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, $\int \limits _{-\frac {\sqrt 3}{2}}^{\sqrt 3/2} \: 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)

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