JEE Main 2022 — Differential Equations Question with Solution

$$\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^4+2x}{\sqrt{1-x^2}}$$

which is first order linear differential equation.

Integrating factor $$(I.F.)=e^{\int \frac{x}{x^2-1}dx}$$

$$=e^{\frac{1}{2}\ln |x^2-1|}=\sqrt{|x^2-1|}$$

$$=\sqrt{1-x^2}$$

$$\because$$ $$x\in (-1,1)$$

Solution of differential equation

$$y\sqrt{1-x^2}=\int (x^4+2x)dx=\frac{x^5}{5}+x^2+c$$

Curve is passing through origin, $$c=0$$

$$y=\frac{x^5+5x^2}{5\sqrt{1-x^2}}$$

$$\underset{\frac{-\sqrt{3}}{2}}{\overset{\frac{\sqrt{3}}{2}}{\int }}\frac{x^5+5x^2}{5\sqrt{1-x^2}}dx=0+2\underset{0}{\overset{\frac{\sqrt{3}}{2}}{\int }}\frac{x^2}{\sqrt{1-x^2}}dx$$

put $$x=\sin \theta$$

$$dx=\cos \theta d\theta$$

$$I=2\underset{0}{\overset{\frac{\pi }{3}}{\int }}\frac{{\sin }^2\theta .\cos \theta d\theta }{\cos \theta }$$

$$=\underset{0}{\overset{\frac{\pi }{3}}{\int }}(1-\cos 2\theta )d\theta$$

$$={\left(\theta -\frac{\sin 2\theta }{2}\right)|}_0^{\frac{\pi }{3}}$$

$$=\frac{\pi }{3}-\frac{\sqrt{3}}{4}$$