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 $f\left(0\right)=0$. Then ${\int }_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}f\left(x\right)dx$ 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. $\frac{\pi }{6}-\frac{\sqrt{3}}{4}$

Answer 1

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

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

This is a linear differential equation

I.F. $=e^{\int \frac{x}{x^2-1}dx}=e^{\frac{1}{2}ln|x^2-1|}=\sqrt{1-x^2}$

$\Rightarrow$ solution is

$y\sqrt{1-x^2}=\int \frac{x(x^3+2)}{\sqrt{1-x^2}}\cdot \sqrt{1-x^2}dx$

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

$f\left(0\right)=0\Rightarrow c=0$

$\Rightarrow f\left(x\right)\sqrt{1-x^2}=\frac{x^5}{5}+x^2$

Now,

${\int }_{-\sqrt{3}/2}^{\sqrt{3}/2}f\left(x\right)dx={\int }_{\sqrt{3}/2}^{\sqrt{3}/2}d\frac{x^2}{\sqrt{1-x^2}}dx$ (Using property)

$=2{\int }_0^{\sqrt{3}/2}\frac{x^2}{\sqrt{1-x^2}}dx=2{\int }_0^{\pi /3}\frac{si{n}^2\theta }{cos\theta }cos\theta d\theta$ (Taking $x=sin\theta$)

$=2{\int }_0^{\pi /3}si{n}^2\theta d\theta =2{[\frac{\theta }{2}-\frac{sin2\theta }{4}]}_0^{\frac{\pi }{3}}=2\left(\frac{\pi }{6}\right)-2\left(\frac{\sqrt{3}}{8}\right)=\frac{\pi }{3}-\frac{\sqrt{3}}{4}$