Question

The function $y=f\left(x\right)$ is the solution of the differential equation $\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^4+2x}{\sqrt{1-x^2}}$ in (-1, 1) satisfying f(0) =0. Then ${\int }_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}f\left(x\right)d\left(x\right)$ is

• A. $\frac{\pi }{3}-\frac{\sqrt{3}}{2}$
• B. $\frac{\pi }{3}-\frac{\sqrt{3}}{4}$
• C. $\frac{\pi }{6}-\frac{\sqrt{3}}{4}$
• D. $\frac{\pi }{6}-\frac{\sqrt{3}}{2}$

Answer 1

The correct option is B $\frac{\pi }{3}-\frac{\sqrt{3}}{4}$

Given D.E. can be written as
$\frac{dy}{dx}-\frac{x}{1-x^2}y=\frac{x^4+2x}{\sqrt{1-x^2}}$
If $e^{\int \frac{-x}{1-x^2}dx}=e^{\frac{+1}{2}log(1-x^2)}=\sqrt{1-x^2}$
$\therefore$ Solution is given by
$y\sqrt{1-x^2}=\int \sqrt{1-x^2}.\frac{x^4+2x}{\sqrt{1-x^2}}dxy\sqrt{1-x^2}=\frac{x^5}{5}+x^2+cf\left(0\right)=0\Rightarrow Atx=0,y=0\therefore c=0\therefore y\sqrt{1-x^2}=\frac{x^5}{5}+x^2$
or $y=f\left(x\right)=\frac{\frac{x^5}{5}+x^2}{\sqrt{1-x^2}}$
$\therefore$ $I={\int }_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}\frac{\frac{x^5}{5}+x^2}{\sqrt{1-x^2}}dx=2{\int }_0^{\frac{\sqrt{3}}{2}}\frac{x^2}{\sqrt{1-x^2}}dx(\because \frac{x^5}{\sqrt{1-x^2}}isodd)$
put $x=sin\theta \Rightarrow dx=cos\theta d\theta$
$\therefore I=2{\int }_0^{\frac{\pi }{3}}si{n}^2\theta d\theta ={\int }_0^{\frac{\pi }{3}}(1-cos2\theta d\theta )$
$={(\theta -\frac{sin2\theta }{2})}_0^{\frac{\pi }{3}}=\frac{\pi }{3}-\frac{\sqrt{3}}{4}$