Consider the differential equation, $$y^2 dx + \left(x - \frac{1}{y}\right) dy = 0$$. If value of y is 1 when x = 1, then the value of x for which y = 2, is

We have the differential equation $$y^{2}\,dx+\left(x-\dfrac1y\right)dy=0.$$

First we isolate $$dx$$ on one side. Dividing every term by $$dy$$, we get

$$y^{2}\dfrac{dx}{dy}+x-\dfrac1y=0.$$

Transposing the last two terms to the right gives

$$y^{2}\dfrac{dx}{dy}= -x+\dfrac1y.$$

Now we divide both sides by $$y^{2}$$ so that the coefficient of $$\dfrac{dx}{dy}$$ becomes $$1$$:

$$\dfrac{dx}{dy}= -\dfrac{x}{y^{2}}+\dfrac{1}{y^{3}}.$$

We can rewrite this as a linear differential equation in the standard form $$\dfrac{dx}{dy}+P(y)\,x=Q(y)$$ by bringing the term with $$x$$ to the left:

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

Here $$P(y)=\dfrac1{y^{2}}$$ and $$Q(y)=\dfrac1{y^{3}}.$$ The integrating factor (I.F.) for such an equation is obtained from the formula $$\text{I.F.}=e^{\displaystyle\int P(y)\,dy}.$$

We compute the integral of $$P(y)$$:

$$\int\dfrac1{y^{2}}\,dy=\int y^{-2}\,dy=-y^{-1}.$$

Thus

$$\text{I.F.}=e^{-y^{-1}}=e^{-1/y}.$$

Multiplying the entire differential equation by this integrating factor produces

$$e^{-1/y}\dfrac{dx}{dy}+e^{-1/y}\dfrac{x}{y^{2}}=e^{-1/y}\dfrac1{y^{3}}.$$

Because $$\dfrac{d}{dy}\!\left(e^{-1/y}\right)=e^{-1/y}\cdot\dfrac1{y^{2}},$$ the left-hand side is precisely the derivative of the product $$x\,e^{-1/y}$$ with respect to $$y$$. Hence we can write

$$\dfrac{d}{dy}\!\left(x\,e^{-1/y}\right)=e^{-1/y}\dfrac1{y^{3}}.$$

Integrating both sides with respect to $$y$$ gives

$$\int\dfrac{d}{dy}\!\left(x\,e^{-1/y}\right)dy=\int e^{-1/y}\dfrac1{y^{3}}\,dy.$$

The integral on the left is straightforward:

$$x\,e^{-1/y}= \int y^{-3}e^{-1/y}\,dy +C,$$

where $$C$$ is the constant of integration. To evaluate the right-hand integral we set a substitution.

Let $$t=-\dfrac1y,$$ so that $$dt=\dfrac1{y^{2}}\,dy$$ and hence $$y^{-2}\,dy=dt.$$ We rewrite the integrand $$y^{-3}e^{-1/y}\,dy$$ as $$y^{-1}(y^{-2}dy)\,e^{t}=-\dfrac1y\,e^{t}\,dt=-t\,e^{t}\,dt.$$

Therefore

$$\int y^{-3}e^{-1/y}\,dy=\int(-t\,e^{t})\,dt=-\int t\,e^{t}\,dt.$$

Using the integration by parts formula $$\int t\,e^{t}\,dt=t\,e^{t}-e^{t},$$ we obtain

$$-\int t\,e^{t}\,dt=-(t\,e^{t}-e^{t})=-t\,e^{t}+e^{t}.$$

Substituting back $$t=-\dfrac1y$$ gives

$$-t\,e^{t}+e^{t}= \left(\dfrac1y+1\right)e^{-1/y}.$$

Placing this result in the earlier equation yields

$$x\,e^{-1/y}= \left(1+\dfrac1y\right)e^{-1/y}+C.$$

Since the factor $$e^{-1/y}$$ is common, we divide through by it to make $$x$$ explicit:

$$x=1+\dfrac1y+C\,e^{1/y}.$$

We now apply the initial condition. When $$x=1$$ we are told $$y=1$$, so

$$1=1+\dfrac11+C\,e^{1/1}\quad\Longrightarrow\quad1=1+1+Ce,$$

which simplifies to

$$Ce=-1\quad\Longrightarrow\quad C=-\dfrac1e.$$

Substituting this value of $$C$$ back into the general solution gives

$$x=1+\dfrac1y-\dfrac1e\,e^{1/y}=1+\dfrac1y-e^{\frac1y-1}.$$

The problem asks for the value of $$x$$ when $$y=2$$. Substituting $$y=2$$ (so $$\dfrac1y=\dfrac12$$) we have

$$x=1+\dfrac12-e^{\frac12-1}=1+\dfrac12-e^{-1/2}.$$

Since $$e^{-1/2}=\dfrac1{\sqrt{e}},$$ the expression becomes

$$x=\dfrac32-\dfrac1{\sqrt{e}}.$$

Hence, the correct answer is Option A.