Question

The function $y=f(x)$ 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, $\underset{-\frac{\sqrt{3}}{2}}{\overset{\sqrt{3}/2}{\int }}f(x)dx$ 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. $\frac{\pi }{6}-\frac{\sqrt{3}}{2}$
  

Answer 1

Answer: (B)

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