{\log v-2}{v(\log v-1)}\,dv=\int\frac{u-1}{u}\,du=\int\Bigl(1-\frac{1}{u}\Bigr)du=u-\log|u|+C.$$

Substitute back $$u=\log v-1$$:

$$\bigl(\log v-1\bigr)-\log\!\bigl|\log v-1\bigr|=-\log|x|+C.$$

4. Restoring $$v=\dfrac{y}{x}$$

Replace $$v$$ by $$y/x$$:

$$\log\!\left(\dfrac{y}{x}\right)-1-\log\!\Bigl|\,\log\!\left(\dfrac{y}{x}\right)-1\Bigr|=-\log|x|+C.$$

Bring $$\log|x|$$ to the left; absorb the constant shift into a new constant $$C$$:

$$\log\!\left(\dfrac{y}{x}\right)+\log|x|-\log\!\Bigl|\,\log\!\left(\dfrac{y}{x}\right)-1\Bigr|=C.$$

Combine the first two logs:

$$\log\!\left(\dfrac{y}{x}\right)+\log|x|=\log|y|,$$

so the integrated relation can be rewritten as

$$\log|y| - \log\!\Bigl|\,\log\!\left(\dfrac{y}{x}\right)-1\Bigr| = C.$$

Exponentiating gives the equivalent implicit solution

$$\frac{y}{\bigl|\,\log\!\left(\dfrac{y}{x}\right)-1\bigr|}=K,$$

where $$K\,(\neq0)$$ is an arbitrary constant.

5. Final answer

The given differential equation is homogeneous, and its general (implicit) solution is

$$\boxed{\dfrac{y}{\log\!\left(\dfrac{y}{x}\right)-1}=C},\qquad C\in\mathbb{R}\setminus\{0\}.$$

Write the given equation as $$M(x, y)\,dx + N(x, y)\,dy = 0$$ with

$$M(x, y) = 1 + e^{x/y}, \qquad N(x, y) = e^{x/y}\!\left(1 - \frac{x}{y}\right).$$

1. Homogeneity check.

Replacing $$x \to tx$$, $$y \to ty$$:

$$M(tx, ty) = 1 + e^{(tx)/(ty)} = 1 + e^{x/y} = M(x, y),$$

$$N(tx, ty) = e^{x/y}\!\left(1 - \frac{x}{y}\right) = N(x, y).$$

Both functions are homogeneous of degree $$0$$, so the differential equation is homogeneous.

2. Substitution.

Because the equation involves $$x/y$$, set $$x = v\,y$$, where $$v = v(y)$$. Then

$$dx = v\,dy + y\,dv.$$

Substitute these in the equation:

$$(1 + e^{v})(v\,dy + y\,dv) + e^{v}(1 - v)\,dy = 0.$$

Expanding:

$$\bigl[v(1 + e^{v}) + e^{v}(1 - v)\bigr]dy + y(1 + e^{v})\,dv = 0.$$

Simplify the coefficient of $$dy$$:

$$v + v\,e^{v} + e^{v} - v\,e^{v} = v + e^{v}.$$

Hence

$$(v + e^{v})\,dy + y(1 + e^{v})\,dv = 0.$$

3. Separate the variables.

$$\frac{1 + e^{v}}{v + e^{v}}\,dv + \frac{dy}{y} = 0.$$

4. Integrate.

Since $$\dfrac{d}{dv}(v + e^{v}) = 1 + e^{v}$$, the first integral is a direct logarithm:

$$\int \frac{1 + e^{v}}{v + e^{v}}\,dv = \ln|v + e^{v}|.$$

The second integral gives $$\ln|y|$$. Therefore

$$\ln|v + e^{v}| + \ln|y| = C_{0} \;\Longrightarrow\; \ln\bigl|y(v + e^{v})\bigr| = C_{0}.$$

Exponentiating and renaming the constant as $$C$$:

$$y(v + e^{v}) = C.$$

5. Back-substitute $$v = x/y$$.

$$y\!\left(\frac{x}{y} + e^{x/y}\right) = C \;\Longrightarrow\; x + y\,e^{x/y} = C.$$

This is the required general solution.

We are given the differential equation

$$ (x + y)\,dy + (x - y)\,dx = 0 $$

and the condition $$y = 1$$ when $$x = 1$$. We have to find the particular solution.

1. Express the equation in the form $$dy/dx = f(x,y)$$

Move the second term to the right:

$$ (x + y)\,dy = -(x - y)\,dx $$

Divide by $$dx\,(x + y)$$ (assuming $$x + y \neq 0$$):

$$ \frac{dy}{dx} = -\frac{x - y}{x + y} = \frac{y - x}{x + y}. $$

2. Show that the equation is homogeneous

The right-hand side can be written wholly in terms of the ratio $$y/x$$:

$$ \frac{y - x}{x + y} = \frac{\dfrac{y}{x} - 1}{1 + \dfrac{y}{x}}. $$

Hence the equation is homogeneous.

3. Use the substitution $$y = vx$$

• Let $$v = y/x \;\;(\implies y = vx).$$
• Then $$\dfrac{dy}{dx} = v + x\,\dfrac{dv}{dx}. $$

Substitute these into the differential equation:

$$ v + x\,\frac{dv}{dx} = \frac{v - 1}{v + 1}. $$

4. Separate the variables

Move $$v$$ to the right:

$$ x\,\frac{dv}{dx} = \frac{v - 1}{v + 1} - v. $$

Simplify the right-hand side:

$$ \frac{v - 1}{v + 1} - v = \frac{v - 1 - v(v + 1)}{v + 1} = \frac{-1 - v^{2}}{v + 1} = -\frac{1 + v^{2}}{v + 1}. $$

Hence

$$ x\,\frac{dv}{dx} = -\frac{1 + v^{2}}{v + 1}. $$

Separate $$x$$ and $$v$$:

$$ \frac{v + 1}{1 + v^{2}}\,dv = -\frac{dx}{x}. $$

5. Integrate both sides

Left-hand side:

$$ \int \frac{v + 1}{1 + v^{2}}\,dv = \int \frac{v}{1 + v^{2}}\,dv + \int \frac{1}{1 + v^{2}}\,dv. $$

• For $$\displaystyle \int \frac{v}{1 + v^{2}}\,dv$$ set $$w = 1 + v^{2} \Rightarrow dw = 2v\,dv \Rightarrow \int = \frac12\ln(1 + v^{2}).$$
• For $$\displaystyle \int \frac{1}{1 + v^{2}}\,dv$$ use the standard form $$\tan^{-1}v.$$

Thus

$$ \int \frac{v + 1}{1 + v^{2}}\,dv = \frac12\ln(1 + v^{2}) + \tan^{-1}v. $$

Right-hand side:

$$ \int -\frac{dx}{x} = -\ln|x| + C, $$

where $$C$$ is the constant of integration.

6. Combine the integrals

$$ \frac12\ln(1 + v^{2}) + \tan^{-1}v = -\ln|x| + C. $$

Replace $$v$$ by $$y/x$$:

$$ \frac12\ln\!\left(1 + \frac{y^{2}}{x^{2}}\right) + \tan^{-1}\!\left(\frac{y}{x}\right) + \ln|x| = C. $$

7. Use the initial condition $$y(1)=1$$

• At $$x = 1,\; y = 1 \Rightarrow \dfrac{y}{x} = 1.$$
• Compute each term:
  • $$\displaystyle \frac12\ln\!(1 + 1^{2}) = \frac12\ln 2.$$
  • $$\displaystyle \tan^{-1}(1) = \frac{\pi}{4}.$$
  • $$\ln|1| = 0.$$

Thus

$$ C = \frac12\ln 2 + \frac{\pi}{4}. $$

8. Write the particular solution

$$ \frac12\ln\!\left(1 + \frac{y^{2}}{x^{2}}\right) + \tan^{-1}\!\left(\frac{y}{x}\right) + \ln|x| = \frac12\ln 2 + \frac{\pi}{4}. $$

This equation represents the required particular solution through the point $$(1,1).$$

Step 1: Convert to \(dydx\) form

The given differential equation is

$$x^{2}\,dy + (xy + y^{2})\,dx = 0$$

Dividing by $$dx$$ and by $$x^{2}$$ gives

$$\dfrac{dy}{dx} = -\dfrac{xy + y^{2}}{x^{2}} = -\dfrac{y}{x} - \dfrac{y^{2}}{x^{2}}.$$

Step 2: Introduce the substitution \(y = vx\)

Let $$y = vx \;(\text{so } v = \dfrac{y}{x}).$$ Then

$$\dfrac{dy}{dx} = v + x\,\dfrac{dv}{dx}.$$

Substituting in the differential equation,

$$v + x\,\dfrac{dv}{dx} = -v - v^{2}.$$

Simplifying,

$$x\,\dfrac{dv}{dx} = -2v - v^{2}.$$

Step 3: Separate the variables

$$\dfrac{dv}{v^{2} + 2v} = -\dfrac{dx}{x}.$$

Step 4: Integrate

First decompose the left–hand side:

$$\frac{1}{v^{2}+2v} = \frac{1}{v(v+2)} = \frac{A}{v} + \frac{B}{v+2}.$$

Solving \(1 = A(v+2)+Bv\) gives $$A = \tfrac12, \; B = -\tfrac12.$$ Hence

$$\int \dfrac{dv}{v^{2}+2v} = \tfrac12 \int \left(\dfrac{1}{v} - \dfrac{1}{v+2}\right)dv = \tfrac12\,\ln\left|\dfrac{v}{v+2}\right| + C.$$

Therefore,

$$\tfrac12\,\ln\left|\dfrac{v}{v+2}\right| = -\ln|x| + C.$$ Multiplying by 2,

$$\ln\left|\dfrac{v}{v+2}\right| = -2\ln|x| + C_{1}.$$

Exponentiating,

$$\dfrac{v}{v+2} = \dfrac{C_{1}}{x^{2}}.$$

Let the combined constant be $$k$$, so

$$\dfrac{v}{v+2} = \dfrac{k}{x^{2}}.$$

Step 5: Return to \(x\) and \(y\)

Because $$v = \dfrac{y}{x},$$

$$\dfrac{y}{x}\Big/\left(\dfrac{y}{x}+2\right) = \dfrac{y}{y+2x} = \dfrac{k}{x^{2}}.$$

Cross-multiplying,

$$y\,x^{2} = k\,(y + 2x).$$

Step 6: Apply the initial condition

Given $$y = 1$$ when $$x = 1$$:

$$1 \cdot 1^{2} = k\,(1 + 2\cdot 1) \;\Rightarrow\; 1 = 3k \;\Rightarrow\; k = \dfrac13.$$

Step 7: Write the particular solution

Substituting $$k = \tfrac13$$:

$$y\,x^{2} = \dfrac13\,(y + 2x).$$

Multiplying by 3,

$$3y\,x^{2} = y + 2x.$$

Solving for $$y$$:

$$y\,(3x^{2} - 1) = 2x \;\Rightarrow\; y = \dfrac{2x}{3x^{2} - 1}.$$

Hence the required particular solution is

$$y = \dfrac{2x}{3x^{2} - 1}.$$

Given differential equation

$$\left[x\,\sin^2\left(\dfrac{y}{x}\right)-y\right]dx+x\,dy=0\;.$$

Write it as $$M(x,y)\,dx+N(x,y)\,dy=0$$ with

$$M=x\sin^2\left(\dfrac{y}{x}\right)-y,\quad N=x.$$

Because both $$M,N$$ are homogeneous of the same degree (each term is of degree 1 in $$x,y$$), use the substitution

$$y=vx\;\;(\text{so }v=\dfrac{y}{x}),\qquad dy=v\,dx+x\,dv.$$

Substituting in the differential equation:

$$\bigl[x\sin^2(v)-vx\bigr]dx+x\bigl(v\,dx+x\,dv\bigr)=0.$$

Simplify the $$dx$$–terms:

$$x\bigl[\sin^2(v)-v\bigr]dx+xv\,dx+x^2\,dv=0\;\Longrightarrow\;x\sin^2(v)\,dx+x^2\,dv=0.$$

Divide by $$x\;(x\neq0)$$:

$$\sin^2(v)\,dx+x\,dv=0\;\Longrightarrow\;x\,dv=-\sin^2(v)\,dx.$$

Separate the variables:

$$\frac{dv}{\sin^2(v)}=-\frac{dx}{x}.$$

Recall $$\dfrac{1}{\sin^2 v}=\csc^2 v$$ and integrate:

$$\int\!\csc^2 v\,dv=-\int\!\frac{dx}{x}\;\Longrightarrow\;-\cot v=-\ln|x|+C_1.$$

Multiply by $$-1$$ and rename the constant $$C$$:

$$\cot v=\ln|x|+C.$$

Replace $$v=\dfrac{y}{x}$$ to obtain the general solution

$$\cot\left(\dfrac{y}{x}\right)=\ln|x|+C.$$

Applying the initial condition $$y=\dfrac{\pi}{4}\;\text{when }x=1$$:

$$\cot\left(\dfrac{\pi/4}{1}\right)=\ln|1|+C\;\Longrightarrow\;1=0+C\;\Longrightarrow\;C=1.$$

Hence the required particular solution is

$$\boxed{\;\cot\left(\dfrac{y}{x}\right)=\ln|x|+1\;}.$$

We have the differential equation

$$\frac{dy}{dx} - \frac{y}{x} + \csc\!\left(\frac{y}{x}\right) = 0,$$

which we rewrite as

$$\frac{dy}{dx} = \frac{y}{x} - \csc\!\left(\frac{y}{x}\right).$$

Step 1: Homogeneous substitution.

Since the right-hand side depends only on $$y/x$$, put $$v = \dfrac{y}{x}$$, so $$y = vx$$ and $$\dfrac{dy}{dx} = v + x\,\dfrac{dv}{dx}$$.

Step 2: Substitute and simplify.

$$v + x\,\frac{dv}{dx} = v - \csc v \;\Longrightarrow\; x\,\frac{dv}{dx} = -\csc v.$$

Step 3: Separate the variables.

$$\sin v\,dv = -\frac{dx}{x}.$$

Step 4: Integrate.

$$\int \sin v\,dv = -\int \frac{dx}{x}.$$

$$-\cos v = -\ln|x| + C_{1} \;\Longrightarrow\; \cos v = \ln|x| + C.$$

Step 5: Back-substitute and apply the initial condition.

$$\cos\!\left(\frac{y}{x}\right) = \ln|x| + C.$$

At $$x = 1$$, $$y = 0$$: $$\cos 0 = \ln 1 + C \;\Longrightarrow\; C = 1$$.

Step 6: Particular solution.

$$\cos\!\left(\frac{y}{x}\right) = \ln|x| + 1.$$

We start with the given differential equation

$$2xy + y^2 - 2x^2\frac{dy}{dx} = 0.$$

Move the derivative term to one side and divide by $$2x^2$$ (which is non-zero for $$x \neq 0$$):

$$2x^2\frac{dy}{dx} = 2xy + y^2 \;\;\Longrightarrow\;\; \frac{dy}{dx} = \frac{2xy + y^2}{2x^2} = \frac{y}{x} + \frac{y^2}{2x^2}. $$

The right–hand side is a function of the single ratio $$y/x$$, so the equation is homogeneous of degree 0. Put

$$y = vx \quad (\text{so } v = y/x).$$

Then

$$\frac{dy}{dx} = v + x\frac{dv}{dx}.$$

Substitute $$y = vx$$ and $$dy/dx = v + x\,dv/dx$$ into the differential equation:

$$v + x\frac{dv}{dx} = v + \frac{v^2}{2}.$$

Cancel the common term $$v$$ from both sides:

$$x\frac{dv}{dx} = \frac{v^2}{2}.$$

Separate the variables:

$$\frac{dv}{v^2} = \frac{dx}{2x}.$$

Integrate both sides:

$$\int v^{-2}\,dv = \int \frac{1}{2x}\,dx \;\;\Longrightarrow\;\; -\frac{1}{v} = \frac12\ln|x| + C.$$

Rewrite in terms of $$y$$ and $$x$$ (remember $$v = y/x$$):

$$-\frac{x}{y} = \frac12\ln|x| + C.$$

Multiply by $$-1$$ and rename the constant $$( -C \to K )$$ for convenience:

$$\frac{x}{y} = K - \frac12\ln|x|.$$

Use the initial condition $$y = 2$$ when $$x = 1$$ to find $$K$$:

$$\frac{1}{2} = K - \frac12\ln|1| \;\;\Longrightarrow\;\; \frac{1}{2} = K - 0 \;\;\Longrightarrow\;\; K = \frac12.$$

Substitute $$K = 1/2$$ back into the relation:

$$\frac{x}{y} = \frac12 - \frac12\ln|x|.$$

Multiply by 2 and solve for $$y$$:

$$\frac{2x}{y} = 1 - \ln|x| \;\;\Longrightarrow\;\; y = \frac{2x}{1 - \ln|x|}.$$

This satisfies both the differential equation and the given initial condition. Hence the required particular solution is written compactly as

$$y = \dfrac{2x}{1 - \ln|x|}.$$

The given differential equation is $$\frac{dx}{dy}=h\!\left(\frac{x}{y}\right).$$ It is called homogeneous because the right–hand side depends only on the ratio $$\frac{x}{y}.$$

For a first-order homogeneous equation written with the independent variable $$y,$$ we make the variable in the numerator (here, $$x$$) proportional to the independent variable. Therefore set

$$x=v\,y\quad\bigl(v=v(y)\bigr).$$

At once $$\dfrac{x}{y}=v,$$ so the ratio becomes a single new variable.

Differentiating $$x=v y$$ with respect to $$y$$ gives

$$\frac{dx}{dy}=v+y\,\frac{dv}{dy}.$$

Substitute this value and $$\tfrac{x}{y}=v$$ in the original equation:

$$v+y\,\frac{dv}{dy}=h(v).$$

Rearrange to obtain a separable form:

$$y\,\frac{dv}{dy}=h(v)-v\;\;\Longrightarrow\;\;\frac{dy}{y}=\frac{dv}{h(v)-v}.$$

This differential equation can now be integrated by separation of variables. Hence the substitution that solves the homogeneous equation is $$x=vy,$$ i.e. option (C).

For a first-order differential equation written as

$$M(x,y)\,dx+N(x,y)\,dy=0$$

it is called homogeneous when both $$M$$ and $$N$$ are homogeneous functions of the same degree, i.e. when every term of each function has the same total power of $$x$$ and $$y$$.

We inspect each option in turn.

1. Option (A)
   $$M(x,y)=-(3y+2x+4),\qquad N(x,y)=4x+6y+5$$ The terms have degrees 1, 1, 0 in each function. Because the constant terms ($$4,5$$) are degree 0, $$M$$ and $$N$$ are not homogeneous.
   Not homogeneous.
2. Option (B)
   $$M(x,y)=xy\;(\text{degree }2),\qquad N(x,y)=-(x^{3}+y^{3})\;(\text{degree }3)$$ $$M$$ and $$N$$ are homogeneous but of different degrees (2 and 3).
   Not homogeneous.
3. Option (C)
   $$M(x,y)=x^{3}+2y^{2},\qquad N(x,y)=2xy$$ Inside $$M$$ the term $$x^{3}$$ is degree 3 whereas $$2y^{2}$$ is degree 2, so $$M$$ itself is not homogeneous.
   Not homogeneous.
4. Option (D)
   $$M(x,y)=y^{2}\;(\text{degree }2),\qquad N(x,y)=x^{2}-xy-y^{2}\;(\text{each term degree }2)$$ Both $$M$$ and $$N$$ are homogeneous of degree 2; hence the differential equation is homogeneous.
   Homogeneous.

Therefore, only option (D) satisfies the condition.

The given differential equation is

$$\frac{dy}{dx}-y=\cos x$$

This is a linear first-order ordinary differential equation of the standard form

$$\frac{dy}{dx}+P(x)\,y = Q(x)$$

with

• $$P(x)=-1$$
• $$Q(x)=\cos x$$

1. Compute the integrating factor (I.F.)

$$\text{I.F.}=e^{\int P(x)\,dx}=e^{\int(-1)\,dx}=e^{-x}$$

2. Multiply every term of the differential equation by the I.F.

$$e^{-x}\,\frac{dy}{dx}-e^{-x}y=e^{-x}\cos x$$

3. Recognise the left side as a single derivative

Because $$\frac{d}{dx}\bigl(e^{-x}y\bigr)=e^{-x}\,\frac{dy}{dx}-e^{-x}y,$$ the equation becomes

$$\frac{d}{dx}\bigl(e^{-x}y\bigr)=e^{-x}\cos x$$

4. Integrate both sides

$$e^{-x}y=\int e^{-x}\cos x\,dx+C$$

Evaluate the integral

Use the standard result $$\int e^{ax}\cos bx\,dx=\frac{e^{ax}\,(a\cos bx+b\sin bx)}{a^{2}+b^{2}}$$ with $$a=-1,\;b=1$$:

$$\int e^{-x}\cos x\,dx=\frac{e^{-x}\,\bigl((-1)\cos x+1\sin x\bigr)}{(-1)^{2}+1^{2}}=\frac{e^{-x}\,(\sin x-\cos x)}{2}$$

Hence

$$e^{-x}y=\frac{e^{-x}}{2}(\sin x-\cos x)+C$$

5. Solve for $$y$$

Multiply by $$e^{x}$$:

$$y=\frac{1}{2}(\sin x-\cos x)+Ce^{x}$$

where $$C$$ is an arbitrary real constant.

Therefore, the general solution is

$$y=Ce^{x}+\tfrac{1}{2}(\sin x-\cos x).$$

The given equation is

$$x\,\frac{dy}{dx}+2y=x^{2},\;\;(x\neq 0).$$

Divide throughout by $$x$$ to write it in the standard linear form $$\displaystyle \frac{dy}{dx}+P(x)y=Q(x)$$:

$$\frac{dy}{dx}+\frac{2}{x}\,y=x.$$

Here $$P(x)=\dfrac{2}{x}$$ and $$Q(x)=x.$$

Step 1: Integrating factor (I.F.).

$$\text{I.F.}=\exp\left(\int P(x)\,dx\right)=\exp\left(\int \frac{2}{x}\,dx\right)=\exp\bigl(2\ln|x|\bigr)=x^{2}.$$(Because $$x\neq 0$$, we drop the absolute value.)

Step 2: Multiply the whole differential equation by the I.F.

$$x^{2}\frac{dy}{dx}+2x\,y=x^{3}.$$

The left-hand side is the derivative of $$y\,x^{2}$$:

$$\frac{d}{dx}(y\,x^{2})=x^{3}.$$

Step 3: Integrate both sides.

$$\int \frac{d}{dx}(y\,x^{2})\,dx=\int x^{3}\,dx\;\;\Longrightarrow\;\;y\,x^{2}=\frac{x^{4}}{4}+C.$$

Step 4: Solve for $$y$$.

$$y=\frac{x^{4}}{4x^{2}}+\frac{C}{x^{2}}=\frac{x^{2}}{4}+\frac{C}{x^{2}}.$$

Thus the required general solution of the differential equation is

$$y=\frac{x^{2}}{4}+\frac{C}{x^{2}},\;\;(x\neq 0).$$

We are given the differential equation

$$y\,dx-(x+2y^2)\,dy=0.$$

First, separate the differentials so that the derivative of one variable with respect to the other appears explicitly. Transposing the second term to the right side gives

$$y\,dx=(x+2y^2)\,dy.$$

Divide by $$dy$$ (and assume $$y\neq0$$ so that division by $$y$$ is permissible):

$$\frac{dx}{dy}=\frac{x+2y^2}{y}.$$

Rewrite to obtain the standard linear form in the dependent variable $$x$$ and the independent variable $$y$$:

$$\frac{dx}{dy}-\frac{x}{y}=2y.$$

This is a first–order linear differential equation of the form

$$\frac{dx}{dy}+P(y)\,x=Q(y)\quad\text{with}\quad P(y)=-\frac1y, \; Q(y)=2y.$$

Step 1 : Find the integrating factor (IF)

$$\text{IF}=\mu(y)=\exp\!\left(\int P(y)\,dy\right)=\exp\!\left(\int-\frac1y\,dy\right)=\exp(-\ln|y|)=\frac1y.$$

Step 2 : Multiply the whole equation by the integrating factor

$$\frac1y\,\frac{dx}{dy}-\frac{x}{y^2}=2.$$

Step 3 : Recognise the left–hand side as a single derivative

Because

$$\frac{d}{dy}\left(\frac{x}{y}\right)=\frac1y\,\frac{dx}{dy}-\frac{x}{y^2},$$

the differential equation becomes

$$\frac{d}{dy}\left(\frac{x}{y}\right)=2.$$

Step 4 : Integrate with respect to $$y$$

$$\int\frac{d}{dy}\left(\frac{x}{y}\right)\,dy=\int2\,dy \;\;\Rightarrow\;\; \frac{x}{y}=2y+C,$$

where $$C$$ is the constant of integration.

Step 5 : Express the solution neatly

Multiply by $$y$$:

$$x=2y^2+Cy.$$

Equivalently one may write

$$x-2y^2=Cy\quad\text{or}\quad\frac{x-2y^2}{y}=C,$$

but the form $$x=2y^2+Cy$$ is perfectly acceptable as the general solution.

Given differential equation

$$\frac{dy}{dx}+y\cot x=2x+x^{2}\cot x \quad (x\neq 0).$$

This is a linear first-order ODE of the form $$\dfrac{dy}{dx}+P(x)\,y=Q(x)$$ with

$$P(x)=\cot x, \qquad Q(x)=2x+x^{2}\cot x.$$

Step 1 : Integrating factor (I.F.)

$$\text{I.F.}=e^{\int P(x)\,dx}=e^{\int \cot x\,dx}=e^{\ln|\sin x|}=\sin x.$$

Step 2 : Multiply the equation by the I.F.

$$\sin x\,\frac{dy}{dx}+y\sin x\,\cot x=(2x+x^{2}\cot x)\sin x.$$

Because $$\sin x\,\cot x=\sin x\,\dfrac{\cos x}{\sin x}=\cos x,$$ the left side becomes

$$\sin x\,\frac{dy}{dx}+y\cos x.$$

But $$\frac{d}{dx}(y\sin x)=y'\sin x+y\cos x,$$ so

$$\frac{d}{dx}(y\sin x)=2x\sin x+x^{2}\cos x.$$

Step 3 : Integrate both sides

$$y\sin x=\int 2x\sin x\,dx+\int x^{2}\cos x\,dx+C.$$

First integral:

$$\int 2x\sin x\,dx=\!-\,2x\cos x+2\sin x.$$

Second integral (two successive parts):

$$\int x^{2}\cos x\,dx=x^{2}\sin x-2\int x\sin x\,dx$$

and $$\int x\sin x\,dx=-x\cos x+\sin x,$$ so

$$\int x^{2}\cos x\,dx=x^{2}\sin x+2x\cos x-2\sin x.$$

Add the two results:

$$(-2x\cos x+2\sin x)+(x^{2}\sin x+2x\cos x-2\sin x)=x^{2}\sin x.$$

Thus

$$y\sin x=x^{2}\sin x+C.$$

Step 4 : Obtain the general solution

$$y=\dfrac{x^{2}\sin x+C}{\sin x}=x^{2}+C\,\csc x.$$

Step 5 : Apply the initial condition

Given $$y=0$$ when $$x=\dfrac{\pi}{2}.$$ Because $$\csc\dfrac{\pi}{2}=1,$$

$$0=\left(\dfrac{\pi}{2}\right)^{2}+C\cdot 1 \;\;\Rightarrow\;\; C=-\dfrac{\pi^{2}}{4}.$$

Step 6 : Particular solution

$$y=x^{2}-\dfrac{\pi^{2}}{4}\,\csc x.$$

This solution is valid for all $$x$$ such that $$\sin x\neq 0,$$ i.e. $$x\neq n\pi$$ where $$n\in\mathbb{Z}.$$

Let the required curve be $$y=f(x)$$. Its slope at any point $$ (x, y) $$ is given to be the sum of the abscissa $$x$$ and the product of the abscissa and ordinate $$xy$$.

Therefore $$ \frac{dy}{dx}=x+xy=x(1+y). $$

1. Separate the variables
$$ \frac{dy}{1+y}=x\,dx. $$

2. Integrate both sides
$$ \int \frac{dy}{1+y}=\int x\,dx. $$ Left side: $$ \int \frac{dy}{1+y}=\ln|1+y|. $$ Right side: $$ \int x\,dx=\frac{x^{2}}{2}+C, $$ where $$C$$ is the constant of integration.

Hence $$ \ln|1+y|=\frac{x^{2}}{2}+C. $$

3. Remove the logarithm
$$ |1+y|=e^{\frac{x^{2}}{2}+C}=e^{C}\,e^{\frac{x^{2}}{2}}. $$ Put $$e^{C}=K$$ (a non-zero constant). Then $$ 1+y=K e^{\frac{x^{2}}{2}}. $$

4. Use the initial condition
The curve passes through $$ (0,1) $$. Substitute: $$ 1+1=K e^{0}\;\Rightarrow\;2=K. $$ Thus $$K=2$$.

5. Equation of the required curve
Substituting $$K=2$$, $$ 1+y=2 e^{\frac{x^{2}}{2}}\;\Longrightarrow\;y=2 e^{\frac{x^{2}}{2}}-1. $$

Therefore, the equation of the curve is $$ \boxed{\;y=2 e^{\frac{x^{2}}{2}}-1\;}.\! $$

Given differential equation: $$\frac{dy}{dx}+2y=\sin x$$

This is a first-order linear ODE of the form $$\frac{dy}{dx}+P(x)y=Q(x)$$ with $$P(x)=2$$ and $$Q(x)=\sin x$$.

1. Integrating factor

$$\text{I.F.}=e^{\int P(x)\,dx}=e^{\int 2\,dx}=e^{2x}$$

2. Multiply by the I.F.

$$e^{2x}\frac{dy}{dx}+2e^{2x}y=e^{2x}\sin x$$

The left side is $$\frac{d}{dx}(ye^{2x})$$ since $$\frac{d}{dx}(ye^{2x})=e^{2x}\frac{dy}{dx}+2e^{2x}y$$.

3. Rewrite as a perfect derivative

$$\frac{d}{dx}(ye^{2x})=e^{2x}\sin x$$

4. Integrate

$$ye^{2x}=\int e^{2x}\sin x\,dx+C$$

5. Evaluate the integral

Using $$\int e^{ax}\sin bx\,dx=\dfrac{e^{ax}(a\sin bx-b\cos bx)}{a^{2}+b^{2}}$$ with $$a=2,\;b=1$$:

$$\int e^{2x}\sin x\,dx=\frac{e^{2x}(2\sin x-\cos x)}{2^{2}+1^{2}}=\frac{e^{2x}(2\sin x-\cos x)}{5}$$

6. Substitute back

$$ye^{2x}=\frac{e^{2x}(2\sin x-\cos x)}{5}+C$$

7. Solve for $$y$$

$$y=\frac{2\sin x-\cos x}{5}+Ce^{-2x}$$

General solution: $$y=\dfrac{2\sin x-\cos x}{5}+Ce^{-2x}$$

The differential equation is

$$\frac{dy}{dx}+3y = e^{-2x}$$

This is a first-order linear differential equation of the standard form

$$\frac{dy}{dx}+P(x)\,y = Q(x),$$

with

$$P(x)=3 \quad\text{and}\quad Q(x)=e^{-2x}.$$

Step 1: Find the integrating factor (I.F.).

The integrating factor is defined as

$$\text{I.F.}=e^{\int P(x)\,dx}=e^{\int 3\,dx}=e^{3x}.$$

Step 2: Multiply the whole differential equation by the integrating factor.

Multiplying each term by $$e^{3x}$$ gives

$$e^{3x}\,\frac{dy}{dx}+3e^{3x}y=e^{3x}\,e^{-2x}.$$

Simplify the right–hand side:

$$e^{3x}\,e^{-2x}=e^{x}.$$

Step 3: Recognise the left side as the derivative of a product.

Observe that

$$\frac{d}{dx}\bigl(e^{3x}\,y\bigr)=e^{3x}\,\frac{dy}{dx}+3e^{3x}y,$$

which is exactly the left–hand side. Therefore, the equation becomes

$$\frac{d}{dx}\bigl(e^{3x}\,y\bigr)=e^{x}.$$

Step 4: Integrate both sides with respect to $$x$$.

$$\int\frac{d}{dx}\bigl(e^{3x}\,y\bigr)\,dx=\int e^{x}\,dx.$$

This gives

$$e^{3x}\,y = e^{x} + C,$$

where $$C$$ is the constant of integration.

Step 5: Solve for $$y$$.

Divide both sides by $$e^{3x}$$:

$$y = \frac{e^{x}+C}{e^{3x}} = e^{x-3x}+C e^{-3x}=e^{-2x}+C e^{-3x}.$$

Hence, the general solution is

$$y = e^{-2x}+C e^{-3x},$$

where $$C$$ is an arbitrary real constant.

We have to solve the differential equation

$$\frac{dy}{dx} + \frac{y}{x} = x^{2}.$$

This is a first-order linear ODE of the standard form $$\dfrac{dy}{dx} + P(x)\,y = Q(x)$$ with

$$P(x) = \frac{1}{x}, \qquad Q(x) = x^{2}.$$

Step 1: Integrating factor (I.F.).

$$\text{I.F.} = e^{\int P(x)\,dx} = e^{\int \frac{1}{x}\,dx} = e^{\ln|x|} = |x|.$$

For convenience we work on $$x > 0$$, where the integrating factor is simply $$x$$. (The analysis for $$x < 0$$ is analogous and yields the same general formula.)

Step 2: Multiply the equation by the I.F.

$$x\,\frac{dy}{dx} + y = x^{3}.$$

Step 3: Recognise the left side as a perfect derivative.

$$\frac{d}{dx}(xy) = x\,\frac{dy}{dx} + y,$$

so the equation becomes

$$\frac{d}{dx}(xy) = x^{3}.$$

Step 4: Integrate with respect to $$x$$.

$$xy = \int x^{3}\,dx = \frac{x^{4}}{4} + C.$$

Step 5: Solve for $$y$$.

$$y = \frac{x^{3}}{4} + \frac{C}{x},$$

where $$C$$ is an arbitrary constant.

The given differential equation is

$$\frac{dy}{dx}+ (\sec x)\,y = \tan x \qquad\left(0 \le x < \frac{\pi}{2}\right).$$

This is a linear first–order equation of the standard form $$\frac{dy}{dx}+P(x)\,y = Q(x)$$ with

$$P(x)=\sec x, \qquad Q(x)=\tan x.$$

Step 1: Integrating factor (I.F.)

The integrating factor is

$$\text{I.F.}=e^{\int P(x)\,dx}=e^{\int \sec x\,dx}.$$

We know that $$\int \sec x\,dx = \ln|\sec x+\tan x| + C.$$

Hence

$$\text{I.F.}=e^{\ln|\sec x+\tan x|}=|\sec x+\tan x|.$$

In the interval $$0\le x<\dfrac{\pi}{2}$$ we have $$\sec x+\tan x>0,$$ so

$$\text{I.F.}=\sec x+\tan x.$$

Step 2: Multiply the equation by the I.F.

$$\bigl(\sec x+\tan x\bigr)\,\frac{dy}{dx}+\bigl(\sec x+\tan x\bigr)(\sec x)\,y = \tan x\,(\sec x+\tan x).$$

Because the left side equals the derivative of $$y(\sec x+\tan x),$$ we have

$$\frac{d}{dx}\Bigl[y(\sec x+\tan x)\Bigr]=\tan x\,(\sec x+\tan x).$$

Step 3: Integrate both sides.

$$y(\sec x+\tan x)=\int \tan x\,\sec x\,dx+\int \tan^2 x\,dx+C.$$

Compute each integral:

• $$\int \tan x\,\sec x\,dx = \sec x;$$
• $$\int \tan^2 x\,dx = \int (\sec^2 x-1)\,dx = \tan x - x.$$

Therefore

$$y(\sec x+\tan x)=\sec x+\tan x - x + C.$$

Step 4: Solve for $$y$$.

$$y=\frac{\sec x+\tan x-x+C}{\sec x+\tan x}$$

or equivalently

$$y=1-\frac{x-C}{\sec x+\tan x},$$ where $$C$$ is an arbitrary constant.

Thus the general solution is

$$y=1-\frac{x-C}{\sec x+\tan x}, \qquad 0\le x<\frac{\pi}{2}.$$

The given differential equation is

$$\cos^{2} x\,\frac{dy}{dx} + y = \tan x, \qquad 0 \le x \lt \frac{\pi}{2}.$$

Divide every term by $$\cos^{2} x$$ to reduce it to standard linear form $$\dfrac{dy}{dx} + P(x)\,y = Q(x)$$:

$$\frac{dy}{dx} + \sec^{2} x\,y = \tan x\,\sec^{2} x.$$

Here $$P(x) = \sec^{2} x$$ and $$Q(x) = \tan x\,\sec^{2} x$$.

Integrating factor.

$$\text{I.F.} = e^{\int P(x)\,dx} = e^{\int \sec^{2} x\,dx} = e^{\tan x}.$$

Multiply by the I.F.

$$e^{\tan x}\,\frac{dy}{dx} + e^{\tan x}\,\sec^{2} x\,y = e^{\tan x}\,\tan x\,\sec^{2} x.$$

The left-hand side is $$\dfrac{d}{dx}\bigl(y\,e^{\tan x}\bigr)$$, so

$$\frac{d}{dx}\bigl(y\,e^{\tan x}\bigr) = e^{\tan x}\,\tan x\,\sec^{2} x.$$

Integrate.

$$y\,e^{\tan x} = \int e^{\tan x}\,\tan x\,\sec^{2} x\,dx + C.$$

Substitute $$t = \tan x$$, so $$dt = \sec^{2} x\,dx$$:

$$\int t\,e^{t}\,dt.$$

Using integration by parts ($$u = t$$, $$dv = e^{t}\,dt$$):

$$\int t\,e^{t}\,dt = t\,e^{t} - \int e^{t}\,dt = (t - 1)\,e^{t}.$$

Re-substituting $$t = \tan x$$:

$$y\,e^{\tan x} = (\tan x - 1)\,e^{\tan x} + C.$$

Dividing by $$e^{\tan x}$$:

$$y = \tan x - 1 + C\,e^{-\tan x}.$$

This is valid on the given interval $$0 \le x \lt \dfrac{\pi}{2}$$, where $$\cos x \neq 0$$.

We have the first-order linear differential equation

$$x\,\frac{dy}{dx} + 2y = x^{2}\,\log x \qquad (x > 0).$$

1. Reduce to standard linear form.

Divide by $$x$$:

$$\frac{dy}{dx} + \frac{2}{x}\,y = x\,\log x.$$

This matches $$\dfrac{dy}{dx} + P(x)\,y = Q(x)$$ with $$P(x) = \dfrac{2}{x}$$ and $$Q(x) = x\,\log x$$.

2. Integrating factor (I.F.).

$$\text{I.F.} = e^{\int P(x)\,dx} = e^{\int \frac{2}{x}\,dx} = e^{2\ln x} = x^{2}.$$

3. Multiply by the I.F.

$$x^{2}\,\frac{dy}{dx} + 2x\,y = x^{3}\,\log x.$$

The left side is $$\dfrac{d}{dx}(x^{2}y)$$. Hence

$$\frac{d}{dx}(x^{2}y) = x^{3}\,\log x.$$

4. Integrate both sides.

$$x^{2}y = \int x^{3}\,\log x\,dx + C,$$

where $$C$$ is the (single) constant of integration.

Evaluate $$\displaystyle\int x^{3}\log x\,dx$$ by parts with $$u = \log x$$, $$dv = x^{3}\,dx$$, so $$du = \dfrac{dx}{x}$$, $$v = \dfrac{x^{4}}{4}$$:

$$\int x^{3}\,\log x\,dx = \frac{x^{4}}{4}\,\log x - \int \frac{x^{4}}{4} \cdot \frac{1}{x}\,dx = \frac{x^{4}}{4}\,\log x - \frac{1}{4}\int x^{3}\,dx = \frac{x^{4}}{4}\,\log x - \frac{x^{4}}{16}.$$

Therefore

$$x^{2}y = \frac{x^{4}}{4}\,\log x - \frac{x^{4}}{16} + C.$$

5. Solve for $$y$$.

$$y = \frac{x^{2}}{4}\,\log x - \frac{x^{2}}{16} + \frac{C}{x^{2}}, \quad C \in \mathbb{R}.$$

We have to solve the first-order linear differential equation

$$x\,\log x\,\frac{dy}{dx}+y=\frac{2}{x}\,\log x.$$

1. Reduce to the standard linear form.

Divide every term by $$x\,\log x$$ (possible for $$x>0,\;x\neq 1$$):

$$\frac{dy}{dx}+\frac{1}{x\,\log x}\,y=\frac{2}{x^{2}}.$$

This is of the form $$\dfrac{dy}{dx}+P(x)\,y=Q(x)$$ with
$$P(x)=\frac{1}{x\,\log x},\qquad Q(x)=\frac{2}{x^{2}}.$$

2. Find the integrating factor (I.F.).

$$\text{I.F.}=e^{\int P(x)\,dx}=e^{\displaystyle\int\frac{1}{x\,\log x}\,dx}.$$

Put $$t=\log x\;\Rightarrow\;dt=\dfrac{1}{x}\,dx.$$ Then

$$\int\frac{1}{x\,\log x}\,dx=\int\frac{1}{t}\,dt=\log t+K=\log(\log x).$$

Hence $$\text{I.F.}=e^{\log(\log x)}=\log x.$$

3. Multiply the whole equation by the I.F.

$$\log x\,\frac{dy}{dx}+\frac{\log x}{x\,\log x}\,y=\frac{2\,\log x}{x^{2}}\;\Longrightarrow\;\log x\,\frac{dy}{dx}+\frac{y}{x}=\frac{2\,\log x}{x^{2}}.$$

The left side is the derivative of the product $$y\,\log x$$ because

$$\frac{d}{dx}(y\,\log x)=\log x\,\frac{dy}{dx}+y\,\frac{1}{x}.$$

Therefore

$$\frac{d}{dx}(y\,\log x)=\frac{2\,\log x}{x^{2}}.$$

4. Integrate.

$$y\,\log x=\int\frac{2\,\log x}{x^{2}}\,dx+C.$$

To evaluate the integral, again put $$t=\log x,\;dx=x\,dt.$$ Then

$$\int\frac{2\,\log x}{x^{2}}\,dx=\int\frac{2t}{x^{2}}\,dx=\int2t\,e^{-t}\,dt.$$

Using integration by parts:
Let $$u=2t\;(du=2\,dt),\;dv=e^{-t}\,dt\;(v=-e^{-t}).$$
$$\Rightarrow\int2t\,e^{-t}\,dt=-2t\,e^{-t}+2\int e^{-t}\,dt=-2t\,e^{-t}-2e^{-t}+K.$$

Because $$e^{-t}=1/x$$ and $$t=\log x,$$

$$\int\frac{2\,\log x}{x^{2}}\,dx=-\frac{2(\log x+1)}{x}+K.$$

Thus

$$y\,\log x=-\frac{2(\log x+1)}{x}+C.$$

5. Solve for $$y$$.

$$y=-\frac{2(\log x+1)}{x\,\log x}+\frac{C}{\log x}.$$

6. General solution.

$$y(x)=\frac{C}{\log x}-\frac{2(\log x+1)}{x\,\log x},\qquad x>0,\;x\neq1.$$

Given differential equation

$$ (1 + x^2)\,dy + 2xy\,dx = \cot x\,dx \qquad (x \neq 0) $$

Re–arrange so that $$dy/dx$$ is explicit:

$$ (1 + x^2)\,\frac{dy}{dx} = \cot x - 2xy $$

$$ \frac{dy}{dx} + \frac{2x}{1 + x^2}\,y = \frac{\cot x}{1 + x^2} $$

This is a first-order linear ODE of the form
$$ \frac{dy}{dx} + P(x)\,y = Q(x) $$
with

$$ P(x) = \frac{2x}{1 + x^2}, \qquad Q(x) = \frac{\cot x}{1 + x^2}. $$

Integrating factor

$$ \text{IF} = e^{\int P(x)\,dx} = e^{\int \frac{2x}{1 + x^2}\,dx}. $$

Compute the integral:

$$ \int \frac{2x}{1 + x^2}\,dx = \ln|1 + x^2| + C_0 $$

(put $$u = 1 + x^2 \Rightarrow du = 2x\,dx$$).

Hence

$$ \text{IF} = e^{\ln|1 + x^2|} = 1 + x^2. $$

Multiply the ODE by the integrating factor

$$ (1 + x^2)\,\frac{dy}{dx} + 2x\,y = \cot x. $$

The left–hand side is the derivative of $$y(1 + x^2)$$:

$$ \frac{d}{dx}\bigl[y(1 + x^2)\bigr] = \cot x. $$

Integrate both sides

$$ y(1 + x^2) = \int \cot x\,dx + C $$

$$ y(1 + x^2) = \ln|\sin x| + C. $$

General solution

$$ y = \frac{\ln|\sin x| + C}{1 + x^2}. $$

Given differential equation

$$x\,\frac{dy}{dx}+y-x+xy\cot x=0 \quad (x\neq 0)$$

1. Convert to standard linear form

Bring the terms involving $$y$$ to the left and the rest to the right:

$$x\,\frac{dy}{dx}+y+xy\cot x = x$$

Divide by $$x$$ (allowed because $$x\neq0$$):

$$\frac{dy}{dx}+\Bigl(\frac1x+\cot x\Bigr)y = 1$$

This is of the form $$\dfrac{dy}{dx}+P(x)y = Q(x)$$ with

• $$P(x)=\dfrac1x+\cot x$$
• $$Q(x)=1$$

2. Find the integrating factor

$$\text{IF}=e^{\int P(x)\,dx}=e^{\int \bigl(\frac1x+\cot x\bigr)dx} =e^{\int \frac{1}{x}\,dx+\int \cot x\,dx} =e^{\ln|x|+\ln|\sin x|}=|x\sin x|$$

For the general solution we may write simply

$$\text{IF}=x\sin x$$

3. Multiply the whole equation by the integrating factor

$$x\sin x\,\frac{dy}{dx}+x\sin x\Bigl(\frac1x+\cot x\Bigr)y = x\sin x$$

But by construction

$$x\sin x\,\frac{dy}{dx}+\bigl(\frac1x+\cot x\bigr)(x\sin x)y = \frac{d}{dx}(x\sin x\,y)$$

Hence

$$\frac{d}{dx}(x\sin x\,y)=x\sin x$$

4. Integrate both sides

$$x\sin x\,y = \int x\sin x\,dx + C$$

Evaluate the integral by parts:

Let $$u=x,\;dv=\sin x\,dx\;\Rightarrow\;du=dx,\;v=-\cos x$$

$$\int x\sin x\,dx = -x\cos x + \int \cos x\,dx = -x\cos x + \sin x + C_1$$

(the constant of integration has been absorbed into the single constant $$C$$ used above).

Therefore

$$x\sin x\,y = -x\cos x + \sin x + C$$

5. Solve for $$y$$

$$y = \frac{-x\cos x + \sin x + C}{x\sin x}$$

Or, separating the constant term if desired,

$$y = -\cot x + \frac1x + \frac{C}{x\sin x}$$

Hence, the general solution of the given differential equation is

$$y = \frac{\sin x - x\cos x + C}{x\sin x}$$

Given differential equation :

$$(x+y)\,\dfrac{dy}{dx}=1$$

Re-write it in the form $$\dfrac{dy}{dx}=\dfrac1{x+y}$$ and invert the derivative so that the independent variable is $$y$$:

$$\dfrac{dx}{dy}=x+y$$

This is a first–order linear ODE in $$x(y)$$:

$$\dfrac{dx}{dy}-x=y$$    (1)

The standard linear form is $$\dfrac{dx}{dy}+P(y)\,x=Q(y)$$ with

• $$P(y)=-1$$
• $$Q(y)=y$$

Integrating factor (I.F.)

$$\text{I.F.}=e^{\int P(y)\,dy}=e^{\int(-1)\,dy}=e^{-y}$$

Multiply equation (1) by this I.F.:

$$e^{-y}\dfrac{dx}{dy}-e^{-y}x=ye^{-y}$$

The left side is the derivative of $$x e^{-y}$$:

$$\dfrac{d}{dy}\bigl(x e^{-y}\bigr)=y e^{-y}$$

Integrate both sides with respect to $$y$$:

$$x e^{-y}=\int y e^{-y}\,dy+C$$

Compute the integral by parts (take $$u=y,\;dv=e^{-y}dy$$):

$$\int y e^{-y}\,dy=-y e^{-y}-e^{-y}+C_1=-(y+1)e^{-y}+C_1$$

Absorbing the constant $$C_1$$ into the arbitrary constant $$C$$, we have

$$x e^{-y}=-(y+1)e^{-y}+C$$

Multiply by $$e^{y}$$ to clear the exponent:

$$x=-(y+1)+C e^{y}$$

Re-arrange for a tidier implicit form:

$$(x+y+1)=C e^{y}$$

Thus, the general solution of the differential equation is

$$ (x+y+1)=C e^{y}, \; C\in\mathbb{R}. $$

For the given differential equation

$$y\,dx + (x - y^{2})\,dy = 0$$

let

$$M(x,y)=y, \qquad N(x,y)=x - y^{2}.$$

1. Test for exactness

Compute the required partial derivatives:

$$\frac{\partial M}{\partial y}=\frac{\partial}{\partial y}(y)=1, \qquad \frac{\partial N}{\partial x}=\frac{\partial}{\partial x}(x-y^{2})=1.$$

Since $$\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x},$$ the differential form is exact.

2. Find the potential function $$\Phi(x,y)$$

Because the equation is exact, there exists a function $$\Phi(x,y)$$ such that

$$d\Phi = M\,dx + N\,dy.$$

(i) Integrate $$M$$ with respect to $$x$$, treating $$y$$ as a constant:

$$\Phi(x,y)=\int y\,dx = xy + g(y),$$

where $$g(y)$$ is an arbitrary function of $$y$$.

(ii) Differentiate $$\Phi$$ with respect to $$y$$ and equate to $$N$$:

$$\frac{\partial \Phi}{\partial y}=x + g'(y)=N = x - y^{2}.$$ Hence, $$g'(y)=-y^{2}.$$

(iii) Integrate $$g'(y)$$ to get $$g(y)$$:

$$g(y)=\int -y^{2}\,dy = -\frac{y^{3}}{3}+C_{1},$$

where the additive constant $$C_{1}$$ can be absorbed later into the final constant.

Therefore

$$\Phi(x,y)=xy - \frac{y^{3}}{3}.$$

3. Implicit general solution

The solution of an exact differential equation is given by $$\Phi(x,y)=C$$, where $$C$$ is an arbitrary constant. Thus,

$$xy - \frac{y^{3}}{3}=C.$$

This represents the required one-parameter family of integral curves.

Given the differential equation

$$ (x + 3y^2) \dfrac{dy}{dx} = y \quad (y>0). $$

Rewrite it in the form

$$ \dfrac{dy}{dx} = \dfrac{y}{x + 3y^2}. $$

This is not directly separable, so treat $$x$$ as a function of $$y$$ by taking the reciprocal derivative:

$$ \dfrac{dx}{dy} = \dfrac{x + 3y^2}{y}. $$

Simplify:

$$ \dfrac{dx}{dy} = \frac{x}{y} + 3y. $$

This is a linear first-order ODE in $$x$$ with independent variable $$y$$:

$$ \dfrac{dx}{dy} - \frac{1}{y}x = 3y. $$

Integrating factor

$$ \mu(y) = \exp\!\left(\int -\tfrac{1}{y}\,dy \right) = \exp(-\ln y) = y^{-1}. $$

Multiply the equation by $$y^{-1}$$:

$$ y^{-1}\dfrac{dx}{dy} - y^{-2}x = 3. $$

The left side is

$$ \dfrac{d}{dy}\left( \tfrac{x}{y} \right), $$

so

$$ \dfrac{d}{dy}\left( \frac{x}{y} \right) = 3. $$

Integrate

$$ \frac{x}{y} = 3y + C. $$

Hence the implicit general solution is

$$ x = 3y^{2} + C y. $$

Because $$y>0$$ is given, no further restriction on $$C$$ is needed.

We are given the first-order linear differential equation

$$\frac{dy}{dx}+2y\tan x=\sin x$$

and the condition $$y=0$$ when $$x=\frac{\pi}{3}$$.

1. Write the equation in standard linear form

The equation already has the form $$\dfrac{dy}{dx}+P(x)\,y=Q(x)$$ with

$$P(x)=2\tan x, \;\; Q(x)=\sin x.$$

2. Find the integrating factor (I.F.)

$$\text{I.F.}=e^{\int P(x)\,dx}=e^{\int 2\tan x\,dx}.$$

Since $$\int \tan x\,dx=-\ln|\cos x|,$$

$$\int 2\tan x\,dx=-2\ln|\cos x|.$$

Therefore

$$\text{I.F.}=e^{-2\ln|\cos x|}=e^{\ln|\cos x|^{-2}}=|\cos x|^{-2}=\sec^{2}x.$$

(On the interval containing $$x=\frac{\pi}{3}$$ we may drop the absolute-value sign.)

3. Multiply the differential equation by the I.F.

$$\sec^{2}x\,\frac{dy}{dx}+2y\tan x\,\sec^{2}x=\sin x\,\sec^{2}x.$$

The left side is the derivative of $$\sec^{2}x\,y$$ because

$$\frac{d}{dx}(\sec^{2}x\,y)=\sec^{2}x\,\frac{dy}{dx}+y\,\frac{d}{dx}(\sec^{2}x)$$

and $$\frac{d}{dx}(\sec^{2}x)=2\sec^{2}x\tan x.$$

Hence

$$\frac{d}{dx}(\sec^{2}x\,y)=\sin x\,\sec^{2}x.$$

4. Integrate both sides

$$\sec^{2}x\,y=\int \sin x\,\sec^{2}x\,dx+C.$$

Evaluate the integral:

$$\int \sin x\,\sec^{2}x\,dx=\int \frac{\sin x}{\cos^{2}x}\,dx.$$

Put $$u=\tan x\;(=\frac{\sin x}{\cos x}),$$ so $$du=\sec^{2}x\,dx.$$
Then $$\sin x\,\sec^{2}x\,dx=(\tan x\cos x)\sec^{2}x\,dx=\tan x\,d(\tan x)=u\,du.$$

Thus

$$\int u\,du=\frac{u^{2}}{2}=\frac{\tan^{2}x}{2}.$$

Alternatively, note directly that

$$\int \tan x\sec x\,dx=\sec x,$$

and since $$\sin x\sec^{2}x=\tan x\sec x,$$ we have

$$\int \sin x\sec^{2}x\,dx=\sec x.$$

Therefore

$$\sec^{2}x\,y=\sec x+C.$$

5. Solve for $$y$$

$$y=\frac{\sec x+C}{\sec^{2}x}=\cos^{2}x(\sec x+C)=\cos x+C\cos^{2}x.$$

6. Apply the initial condition

At $$x=\dfrac{\pi}{3},$$ $$\cos\frac{\pi}{3}=\tfrac12, \; \cos^{2}\frac{\pi}{3}=\tfrac14.$$ Hence

$$0=y\Big|_{x=\pi/3}=\frac12+C\cdot\frac14.$$

Solving, $$C=-2.$$

7. Particular solution

Substitute $$C=-2$$ into $$y=\cos x+C\cos^{2}x$$:

$$y=\cos x-2\cos^{2}x.$$

This satisfies both the differential equation and the given condition.

The differential equation is

$$ (1 + x^2) \dfrac{dy}{dx} + 2x y = \dfrac{1}{1 + x^2}. $$

This is a first-order linear differential equation in the variable $$y$$.

1. Put the equation in standard linear form

Divide every term by $$1 + x^2$$ (which is never zero):

$$ \dfrac{dy}{dx} + \dfrac{2x}{1 + x^2} \, y = \dfrac{1}{(1 + x^2)^2}. $$

So

• $$P(x) = \dfrac{2x}{1 + x^2}$$,
• $$Q(x) = \dfrac{1}{(1 + x^2)^2}.$$

2. Find the Integrating Factor (I.F.)

$$\text{I.F.} = e^{\int P(x)\,dx} = e^{\int \dfrac{2x}{1 + x^2}\,dx}.$$

Substitute $$u = 1 + x^2 \;\Rightarrow\; du = 2x\,dx$$:

$$ \int \dfrac{2x}{1 + x^2}\,dx = \int \dfrac{1}{u}\,du = \ln|u| = \ln(1 + x^2). $$

Therefore

$$ \text{I.F.} = e^{\ln(1 + x^2)} = 1 + x^2. $$

3. Multiply the whole differential equation by the I.F.

$$ (1 + x^2)\,\dfrac{dy}{dx} + 2x y = \dfrac{1}{1 + x^2}. $$

4. Recognise the left side as a single derivative

Because

$$ \dfrac{d}{dx}\bigl[ y(1 + x^2) \bigr] = (1 + x^2)\,\dfrac{dy}{dx} + 2x y, $$

the equation becomes

$$ \dfrac{d}{dx}\bigl[ y(1 + x^2) \bigr] = \dfrac{1}{1 + x^2}. $$

5. Integrate both sides

$$ y(1 + x^2) = \int \dfrac{1}{1 + x^2}\,dx + C. $$

But

$$ \int \dfrac{1}{1 + x^2}\,dx = \tan^{-1} x, $$

so

$$ y(1 + x^2) = \tan^{-1} x + C. $$

6. Express $$y$$ explicitly

$$ y = \dfrac{\tan^{-1} x + C}{1 + x^2}. $$

7. Use the initial condition

Given $$y = 0$$ when $$x = 1$$:

$$ 0 = \dfrac{\tan^{-1} 1 + C}{1 + 1^2} \;\Longrightarrow\; \tan^{-1} 1 + C = 0. $$

Since $$\tan^{-1} 1 = \dfrac{\pi}{4}$$,

$$ C = -\dfrac{\pi}{4}. $$

8. Particular solution

Substitute $$C$$ back:

$$ y = \dfrac{\tan^{-1} x - \dfrac{\pi}{4}}{1 + x^2}. $$

This satisfies both the differential equation and the given initial condition.

The differential equation is

$$\frac{dy}{dx}-3y\cot x = \sin 2x.$$

1. Write it in the standard linear form.

$$\frac{dy}{dx}+P(x)\,y = Q(x), \;\;\text{where} \; P(x)=-3\cot x, \; Q(x)=\sin 2x.$$

2. Find the integrating factor (I.F.).

$$\text{I.F.}=e^{\int P(x)\,dx}=e^{\int -3\cot x\,dx}=e^{-3\int \cot x\,dx} =e^{-3\ln|\sin x|}=|\sin x|^{-3}.$$

Near $$x=\tfrac{\pi}{2}$$ we have $$\sin x>0,$$ so we may simply write

$$\text{I.F.}=(\sin x)^{-3}.$$

3. Multiply the whole equation by the integrating factor.

$$ (\sin x)^{-3}\,\frac{dy}{dx}-3(\sin x)^{-3}y\cot x=(\sin x)^{-3}\sin 2x. $$

The left-hand side is the derivative of $$y(\sin x)^{-3}$$ because

$$\frac{d}{dx}\bigl[y(\sin x)^{-3}\bigr]=(\sin x)^{-3}\,\frac{dy}{dx}+y\,\frac{d}{dx}(\sin x)^{-3}$$

and $$\frac{d}{dx}(\sin x)^{-3}=-3(\sin x)^{-3}\cot x.$$

Thus

$$\frac{d}{dx}\bigl[y(\sin x)^{-3}\bigr]=(\sin x)^{-3}\sin 2x.$$

4. Integrate both sides.

$$y(\sin x)^{-3}=\int (\sin x)^{-3}\sin 2x\,dx+C.$$

Because $$\sin 2x=2\sin x\cos x,$$

$$ (\sin x)^{-3}\sin 2x=\frac{2\cos x}{(\sin x)^2}. $$

Put $$u=\sin x \Rightarrow du=\cos x\,dx$$ to obtain

$$\int\frac{2\cos x}{(\sin x)^2}dx=2\int u^{-2}\,du=2\left(-u^{-1}\right)=-\frac{2}{\sin x}+C_1.$$

Absorbing $$C_1$$ into the constant of integration $$C,$$ we have

$$y(\sin x)^{-3}=-\frac{2}{\sin x}+C.$$

5. Solve for $$y.$

Multiply throughout by $$\sin^3 x$$:

$$y=-2\sin^2 x+C\sin^3 x.$$

6. Use the initial condition $$y=2$$ when $$x=\tfrac{\pi}{2}.$$

At $$x=\tfrac{\pi}{2},$$ $$\sin\tfrac{\pi}{2}=1,$$ so

$$2=-2(1)^2+C(1)^3\;\Rightarrow\;2=-2+C\;\Rightarrow\;C=4.$$

7. Particular solution.

Substituting $$C=4$$ gives

$$y=-2\sin^2 x+4\sin^3 x.$$

This satisfies both the differential equation and the given initial condition.

Given condition

The slope of the tangent at any point $$(x,y)$$ equals the sum of its coordinates:

$$\dfrac{dy}{dx}=x+y.$$

This is a first–order linear differential equation.

Step 1 : Write in standard linear form

Bring the $$y$$-term to the left:

$$\dfrac{dy}{dx}-y=x.$$

The equation now has the standard form $$\dfrac{dy}{dx}+P(x)y=Q(x)$$ with

• $$P(x)=-1,$$
• $$Q(x)=x.$$

Step 2 : Compute the integrating factor (I.F.)

$$\text{I.F.}=e^{\int P(x)\,dx}=e^{\int (-1)\,dx}=e^{-x}. $$

Step 3 : Multiply the differential equation by the I.F.

$$e^{-x}\dfrac{dy}{dx}-ye^{-x}=x e^{-x}.$$

Step 4 : Recognise the left side as a single derivative

The left side is $$\dfrac{d}{dx}\bigl(y e^{-x}\bigr)$$ because

$$\dfrac{d}{dx}\bigl(y e^{-x}\bigr)=e^{-x}\dfrac{dy}{dx}+y\bigl(-e^{-x}\bigr)=e^{-x}\dfrac{dy}{dx}-y e^{-x}.$$

Hence

$$\dfrac{d}{dx}\bigl(y e^{-x}\bigr)=x e^{-x}. $$

Step 5 : Integrate both sides

$$\int \dfrac{d}{dx}\bigl(y e^{-x}\bigr)\,dx=\int x e^{-x}\,dx.$$

Left side integrates directly:

$$y e^{-x}=\int x e^{-x}\,dx+C.$$

Integral on the right (by parts)

Let $$u=x,\; dv=e^{-x}dx;\; du=dx,\; v=-e^{-x}.$$

Then

$$\int x e^{-x}\,dx = -x e^{-x}+\int e^{-x}\,dx=-x e^{-x}-e^{-x}+C_1.$$

(The constant $$C_1$$ will merge with $$C$$.)

Therefore

$$y e^{-x} = -x e^{-x}-e^{-x}+C.$$

Step 6 : Solve for $$y$$

Multiply by $$e^{x}$$ throughout:

$$y = -x-1+Ce^{x}.$$

Step 7 : Use the initial condition $$y(0)=0$$ (curve passes through the origin)

Substitute $$x=0,\,y=0$$:

$$0=-0-1+C e^{0}\;\Rightarrow\;0=-1+C\;\Rightarrow\;C=1.$$

Step 8 : Final equation of the curve

Substitute $$C=1$$:

$$y=-x-1+e^{x}.$$

Re-ordering terms gives the required equation

$$y=e^{x}-x-1.$$

Step 1 – Translate the given condition into a differential equation
Let any point on the required curve be $$(x , y)$$ and let the slope of the tangent at that point be $$\dfrac{dy}{dx}$$.
According to the statement, “the sum of the coordinates of any point on the curve exceeds the (signed) slope of the tangent at that point by 5”. That is

$$x + y = \frac{dy}{dx} + 5$$

Re-arranging we obtain the first–order differential equation

$$\frac{dy}{dx} = x + y - 5 \;\;\;\;\; (1)$$

Step 2 – Write the equation in the standard linear form
Equation (1) may be written as

$$\frac{dy}{dx} - y = x - 5$$

This is a linear differential equation of the form $$\dfrac{dy}{dx}+P(x)\,y = Q(x)$$ with
$$P(x) = -1 , \; Q(x)=x-5.$$

Step 3 – Find the integrating factor (I.F.)
$$\text{I.F.}=e^{\int P(x)\,dx}=e^{\int (-1)\,dx}=e^{-x}.$$

Step 4 – Multiply throughout by the integrating factor

$$e^{-x}\,\frac{dy}{dx}-e^{-x}y=(x-5)e^{-x}$$

The left–hand side is the derivative of $$e^{-x}y$$, because

$$\frac{d}{dx}\bigl(e^{-x}y\bigr)=e^{-x}\,\frac{dy}{dx}-e^{-x}y.$$

Hence

$$\frac{d}{dx}\bigl(e^{-x}y\bigr)=(x-5)e^{-x}$$

Step 5 – Integrate both sides

$$\int\frac{d}{dx}\bigl(e^{-x}y\bigr)\,dx=\int (x-5)e^{-x}\,dx$$

$$e^{-x}y = \int x e^{-x}\,dx-5\int e^{-x}\,dx + C$$

Evaluate the two elementary integrals one by one.

• Using integration by parts, $$\displaystyle\int x e^{-x}\,dx = - (x+1)e^{-x}.$$
• $$\displaystyle\int e^{-x}\,dx = -e^{-x}. $$

Therefore

$$e^{-x}y = - (x+1)e^{-x}-5(-e^{-x})+C$$

$$e^{-x}y = -(x+1)e^{-x}+5e^{-x}+C$$

$$e^{-x}y = (4-x)e^{-x}+C$$

Step 6 – Solve for $$y$$

Multiply by $$e^{x}$$:

$$y = 4 - x + C e^{x} \;\;\;\;\; (2)$$

Step 7 – Determine the constant with the initial condition
The required curve passes through $$(0,2)$$, so substitute $$x=0,\;y=2$$ in (2):

$$2 = 4 - 0 + C e^{0} \Longrightarrow 2 = 4 + C \Longrightarrow C = -2.$$

Step 8 – Write the particular solution

Substitute $$C=-2$$ in (2):

$$y = 4 - x - 2 e^{x}.$$

This curve passes through $$(0,2)$$ and, by construction, satisfies the required condition.

The given differential equation is $$x \frac{dy}{dx} - y = 2x^2$$.

Divide every term by $$x$$ to cast it in the linear form:

$$\frac{dy}{dx} - \frac{y}{x} = 2x.$$

Comparing with $$\frac{dy}{dx} + P(x)\,y = Q(x)$$ gives $$P(x) = -\frac{1}{x}.$$

The integrating factor (I.F.) is therefore

$$\text{I.F.} = e^{\int P(x)\,dx} = e^{\int -\frac{1}{x}\,dx} = e^{-\ln|x|} = |x|^{-1}.$$

For $$x \neq 0$$ we may write $$\text{I.F.} = \dfrac{1}{x}.$$

Thus the required integrating factor is $$\frac{1}{x}$$, i.e. option (C).

We have a first-order linear differential equation in the dependent variable $$x$$ and the independent variable $$y$$:

$$(1 - y^2) \frac{dx}{dy} + yx = ay, \qquad -1 < y < 1.$$

Rewrite it in the standard linear form
$$\frac{dx}{dy} + P(y)\,x = Q(y),$$
by dividing throughout by $$1 - y^2\;(\neq 0\text{ for }|y|<1):$$

$$\frac{dx}{dy} + \frac{y}{1 - y^2}\,x = \frac{ay}{1 - y^2}.$$

Thus $$P(y) = \dfrac{y}{1 - y^2}.$$

The integrating factor (I.F.) is defined by

$$\text{I.F.} = \exp\Bigl(\int P(y)\,dy\Bigr).$$

Compute the integral:
$$\int \frac{y}{1 - y^2}\,dy.$$

Put $$u = 1 - y^2 \;\Rightarrow\; du = -2y\,dy \;\Rightarrow\; y\,dy = -\tfrac12 du.$$

Then
$$\int \frac{y}{1 - y^2}\,dy = \int \frac{-\tfrac12 du}{u} = -\frac12 \int \frac{du}{u} = -\frac12 \ln|u| + C$$
$$= -\frac12 \ln|1 - y^2| + C.$$

Ignoring the integration constant (it multiplies the integrating factor by a non-zero constant and does not affect the solution), we get

$$\text{I.F.} = \exp\Bigl(-\tfrac12 \ln|1 - y^2|\Bigr).$$

Use the property $$e^{\ln a^k}=a^k$$:

$$\text{I.F.} = |1 - y^2|^{-1/2}.$$

Because $$-10,$$ the absolute value can be removed:

$$\boxed{\text{I.F.}=\dfrac{1}{\sqrt{1 - y^2}}}.$$

Comparing with the given options, this corresponds to option (D).

Let

$$y = c_{1}\,e^{ax}\cos bx + c_{2}\,e^{ax}\sin bx = e^{ax}(c_{1}\cos bx + c_{2}\sin bx).$$

1. First derivative.

By the product rule,

$$\frac{dy}{dx} = e^{ax}\bigl[a(c_{1}\cos bx + c_{2}\sin bx) + b(-c_{1}\sin bx + c_{2}\cos bx)\bigr].$$

For brevity let

$$F = a(c_{1}\cos bx + c_{2}\sin bx) + b(-c_{1}\sin bx + c_{2}\cos bx),$$

so that $$\dfrac{dy}{dx} = e^{ax}F$$.

2. Second derivative.

$$\frac{d^{2}y}{dx^{2}} = e^{ax}\frac{dF}{dx} + a\,e^{ax}F = e^{ax}\!\left(\frac{dF}{dx} + aF\right).$$

Differentiating $$F$$ term by term:

$$\frac{dF}{dx} = a\bigl(-b\,c_{1}\sin bx + b\,c_{2}\cos bx\bigr) + b\bigl(-b\,c_{1}\cos bx - b\,c_{2}\sin bx\bigr).$$

That is,

$$\frac{dF}{dx} = ab(-c_{1}\sin bx + c_{2}\cos bx) + b^{2}(-c_{1}\cos bx - c_{2}\sin bx).$$

3. Substitute into the differential equation.

The left-hand side is

$$\frac{d^{2}y}{dx^{2}} - 2a\,\frac{dy}{dx} + (a^{2} + b^{2})\,y.$$

Substituting the expressions above:

$$e^{ax}\!\left(\frac{dF}{dx} + aF\right) - 2a\,e^{ax}F + (a^{2} + b^{2})\,e^{ax}(c_{1}\cos bx + c_{2}\sin bx)$$

$$= e^{ax}\!\left[\frac{dF}{dx} - aF + (a^{2} + b^{2})(c_{1}\cos bx + c_{2}\sin bx)\right].$$

Now compute $$aF$$:

$$aF = a^{2}(c_{1}\cos bx + c_{2}\sin bx) + ab(-c_{1}\sin bx + c_{2}\cos bx).$$

So

$$\frac{dF}{dx} - aF = b^{2}(-c_{1}\cos bx - c_{2}\sin bx) - a^{2}(c_{1}\cos bx + c_{2}\sin bx) = -(a^{2} + b^{2})(c_{1}\cos bx + c_{2}\sin bx).$$

Therefore

$$\frac{dF}{dx} - aF + (a^{2} + b^{2})(c_{1}\cos bx + c_{2}\sin bx) = 0,$$

and the entire left-hand side equals $$e^{ax} \cdot 0 = 0$$.

Hence the function satisfies the given differential equation for all real $$x$$ and all arbitrary constants $$c_{1}, c_{2}$$. Verified.

We are given the differential equation

$$\log\left(\frac{dy}{dx}\right)=3x+4y\;,$$

where log denotes the natural logarithm (base e). The initial condition is

$$y=0 \;\text{when}\; x=0.$$

Step 1: Remove the logarithm.

Exponentiating both sides gives

$$\frac{dy}{dx}=e^{3x+4y}=e^{3x}\,e^{4y}.$$

Step 2: Separate the variables.

Bring all terms in y to the left and those in x to the right:

$$e^{-4y}\,dy=e^{3x}\,dx.$$

Step 3: Integrate both sides.

Left side:

$$\int e^{-4y}\,dy=-\tfrac14 e^{-4y}+C_1.$$

Right side:

$$\int e^{3x}\,dx=\tfrac13 e^{3x}+C_2.$$

Combine the two arbitrary constants into a single constant C:

$$-\tfrac14 e^{-4y}=\tfrac13 e^{3x}+C.$$

Step 4: Isolate the exponential term.

Multiply by −4:

$$e^{-4y}=-\tfrac43 e^{3x}-4C.$$

Rename $$-4C$$ as a fresh constant $$K$$ (still arbitrary):

$$e^{-4y}=K-\tfrac43 e^{3x}.$$

Step 5: Use the initial condition to find K.

When $$x=0,\;y=0$$, so

$$e^{0}=1=K-\tfrac43 e^{0}=K-\tfrac43.$$

Hence

$$K=1+\tfrac43=\tfrac73.$$

Step 6: Substitute K and solve for y.

$$e^{-4y}=\tfrac73-\tfrac43 e^{3x}=\tfrac13\,(7-4e^{3x}).$$

Take natural logarithms:

$$-4y=\log\!\left(\tfrac13\,(7-4e^{3x})\right).$$

Therefore

$$y=-\tfrac14\,\log\!\left(\frac{7-4e^{3x}}{3}\right).$$

Step 7: Present the particular solution.

The required particular solution satisfying $$y(0)=0$$ is

$$y=-\frac14\,\log\left(\frac{7-4e^{3x}}{3}\right).$$

The given differential equation is

$$ (x\,dy-y\,dx)\,y\sin\left(\frac{y}{x}\right)=(y\,dx+x\,dy)\,x\cos\left(\frac{y}{x}\right). $$

Divide every term by $$dx\;(\neq 0)$$ and write $$y'=\dfrac{dy}{dx}$$, then

$$ (x y'-y)\,y\sin\left(\frac{y}{x}\right)=(y+x y')\,x\cos\left(\frac{y}{x}\right). $$

Introduce the new dependent variable $$v=\dfrac{y}{x}\;\Longrightarrow\;y=vx\;. $$

Differentiate: $$y'=\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}. $$

Substitute $$y,\;y'$$ in the equation.

1. Evaluate the two brackets

• $$ x y'-y=x\bigl(v+x\dfrac{dv}{dx}\bigr)-vx=x^2\dfrac{dv}{dx}. $$
• $$ y+x y'=vx+x\bigl(v+x\dfrac{dv}{dx}\bigr)=2vx+x^2\dfrac{dv}{dx}. $$

2. Substitute in the differential equation

$$ (x^2\dfrac{dv}{dx})\,(vx)\,\sin v-\bigl(2vx+x^2\dfrac{dv}{dx}\bigr)\,x\cos v=0. $$

Simplify (divide by $$x^2\;(x\neq 0)$$):

$$ v x\dfrac{dv}{dx}\sin v-(2v+x\dfrac{dv}{dx})\cos v=0. $$

Group the terms containing $$\dfrac{dv}{dx}$$:

$$ x\dfrac{dv}{dx}(v\sin v-\cos v)=2v\cos v. $$

Hence

$$ \dfrac{dv}{dx}=\dfrac{2v\cos v}{x\,(v\sin v-\cos v)}. $$

3. Separate the variables

$$ \left(\tfrac{v\sin v-\cos v}{v\cos v}\right)dv=\tfrac{2}{x}\,dx. $$

Simplify the left factor:

$$ \frac{v\sin v}{v\cos v}-\frac{\cos v}{v\cos v}=\tan v-\frac{1}{v}. $$

Therefore the separable form is

$$ \bigl(\tan v-\tfrac1v\bigr)dv=\tfrac{2}{x}dx. $$

4. Integrate

\(\displaystyle\int\tan v\,dv=-\ln|\cos v|\) and \(\displaystyle\int -\dfrac1v\,dv=-\ln|v|\). Hence

$$ -\ln|\cos v|-\ln|v| = 2\ln|x|+C, $$

or

$$ \ln|v\cos v| = -2\ln|x|+C. $$

5. Remove logarithms

$$ v\cos v = C_1 x^{-2}, $$

where $$C_1$$ is a non–zero arbitrary constant.

Finally substitute $$v=\dfrac{y}{x}$$:

$$ \frac{y}{x}\cos\left(\frac{y}{x}\right)=\frac{C_1}{x^{2}}. $$

Multiplying by $$x^{2}$$ gives the implicit solution in its simplest form

$$ x y \cos\left(\frac{y}{x}\right)=C, $$

where $$C$$ is an arbitrary constant.

We are required to solve the differential equation

$$ (\tan^{-1}y-x)\,dy=(1+y^2)\,dx. $$

1. Express $$x$$ as a function of $$y$$.

Divide by $$dy$$ (and assume $$dy\neq0$$):

$$ \tan^{-1}y-x=(1+y^2)\,\dfrac{dx}{dy}. $$

Hence

$$ \dfrac{dx}{dy}=\dfrac{\tan^{-1}y}{1+y^2}-\dfrac{x}{1+y^2}. $$

2. Identify the linear ODE.

This is a linear first–order ODE in $$x$$ (dependent variable) with independent variable $$y$$:

$$ \dfrac{dx}{dy}+\dfrac{1}{1+y^2}\;x=\dfrac{\tan^{-1}y}{1+y^2}. $$

Compare with $$\dfrac{dx}{dy}+P(y)\,x=Q(y)$$, where

$$ P(y)=\dfrac{1}{1+y^2},\qquad Q(y)=\dfrac{\tan^{-1}y}{1+y^2}. $$

3. Compute the integrating factor.

$$ \mu(y)=\exp\!\left(\int P(y)\,dy\right)=\exp\!\left(\int\dfrac{1}{1+y^2}\,dy\right)=\exp(\tan^{-1}y). $$

4. Multiply the ODE by $$\mu(y)$$.

$$ e^{\tan^{-1}y}\,\dfrac{dx}{dy}+e^{\tan^{-1}y}\,\dfrac{1}{1+y^2}\,x=e^{\tan^{-1}y}\,\dfrac{\tan^{-1}y}{1+y^2}. $$

The left‐hand side is the derivative of $$e^{\tan^{-1}y}x$$ with respect to $$y$$ since

$$ \dfrac{d}{dy}\bigl(e^{\tan^{-1}y}x\bigr)=e^{\tan^{-1}y}\dfrac{dx}{dy}+e^{\tan^{-1}y}\,\dfrac{1}{1+y^2}\,x. $$

5. Integrate.

$$ \dfrac{d}{dy}\bigl(e^{\tan^{-1}y}x\bigr)=e^{\tan^{-1}y}\,\dfrac{\tan^{-1}y}{1+y^2}. $$

Integrate both sides with respect to $$y$$:

$$ e^{\tan^{-1}y}x=\int e^{\tan^{-1}y}\,\dfrac{\tan^{-1}y}{1+y^2}\,dy+C. $$

6. Evaluate the integral.

Put $$t=\tan^{-1}y\;\;(\Rightarrow\;y=\tan t)$$. Then $$dt=\dfrac{dy}{1+y^2}$$.

The integral becomes

$$ \int e^{t}\,t\,dt. $$

Using integration by parts:

$$ \int t\,e^{t}dt=t\,e^{t}-\int e^{t}dt=t\,e^{t}-e^{t}+C. $$

Returning to $$t=\tan^{-1}y$$, we obtain

$$ \int e^{\tan^{-1}y}\,\dfrac{\tan^{-1}y}{1+y^2}\,dy=e^{\tan^{-1}y}\bigl(\tan^{-1}y-1\bigr)+C. $$

7. General solution.

Substituting this in the equality from step 5, we have

$$ e^{\tan^{-1}y}x=e^{\tan^{-1}y}\bigl(\tan^{-1}y-1\bigr)+C. $$

Divide by $$e^{\tan^{-1}y}(\neq0)$$:

$$ x=\tan^{-1}y-1+Ce^{-\tan^{-1}y}. $$

Rearranging, an equivalent implicit form is

$$ e^{\tan^{-1}y}\bigl(x-\tan^{-1}y+1\bigr)=C. $$

This represents the required one‐parameter family of solutions.

Given differential equation

\[(x^{3}-3xy^{2})\,dx=(y^{3}-3x^{2}y)\,dy\tag{1}\]

We have to show that the one–parameter family of curves

\[x^{2}-y^{2}=c\,(x^{2}+y^{2})^{2}\tag{2}\]

is its general solution.

Step 1 : Implicitly differentiate (2)

Differentiate both sides of (2) with respect to $$x$$, treating $$y$$ as a function of $$x$$:

$$2x-2y\,\dfrac{dy}{dx}=c\,2(x^{2}+y^{2})\,2\bigl(x+y\,\dfrac{dy}{dx}\bigr).$$

Cancel the common factor $$2$$:

\[x-y\,\dfrac{dy}{dx}=2c(x^{2}+y^{2})\Bigl(x+y\,\dfrac{dy}{dx}\Bigr).\tag{3}\]

Step 2 : Collect the terms containing $$\dfrac{dy}{dx}$$

Expand the right–hand side of (3):

$$x-y\,\dfrac{dy}{dx}=2c(x^{2}+y^{2})x+2c(x^{2}+y^{2})y\,\dfrac{dy}{dx}.$$

Move all the $$\dfrac{dy}{dx}$$ terms to the right and the others to the left:

$$x-2c(x^{2}+y^{2})x=\dfrac{dy}{dx}\Bigl[y+2c(x^{2}+y^{2})y\Bigr].$$

Hence

\[\dfrac{dy}{dx}=\dfrac{x\,[1-2c(x^{2}+y^{2})]}{y\,[1+2c(x^{2}+y^{2})]}.\tag{4}\]

Step 3 : Express the constant $$c$$ in terms of $$x$$ and $$y$$

From (2):

\[c=\dfrac{x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}.\tag{5}\]

Put $$R=x^{2}+y^{2}$$ for brevity. Then (5) gives $$c=(x^{2}-y^{2})/R^{2}$$ and

$$1\mp2cR=1\mp2\dfrac{x^{2}-y^{2}}{R}=\dfrac{R\mp2(x^{2}-y^{2})}{R}.$$

Compute the two brackets that occur in (4):

• $$1-2cR=\dfrac{R-2(x^{2}-y^{2})}{R}=\dfrac{3y^{2}-x^{2}}{R}= -\dfrac{x^{2}-3y^{2}}{R},$$
• $$1+2cR=\dfrac{R+2(x^{2}-y^{2})}{R}=\dfrac{3x^{2}-y^{2}}{R}= -\dfrac{y^{2}-3x^{2}}{R}.$$

Insert these expressions in (4). The common factor $$-1/R$$ cancels from numerator and denominator, giving

\[\dfrac{dy}{dx}=\dfrac{x\,(x^{2}-3y^{2})}{y\,(y^{2}-3x^{2})}.\tag{6}\]

Step 4 : Rewrite (1) as a first-order derivative

Divide equation (1) by $$dx$$ to obtain

\[\dfrac{dy}{dx}=\dfrac{x^{3}-3xy^{2}}{y^{3}-3x^{2}y}=\dfrac{x\,(x^{2}-3y^{2})}{y\,(y^{2}-3x^{2})}.\tag{7}\]

Step 5 : Identity of (6) and (7)

The right–hand sides of (6) and (7) are identical, so the derivative produced by (2) satisfies the given differential equation. Hence every curve of the family (2) is a solution of (1).

Step 6 : Presence of one arbitrary constant

Equation (2) contains exactly one arbitrary constant $$c$$. Because the given differential equation is of first order, its general integral must involve one arbitrary constant. Therefore (2) is not just a solution; it is the general solution of (1).

Conclusion : The family $$x^{2}-y^{2}=c\,(x^{2}+y^{2})^{2}$$ is the general solution of the differential equation $$(x^{3}-3xy^{2})\,dx=(y^{3}-3x^{2}y)\,dy.$$

The given differential equation is

$$\frac{dy}{dx}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0.$$

Bring the second term to the right–hand side so that the derivative is isolated:

$$\frac{dy}{dx}=-\sqrt{\frac{1-y^{2}}{1-x^{2}}}.$$

This equation is separable. Move all expressions in $$y$$ to the left and all expressions in $$x$$ to the right:

$$\frac{dy}{\sqrt{1-y^{2}}}=-\frac{dx}{\sqrt{1-x^{2}}}.$$

Integrate both sides:

$$\int\frac{dy}{\sqrt{1-y^{2}}}= -\int\frac{dx}{\sqrt{1-x^{2}}}+C,$$

where $$C$$ is the constant of integration.

The standard integral $$\displaystyle\int \frac{du}{\sqrt{1-u^{2}}}=\sin^{-1}u$$ gives

$$\sin^{-1}y = -\sin^{-1}x + C.$$

Rearranging, the general solution may be written in the symmetric form

$$\boxed{\sin^{-1}y+\sin^{-1}x=C}.$$

(Equivalently, $$y=\sin\!(C-\sin^{-1}x)$$.) The solution is valid for those $$x$$ and $$y$$ that satisfy $$|x|<1$$ and $$|y|<1$$ so that the square–root and inverse–sine functions remain real.

We have to solve the differential equation

$$\frac{dy}{dx}+\frac{y^{2}+y+1}{x^{2}+x+1}=0\;.$$

Rewrite it so that the derivative stands alone:

$$\frac{dy}{dx}=-\frac{y^{2}+y+1}{x^{2}+x+1}\;.$$

This is already of the separable type because the right-hand side is the quotient of a function of y and a function of x.

Separate the variables:

$$\frac{dy}{y^{2}+y+1}=-\frac{dx}{x^{2}+x+1}\;.$$

Integrate both sides:

$$\int \frac{dy}{y^{2}+y+1}= - \int \frac{dx}{x^{2}+x+1}+C\;.$$

To evaluate the integral on the left, complete the square in the denominator:

$$y^{2}+y+1=\Bigl(y+\tfrac12\Bigr)^{2}+\tfrac34\;.$$

Put $$u=y+\tfrac12\;(\,du=dy\,)$$; then

$$\int \frac{dy}{y^{2}+y+1}=\int \frac{du}{u^{2}+\left(\tfrac{\sqrt3}{2}\right)^{2}}=\frac{2}{\sqrt3}\tan^{-1}\!\left(\frac{2u}{\sqrt3}\right)+C =\frac{2}{\sqrt3}\tan^{-1}\!\left(\frac{2y+1}{\sqrt3}\right)+C.$$

The x-integral is handled in exactly the same way:

$$\int \frac{dx}{x^{2}+x+1}=\frac{2}{\sqrt3}\tan^{-1}\!\left(\frac{2x+1}{\sqrt3}\right)+C.$$

Substituting these results in the equality of the two integrals gives

$$\frac{2}{\sqrt3}\tan^{-1}\!\left(\frac{2y+1}{\sqrt3}\right) +\frac{2}{\sqrt3}\tan^{-1}\!\left(\frac{2x+1}{\sqrt3}\right)=C_1\;.$$

Multiply by $$\tfrac{\sqrt3}{2}$$ to clear the factor:

$$\tan^{-1}\!\left(\frac{2y+1}{\sqrt3}\right)+\tan^{-1}\!\left(\frac{2x+1}{\sqrt3}\right)=C_2\;(\text{a constant}).$$

Let

$$\alpha=\tan^{-1}\!\left(\frac{2y+1}{\sqrt3}\right),\quad \beta =\tan^{-1}\!\left(\frac{2x+1}{\sqrt3}\right),\quad \alpha+\beta=C_2.$$ Using the tangent–addition formula $$\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$$ we get

$$\tan C_2=\frac{\tfrac{2y+1}{\sqrt3}+\tfrac{2x+1}{\sqrt3}} {1-\tfrac{(2y+1)(2x+1)}{3}}\;.$$

Simplify the numerator and denominator separately.

• Numerator: $$\tfrac{2y+1+2x+1}{\sqrt3}=\tfrac{2(x+y+1)}{\sqrt3}\;.$$
• Denominator: $$1-\frac{(2y+1)(2x+1)}{3}=\frac{3-(2y+1)(2x+1)}{3} =\frac{3-(4xy+2x+2y+1)}{3} =\frac{2-2x-2y-4xy}{3}=\frac{2\bigl(1-x-y-2xy\bigr)}{3}\;.$$

Hence

$$\tan C_2=\frac{\tfrac{2(x+y+1)}{\sqrt3}}{\tfrac{2(1-x-y-2xy)}{3}}= \frac{3}{\sqrt3}\,\frac{x+y+1}{1-x-y-2xy}=\sqrt3\,\frac{x+y+1}{1-x-y-2xy}\;.$$

The left-hand side is a constant; call $$\tan C_2=\sqrt3\,A.$$ With this notation the last equation reduces to

$$\sqrt3\,A=\sqrt3\,\frac{x+y+1}{1-x-y-2xy}\;,$$

or, after cancelling $$\sqrt3$$,

$$x+y+1=A\,(1-x-y-2xy).$$

Because $$A=\dfrac{\tan C_2}{\sqrt3}$$ can take every real value (arising from the arbitrary constant of integration), it serves as the required parameter. Consequently, the general solution of the given differential equation is

$$ (x+y+1)=A(1-x-y-2xy),$$

which is exactly what we had to prove.

Given differential equation
$$\sin x\,\cos y\,dx+\cos x\,\sin y\,dy=0$$

1. Rewrite so that the variables separate

Divide every term by $$\cos x\,\cos y$$ (which is legitimate whenever both cosines are non-zero): $$ \frac{\sin x}{\cos x}\,dx+\frac{\sin y}{\cos y}\,dy=0 $$ Since $$\dfrac{\sin u}{\cos u}=\tan u$$, this becomes $$ \tan x\,dx+\tan y\,dy=0. $$

2. Separate and integrate

Move the second term to the right: $$ \tan x\,dx=-\tan y\,dy. $$ Integrate both sides: $$ \int\tan x\,dx=-\int\tan y\,dy. $$ Recall the standard integral $$\int\tan u\,du=-\ln|\cos u|$$, so $$ -\ln|\cos x|=-\bigl(-\ln|\cos y|\bigr)=\ln|\cos y|+C_1. $$

3. Simplify the constant

Multiply by $$-1$$: $$ \ln|\cos x|+\ln|\cos y|=C_2. $$ Use the property $$\ln a+\ln b=\ln(ab)$$: $$ \ln\bigl(|\cos x|\,|\cos y|\bigr)=C_2. $$ Exponentiate: $$ |\cos x|\,|\cos y|=C,\qquad C>0. $$ For simplicity (and because a sign can be absorbed into the constant) we write $$ \cos x\,\cos y=C. $$

4. Determine the constant using the given point

The required curve passes through $$\bigl(0,\,\tfrac{\pi}{4}\bigr)$$. Substitute $$x=0,\;y=\tfrac{\pi}{4}$$: $$ \cos0\,\cos\frac{\pi}{4}=C\quad\Longrightarrow\quad 1\times\frac{\sqrt2}{2}=C. $$ Hence $$C=\dfrac1{\sqrt2}.$$

5. Equation of the required curve

$$ \boxed{\cos x\,\cos y=\dfrac1{\sqrt2}}. $$ Any equivalent form, such as $$\cos y=\dfrac1{\sqrt2\,\cos x}$$, also represents the required curve.

Given differential equation:

$$ (1 + e^{2x})\,dy + (1 + y^2)\,e^{x}\,dx = 0 $$

Step 1   Make the equation separable

Move the terms in $$dx$$ to the right and those in $$dy$$ to the left:

$$ (1 + e^{2x})\,dy = -(1 + y^2)\,e^{x}\,dx $$

Divide by $$(1 + y^2)(1 + e^{2x})$$:

$$ \dfrac{dy}{1 + y^2} = -\dfrac{e^{x}}{1 + e^{2x}}\,dx $$

The variables are now separated.

Step 2   Integrate both sides

Left side:

$$ \int \dfrac{dy}{1 + y^2} = \arctan y + C_1 $$

Right side: set $$t = e^{x}\;\Rightarrow\;dt = e^{x}\,dx$$

Then

$$ -\int \dfrac{e^{x}}{1 + e^{2x}}\,dx = -\int \dfrac{dt}{1 + t^{2}} = -\arctan t + C_2 = -\arctan\bigl(e^{x}\bigr) + C_2 $$

Combine the two constants $$C_1 - C_2 = C$$:

$$ \arctan y = -\arctan\bigl(e^{x}\bigr) + C $$

Step 3   Apply the initial condition

Given $$y = 1$$ when $$x = 0$$.

At $$x = 0$$, $$e^{x} = 1$$. Thus

$$ \arctan(1) = -\arctan(1) + C \;\Longrightarrow\; \frac{\pi}{4} = -\frac{\pi}{4} + C $$

Hence $$C = \frac{\pi}{2}$$.

Step 4   Write the particular solution

Substitute $$C$$ back:

$$ \arctan y = -\arctan\bigl(e^{x}\bigr) + \frac{\pi}{2} $$

Take tangent of both sides:

$$ y = \tan\!\Bigl(\frac{\pi}{2} - \arctan(e^{x})\Bigr) $$

Using the identity $$\tan\bigl(\frac{\pi}{2} - \theta\bigr) = \cot\theta = \dfrac{1}{\tan\theta}$$:

$$ y = \dfrac{1}{\tan\bigl(\arctan(e^{x})\bigr)} = \dfrac{1}{e^{x}} = e^{-x} $$

Therefore, the required particular solution is

$$ y = e^{-x}. $$

Given differential equation

$$y\,e^{x/y}\,dx=(x\,e^{x/y}+y^{2})\,dy\qquad(y\neq 0).$$

Bring everything to one side and identify

$$y\,e^{x/y}\,dx-(x\,e^{x/y}+y^{2})\,dy=0$$ ⇒ $$M\,dx+N\,dy=0,$$ where $$M=y\,e^{x/y},\;\;N=-(x\,e^{x/y}+y^{2}).$$

1. Test for an integrating factor depending only on y

Compute the partial derivatives:

$$\frac{\partial M}{\partial y}=e^{x/y}(1-\tfrac{x}{y})=e^{x/y}(1-t),\;\;t=\tfrac{x}{y},$$ $$\frac{\partial N}{\partial x}=-e^{x/y}(1+t).$$

Therefore

$$\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} =-e^{x/y}(1+t)-e^{x/y}(1-t)=-2e^{x/y}.$$

The quotient

$$\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M} =\frac{-2e^{x/y}}{y\,e^{x/y}}=-\frac{2}{y}$$

is a function of $$y$$ alone; hence an integrating factor depending only on $$y$$ exists and is found from

$$\frac{\mu'(y)}{\mu(y)}=-\frac{2}{y}\;\Longrightarrow\; \mu(y)=y^{-2}.$$

2. Multiply by the integrating factor

$$y^{-2}[y\,e^{x/y}\,dx-(x\,e^{x/y}+y^{2})\,dy]=0$$

i.e.

$$\frac{e^{x/y}}{y}\,dx-\Bigl(\frac{x\,e^{x/y}}{y^{2}}+1\Bigr)dy=0.$$

Now $$M_1=\dfrac{e^{x/y}}{y},\;N_1=-\Bigl(\dfrac{x\,e^{x/y}}{y^{2}}+1\Bigr).$$

3. Verify exactness

$$\frac{\partial M_1}{\partial y}=-\frac{e^{x/y}(1+t)}{y^{2}},\; \frac{\partial N_1}{\partial x}=-\frac{e^{x/y}(1+t)}{y^{2}};$$ the two are equal, so the new equation is exact.

4. Find the potential function $$\psi(x,y)$$

• Integrate $$M_1$$ with respect to $$x$$ (keeping $$y$$ constant):

$$\psi(x,y)=\int\frac{e^{x/y}}{y}\,dx+h(y) =\frac{1}{y}\Bigl(y\,e^{x/y}\Bigr)+h(y)=e^{x/y}+h(y).$$

• Differentiate this expression with respect to $$y$$ and equate to $$N_1$$ to determine $$h(y)$$:

$$\frac{\partial \psi}{\partial y}=-\frac{x\,e^{x/y}}{y^{2}}+h'(y)=-\Bigl(\frac{x\,e^{x/y}}{y^{2}}+1\Bigr).$$

Hence $$h'(y)=-1\;\Longrightarrow\;h(y)=-y+C.$$(The constant $$C$$ can be absorbed later.)

Thus

$$\psi(x,y)=e^{x/y}-y.$$

5. Implicit solution

For an exact differential equation, $$\psi(x,y)=K$$ (constant). Therefore

$$e^{x/y}-y=C,$$ or equivalently $$e^{x/y}=y+C.$$

This is the required general solution (with $$y\neq 0$$ as given).

We have to solve

$$(x - y)(dx + dy) = dx - dy,$$

subject to $$y = -1$$ when $$x = 0$$.

1. Convert to $$dy/dx$$ form.

Dividing by $$dx$$:

$$(x - y)\!\left(1 + \frac{dy}{dx}\right) = 1 - \frac{dy}{dx}.$$

With $$p = \dfrac{dy}{dx}$$:

$$(x - y)(1 + p) = 1 - p \;\Longrightarrow\; p\bigl[(x - y) + 1\bigr] = 1 - (x - y).$$

Hence

$$\frac{dy}{dx} = \frac{1 - (x - y)}{(x - y) + 1}. \quad (1)$$

2. Substitution $$t = x - y$$.

Then $$y = x - t$$ and $$\dfrac{dy}{dx} = 1 - \dfrac{dt}{dx}$$. Substituting in (1):

$$1 - \frac{dt}{dx} = \frac{1 - t}{t + 1} \;\Longrightarrow\; \frac{dt}{dx} = 1 - \frac{1 - t}{t + 1} = \frac{2t}{t + 1}.$$

Therefore

$$\frac{dx}{dt} = \frac{t + 1}{2t}. \quad (2)$$

3. Integrate (2).

$$dx = \frac{1}{2}\!\left(1 + \frac{1}{t}\right)dt.$$

Integrating,

$$x = \frac{t}{2} + \frac{1}{2}\ln|t| + C_{0}.$$

Multiplying by $$2$$,

$$2x = t + \ln|t| + C_{1}.$$

Replace $$t = x - y$$:

$$2x = (x - y) + \ln|x - y| + C_{1},$$

which rearranges to

$$x + y = \ln|x - y| + C. \quad (3)$$

4. Apply the initial condition.

At $$x = 0$$, $$y = -1$$:

$$0 + (-1) = \ln|0 - (-1)| + C \;\Longrightarrow\; -1 = \ln 1 + C \;\Longrightarrow\; C = -1.$$

5. Particular solution.

$$x + y = \ln|x - y| - 1, \quad\text{i.e.}\quad x + y + 1 = \ln|x - y|.$$

The given differential equation is

$$\left[\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right]\,\frac{dx}{dy}=1 \;\;(x\neq0).$$

Because the derivative that appears is $$\dfrac{dx}{dy}$$, the dependent variable is $$x$$ and the independent variable is $$y$$. We first rewrite the equation so that it involves $$\dfrac{dy}{dx}$$, the form in which the standard linear-equation method is stated.

Multiply both sides by $$\dfrac{dy}{dx}$$ (the reciprocal of $$\dfrac{dx}{dy}$$):

$$\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}=\frac{dy}{dx}.$$

Re-arranging gives

$$\frac{dy}{dx}+\frac{1}{\sqrt{x}}\,y=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}.$$

This is a first-order linear differential equation of the form

$$\frac{dy}{dx}+P(x)\,y=Q(x)$$

with

$$P(x)=\frac{1}{\sqrt{x}},\qquad Q(x)=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}.$$

Step 1 — Integrating factor

$$\text{I.F.}=e^{\int P(x)\,dx}=e^{\int \!x^{-1/2}\,dx}=e^{2\sqrt{x}}.$$

Step 2 — Multiply the whole equation by the I.F.

$$e^{2\sqrt{x}}\,\frac{dy}{dx}+\frac{e^{2\sqrt{x}}}{\sqrt{x}}\,y=\frac{e^{2\sqrt{x}}\,e^{-2\sqrt{x}}}{\sqrt{x}}=\frac{1}{\sqrt{x}}.$$

The left side is the derivative of $$e^{2\sqrt{x}}y$$, because

$$\frac{d}{dx}\big(e^{2\sqrt{x}}y\big)=e^{2\sqrt{x}}\frac{dy}{dx}+y\,\frac{d}{dx}\big(e^{2\sqrt{x}}\big).$$

But $$\dfrac{d}{dx}\big(e^{2\sqrt{x}}\big)=e^{2\sqrt{x}}\,\dfrac{1}{\sqrt{x}},$$ so the expression matches.

Hence the equation becomes

$$\frac{d}{dx}\big(e^{2\sqrt{x}}y\big)=\frac{1}{\sqrt{x}}.$$

Step 3 — Integrate

$$\int \!\frac{d}{dx}\big(e^{2\sqrt{x}}y\big)\,dx=e^{2\sqrt{x}}y=\int \!x^{-1/2}\,dx=2\sqrt{x}+C,$$

where $$C$$ is the constant of integration.

Step 4 — Solve for $$y$$

$$y=e^{-2\sqrt{x}}\,(2\sqrt{x}+C).$$

This gives the required one-parameter family of solutions (valid for $$x\neq0$$).

The given differential equation is

$$\frac{dy}{dx}+y\cot x = 4x\csc x \qquad (x \neq 0).$$

This is a linear first–order differential equation of the standard form

$$\frac{dy}{dx}+P(x)\,y = Q(x),$$

where

• $$P(x)=\cot x,$$
• $$Q(x)=4x\csc x.$$

Step 1 Find the integrating factor (I.F.).

$$\text{I.F.}=e^{\int P(x)\,dx}=e^{\int \cot x\,dx}=e^{\ln|\sin x|}=\sin x.$$ Because \(\sin x>0\) near the initial point \(x=\pi/2\), we may write simply $$\text{I.F.}=\sin x.$$

Step 2 Multiply the whole equation by the integrating factor.

$$\sin x\,\frac{dy}{dx}+y\sin x\,\cot x=4x\sin x\,\csc x.$$

Simplify:

• \(\sin x\,\cot x = \sin x\,(\cos x/\sin x)=\cos x\),
• \(4x\sin x\,\csc x =4x\).

Therefore

$$\sin x\,\frac{dy}{dx}+y\cos x = 4x.$$

Step 3 Recognise the left side as a derivative.

Because $$\frac{d}{dx}(y\sin x)=y\cos x+\sin x\,\frac{dy}{dx},$$ the equation becomes

$$\frac{d}{dx}(y\sin x)=4x.$$

Step 4 Integrate both sides.

$$\int \frac{d}{dx}(y\sin x)\,dx = \int 4x\,dx$$

$$y\sin x = 2x^{2}+C,$$

where \(C\) is the constant of integration.

Step 5 Use the initial condition to find \(C\).

Given: $$y=0 \text{ when } x=\dfrac{\pi}{2}.$$

Substitute in $$y\sin x = 2x^{2}+C$$:

$$0 \times \sin\left(\dfrac{\pi}{2}\right)=2\left(\dfrac{\pi}{2}\right)^{2}+C \;\Rightarrow\; 0 = 2\cdot\dfrac{\pi^{2}}{4}+C \;\Rightarrow\; C=-\dfrac{\pi^{2}}{2}.$$

Step 6 Write the required particular solution.

Substituting \(C\) back,

$$y\sin x = 2x^{2}-\dfrac{\pi^{2}}{2}$$

$$\boxed{\displaystyle y = \frac{2x^{2}-\dfrac{\pi^{2}}{2}}{\sin x}}.$$

One may also write the numerator with a common factor:

$$y = \dfrac{4x^{2}-\pi^{2}}{2\sin x}.$$

This satisfies both the differential equation and the initial condition, so it is the required particular solution.

The given differential equation is

$$(x + 1)\,\frac{dy}{dx} = 2\,e^{-y} - 1.$$

It is separable. Divide by $$x + 1$$ and bring the $$y$$-terms to the left:

$$\frac{dy}{2\,e^{-y} - 1} = \frac{dx}{x + 1}.$$

Left-hand integral.

Multiplying numerator and denominator by $$e^{y}$$:

$$\int \frac{e^{y}\,dy}{2 - e^{y}}.$$

Let $$t = 2 - e^{y}$$, so $$dt = -e^{y}\,dy$$. Then

$$\int \frac{e^{y}\,dy}{2 - e^{y}} = \int \frac{-dt}{t} = -\ln|t| = -\ln|2 - e^{y}|.$$

Right-hand integral.

$$\int \frac{dx}{x + 1} = \ln|x + 1|.$$

Equating (with a single constant $$C$$):

$$-\ln|2 - e^{y}| = \ln|x + 1| + C.$$

That is, $$\ln\bigl|(2 - e^{y})(x + 1)\bigr| = -C$$, so

$$(2 - e^{y})(x + 1) = A,$$

where $$A$$ is a non-zero arbitrary constant.

Apply the initial condition $$y(0) = 0$$.

At $$x = 0$$, $$y = 0$$: $$(2 - 1)(0 + 1) = A \;\Longrightarrow\; A = 1$$.

Therefore

$$(2 - e^{y})(x + 1) = 1 \;\Longrightarrow\; 2 - e^{y} = \frac{1}{x + 1}.$$

Solving for $$y$$:

$$e^{y} = 2 - \frac{1}{x + 1}, \qquad y = \ln\!\left(2 - \frac{1}{x + 1}\right).$$

Given differential equation

$$\frac{y\,dx-x\,dy}{y}=0$$

Multiply both sides by $$y$$ to clear the denominator:

$$y\,dx-x\,dy=0$$

Rearrange to separate the differentials:

$$y\,dx=x\,dy$$

Divide both sides by $$xy$$ (which is permissible as we assume $$x\neq0,\;y\neq0$$ while solving):

$$\frac{dx}{x}=\frac{dy}{y}$$

Integrate both sides:

$$\int\frac{dx}{x}=\int\frac{dy}{y}$$

$$\ln|x|=\ln|y|+C$$

Collect the logarithms on one side:

$$\ln|x|-\ln|y|=C$$

Using the property $$\ln a-\ln b=\ln\left(\dfrac{a}{b}\right)$$:

$$\ln\left|\dfrac{x}{y}\right|=C$$

Exponentiate to remove the logarithm. Let $$e^{C}=C_1$$ (where $$C_1>0$$ absorbs the constant):

$$\left|\dfrac{x}{y}\right|=C_1\;\;\Longrightarrow\;\;\dfrac{x}{y}=\pm C_1$$

The “$$\pm$$” can be merged into a single arbitrary non-zero constant, say $$C$$. Hence

$$\dfrac{x}{y}=C\quad\Longrightarrow\quad y=Cx$$

Thus the general solution is

$$y=Cx$$

Among the given alternatives, this corresponds to option (C).

We are given the first-order linear differential equation

$$\frac{dx}{dy}+\mathrm P_1\,x=\mathrm Q_1,$$

in which $$x$$ is the dependent variable and $$y$$ is the independent variable. Here $$\mathrm P_1$$ and $$\mathrm Q_1$$ are functions of $$y$$ alone.

Step 1: Write the equation in standard linear form.

The equation is already in the required form $$\dfrac{dx}{dy}+P(y)\,x=Q(y)$$ with $$P(y)=\mathrm P_1$$ and $$Q(y)=\mathrm Q_1.$$

Step 2: Find the Integrating Factor (I.F.).

The integrating factor for $$\dfrac{dx}{dy}+P(y)\,x=Q(y)$$ is

$$\text{I.F.}=e^{\int P(y)\,dy}=e^{\int \mathrm P_1\,dy}.$$

Step 3: Multiply the differential equation by the I.F.

$$e^{\int \mathrm P_1\,dy}\,\frac{dx}{dy}+\mathrm P_1\,e^{\int \mathrm P_1\,dy}\,x=\mathrm Q_1\,e^{\int \mathrm P_1\,dy}.$$

Step 4: Recognise the left–hand side as a derivative.

Because $$\dfrac{d}{dy}\Bigl(x\,e^{\int \mathrm P_1\,dy}\Bigr)=e^{\int \mathrm P_1\,dy}\,\dfrac{dx}{dy}+\mathrm P_1\,e^{\int \mathrm P_1\,dy}\,x,$$ the equation becomes

$$\frac{d}{dy}\Bigl(x\,e^{\int \mathrm P_1\,dy}\Bigr)=\mathrm Q_1\,e^{\int \mathrm P_1\,dy}.$$

Step 5: Integrate with respect to $$y$$.

$$x\,e^{\int \mathrm P_1\,dy}=\int \Bigl(\mathrm Q_1\,e^{\int \mathrm P_1\,dy}\Bigr)dy+C,$$

where $$C$$ is the constant of integration.

Step 6: State the general solution and identify the correct option.

The general solution is therefore

$$x\,e^{\int \mathrm P_1\,dy}=\int \bigl(\mathrm Q_1\,e^{\int \mathrm P_1\,dy}\bigr)dy+C.$$

Comparing with the options, this matches (C).

Hence, the correct alternative is (C).

Given differential equation:

$$e^{x}\,dy + (y\,e^{x} + 2x)\,dx = 0.$$

Write it as $$M(x, y)\,dx + N(x, y)\,dy = 0$$ with

$$M(x, y) = y\,e^{x} + 2x, \qquad N(x, y) = e^{x}.$$

1. Check for exactness.

$$\frac{\partial M}{\partial y} = e^{x}, \qquad \frac{\partial N}{\partial x} = e^{x}.$$

Since $$\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$$, the equation is exact.

2. Find the potential function $$F(x, y)$$.

Because the equation is exact, there exists $$F$$ with $$\dfrac{\partial F}{\partial x} = M$$ and $$\dfrac{\partial F}{\partial y} = N$$.

(i) Integrate $$M$$ with respect to $$x$$ (treating $$y$$ as a constant):

$$F(x, y) = \int (y\,e^{x} + 2x)\,dx = y\,e^{x} + x^{2} + g(y),$$

where $$g(y)$$ is an arbitrary function of $$y$$.

(ii) Differentiate with respect to $$y$$ and equate to $$N$$:

$$\frac{\partial F}{\partial y} = e^{x} + g'(y) = e^{x} \;\Longrightarrow\; g'(y) = 0.$$

Hence $$g(y)$$ is a constant, absorbed into the final constant of integration.

3. General solution.

$$y\,e^{x} + x^{2} = C.$$

This matches option (C).