Question

Let the solution curve $y=f\left(x\right)$ of the differential equation $\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^4+2x}{\sqrt{1-x^2}},x\in (-1,1)$ pass through the origin. Then ${\int }_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}f\left(x\right)dx$ is

Answer 1

$\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^4+2x}{\sqrt{1-x^2}}$
which is first order linear differential equation. Integrating factor (I.F.) $=e^{\int \frac{x}{x^2-1}dx}$
$=e^{(1/2\ln |x^2-1|})=\sqrt{|x^2-1|}=\sqrt{1-x^2}\because x\in (-1,1)$
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, $c=0$
$y=\frac{x^5+5x^2}{5\sqrt{1-x^2}}$
${\int }_{-\sqrt{3}/2}^{\sqrt{3}/2}\frac{x^5+5x^2}{5\sqrt{\sqrt{1-x^2}}}dx=0+2{\int }_0^{\frac{\sqrt{3}}{2}}\frac{x^2}{\sqrt{1-x^2}}dx$
$(\because \frac{x^5}{\sqrt{1-x^2}}\text{is odd fucntion})$
$2{\int }_0^{\sqrt{3}/2}(-\sqrt{1-x^2})+\frac{1}{\sqrt{1-x^2}}dx={[2{\sin }^{-1}x-({\sin }^{-1}x+x\sqrt{1-x^2})]}_0^{\sqrt{3}/2}=\frac{\pi }{3}-\frac{\sqrt{3}}{4}$