Integrating Factor

This is a first-order linear differential equation of the form $$\frac{dy}{dx} + P(x)y = Q(x)$$

where $$P(x) = -2\tan x$$ and $$Q(x) = \sin x.$$

The integrating factor is:
$$\mu(x) = e^{\int P(x)\,dx} = e^{\int -2\tan x\,dx}$$

$$\int -2\tan x\,dx = -2\int \frac{\sin x}{\cos x}\,dx = 2\ln|\cos x| = \ln(\cos^2 x)$$

$$\mu(x) = e^{\ln(\cos^2 x)} = \cos^2 x$$

Multiplying the equation by the integrating factor:
$$\cos^2 x \cdot \frac{dy}{dx} - 2\cos^2 x \cdot y\tan x = \cos^2 x \sin x$$

The left side is the derivative of a product:
$$\frac{d}{dx}(\cos^2 x \cdot y) = \cos^2 x \sin x$$

Integrating both sides:
$$\cos^2 x \cdot y = \int \cos^2 x \sin x\,dx$$

Using substitution $$u = \cos x$$, so $$du = -\sin x\,dx$$:
$$\int \cos^2 x \sin x\,dx = -\int u^2\,du = -\frac{u^3}{3} = -\frac{\cos^3 x}{3} + C$$

The general solution is:
$$\cos^2 x \cdot y = -\frac{\cos^3 x}{3} + C$$
$$y = -\frac{\cos x}{3} + \frac{C}{\cos^2 x}$$

Applying the initial condition $$y(\pi/6) = 1/2$$:
$$\frac{1}{2} = -\frac{\cos(\pi/6)}{3} + \frac{C}{\cos^2(\pi/6)}$$
$$\frac{1}{2} = -\frac{\sqrt{3}/2}{3} + \frac{C}{3/4}$$
$$\frac{1}{2} = -\frac{\sqrt{3}}{6} + \frac{4C}{3}$$
$$\frac{1}{2} + \frac{\sqrt{3}}{6} = \frac{4C}{3}$$
$$\frac{3 + \sqrt{3}}{6} = \frac{4C}{3}$$
$$C = \frac{3+\sqrt{3}}{8}$$

Finding $$y(\pi/3)$$:
$$y(\pi/3) = -\frac{\cos(\pi/3)}{3} + \frac{C}{\cos^2(\pi/3)}$$
$$= -\frac{1/2}{3} + \frac{(3+\sqrt{3})/8}{1/4}$$
$$= -\frac{1}{6} + \frac{4(3+\sqrt{3})}{8}$$

The differential equation is linear:

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

where

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

The integrating factor is

$$I.F.=e^{\int \frac{2x}{x^2+1}\,dx}$$

$$=e^{\ln(x^2+1)}$$

$$=x^2+1$$

Multiplying the differential equation by

$$x^2+1$$

gives

$$\left(x^2+1\right)\frac{dy}{dx}+2xy=6x(x^2+1)^2$$

The left-hand side becomes

$$\frac{d}{dx}\Big[(x^2+1)y\Big]$$

Therefore,

$$\frac{d}{dx}\Big[(x^2+1)y\Big]=6x(x^2+1)^2$$

Integrating,

$$ (x^2+1)y=\int 6x(x^2+1)^2\,dx+C $$

Let

$$u=x^2+1$$

Then

$$du=2x\,dx$$

and

$$6x\,dx=3\,du$$

Hence

$$\int 6x(x^2+1)^2\,dx=3\int u^2\,du$$

$$=u^3+C$$

$$=(x^2+1)^3+C$$

Thus

$$ (x^2+1)y=(x^2+1)^3+C $$

Using

$$y(1)=9$$

$$2(9)=2^3+C$$

$$18=8+C$$

$$C=10$$

Therefore,

$$y=\frac{(x^2+1)^3+10}{x^2+1}$$

Now substitute

$$x=3$$

$$y(3)=\frac{(10)^3+10}{10}$$

$$=\frac{1010}{10}$$

$$=101$$

Final Answer :

$$101$$