Question

Suppose $y=f\left(x\right)$ is the solution of the differential equation $(1-x^2)\frac{dy}{dx}=\left(\sqrt{1-x^2}\right)(x^4+2x)+xy$ for $x\in (a,b)$, where $f\left(0\right)=0$. Then

• A. $f\left(x\right)$ is an even function.
• B. $f\left(x\right)$ is an odd function.
• C. $b-a=2$
• D. $\underset{-1/2}{\overset{1/2}{\int }}f\left(x\right)dx=\frac{\pi }{6}-\frac{\sqrt{3}}{4}$

Answer 1

Answer: (C), (D)

$\underset{-1/2}{\overset{1/2}{\int }}f\left(x\right)dx=\frac{\pi }{6}-\frac{\sqrt{3}}{4}$

Given : $(1-x^2)\frac{dy}{dx}=\left(\sqrt{1-x^2}\right)(x^4+2x)+xy$
$\Rightarrow \frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^4+2x}{\sqrt{1-x^2}}$
For the square root to be defined,
$1-x^2>0\Rightarrow x\in (-1,1)$

Comparing with $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$
$I.F.=e^{\int P\left(x\right)dx}$
$\Rightarrow \int P\left(x\right)dx=\int \frac{x}{x^2-1}dx$
Assuming $x^2-1=t\Rightarrow 2xdx=dt$, then
$\int P\left(x\right)dx=\frac{1}{2}\int \frac{dt}{t}\Rightarrow \int P\left(x\right)dx=\frac{1}{2}\ln |x^2-1|$
As $x\in (-1,1)$, so
$\int P\left(x\right)dx=\frac{1}{2}\ln (1-x^2)\Rightarrow I.F.=\sqrt{1-x^2}$

Solution is :
$y\sqrt{1-x^2}=\int (x^4+2x)dx\Rightarrow y\sqrt{1-x^2}=\frac{x^5}{5}+x^2+C\because f\left(0\right)=0,\therefore C=0\Rightarrow y=f\left(x\right)=\frac{\frac{x^5}{5}+x^2}{\sqrt{1-x^2}}$
Clearly, $f\left(x\right)$ is neither even nor odd function.
$f\left(x\right)$ is defined $\forall x\in (-1,1)$
Therefore, $b-a=2$

$I=\underset{-1/2}{\overset{1/2}{\int }}\frac{\frac{x^5}{5}+x^2}{\sqrt{1-x^2}}dx\Rightarrow I=\underset{-1/2}{\overset{1/2}{\int }}\frac{x^2}{\sqrt{1-x^2}}dx(\because x^5\text{is odd function})\Rightarrow I=2\underset{0}{\overset{1/2}{\int }}\frac{x^2}{\sqrt{1-x^2}}dx$

Taking $x=\sin \theta \Rightarrow dx=\cos \theta d\theta$
$I=2\underset{0}{\overset{\pi /6}{\int }}{\sin }^2\theta d\theta \Rightarrow I=\underset{0}{\overset{\pi /6}{\int }}(1-\cos 2\theta )d\theta \Rightarrow I={[\theta -\frac{\sin 2\theta }{2}]}_0^{\pi /6}\therefore I=\frac{\pi }{6}-\frac{\sqrt{3}}{4}$