ISI2015-MMA-87

Question

If $x(t)$ is a solution of $$(1-t^2) dx -tx\: dt =dt$$ and  $x(0)=1$, then $x\big(\frac{1}{2}\big)$ is equal to

• A. $\frac{2}{\sqrt{3}} (\frac{\pi}{6}+1)$
• B. $\frac{2}{\sqrt{3}} (\frac{\pi}{6}-1)$
• C. $\frac{\pi}{3 \sqrt{3}}$
• D. $\frac{\pi}{\sqrt{3}}$

Answer 1

Rearranging the terms of the differential equation, we get    $\frac{dx}{dt}+x\left ( \frac{-t}{1-t^{2}} \right )=\frac{1}{1-t^{2}}$

This is the standard linear Differential equation form for which the integrating factor is:   $e^{\int \frac{-t}{1-t^{2}}dt}=e^{\frac{1}{2}\times \ln_{e}\left ( 1-t^{2} \right )}=\sqrt{1-t^{2}}$

So, solution to this D.E. is :   $x\sqrt{1-t^{2}}=\int \frac{1}{1-t^{2}}\times \sqrt{1-t^{2}}dt$

$\Rightarrow$    let $t=\sin \theta$   $\Rightarrow$   $dt=\cos\theta d\theta$

$\Rightarrow$   $x\sqrt{1-t^{2}}=\int \frac{\cos\theta }{\cos\theta }d\theta$    $\Rightarrow$    $x\sqrt{1-t^{2}}=\sin^{-1}t+c$

Given that at   $t=0, x=1$.  So,  $1\left ( 1 \right )=\sin^{-1}\left ( 0 \right )+c \Rightarrow c=1$

$\therefore$   Solution to the D.E. is $x\sqrt{1-t^{2}}=1+\sin^{-1}(t)$

For $t=\frac{1}{2}$,   $x\left ( \frac{\sqrt{3}}{2} \right )=\sin^{-1}(\frac{1}{2})+1$    $\Rightarrow$    $x=\frac{2}{\sqrt{3}}\left ( \frac{\pi }{6}+1 \right )$

Option A is correct.