Mathematics - Differential Equation Question with Solution | TestHub

$\frac{dy}{dx}+\frac{x}{x^2-1}y=\frac{x^4+2x}{\sqrt{1-x^2}}$
This is a linear differential equation
I.F.$=e^{\int \frac{x}{x^2-1} dx}=e^{\frac{1}{2} In x^2-1}= \sqrt{1-x^2}$
$\Rightarrow solution is$
$y \sqrt{1-x^2}=\int \frac{x\left(x^3+2\right)}{\sqrt{1-x^2}} . \sqrt{1-x^2} dx$
Or$y\sqrt{1-x^2}=\int \left(x^4+2x\right) dx=\frac{x^5}{5}+x^2+c$
$f\left(0\right)=0 \Rightarrow c=0$
$\Rightarrow f\left(x\right) \sqrt{1-x^2}=\frac{x^5}{5}+x^2$
Now,$\underset{-\frac{\sqrt{3}}{2}}{\overset{\frac{\sqrt{3}}{2}}{\int }}f\left(x\right)dx=\underset{-\frac{\sqrt{3}}{2}}{\overset{\frac{\sqrt{3}}{2}}{\int }}\frac{x^2}{\sqrt{1-x^2}} dx$(Using property)
$=2\underset{0}{\overset{\frac{\sqrt{3}}{2}}{\int }}\frac{x^2}{\sqrt{1-x^2}} dx=2 \underset{0}{\overset{\frac{\pi }{3}}{\int }}\frac{{\sin }^2\theta }{\cos \theta }\cos \theta d\theta$(Taking$x=\sin \theta$)
$=2\underset{0}{\overset{\frac{\pi }{3}}{\int }}{\sin }^2\theta d\theta =2{{\left[\frac{\theta }{2}-\frac{\mathit{sin}2\theta }{4}\right]}_0}^{\frac{\pi }{3}}=2\left(\frac{\pi }{6}\right)-2\left(\frac{\sqrt{3}}{8}\right)=\frac{\pi }{3}-\frac{\sqrt{3}}{4}$