Q) If $x(t)$ is a solution of
$(1-t^{2})\;dx - tx\;dt = dt$
and $x(0)=1$, then $x\left ( \frac{1}{2} \right )$ is equal to
(A) $\frac{2}{\sqrt{3}}\left ( \frac{\pi }{6}+1 \right )$
(B) $\frac{2}{\sqrt{3}}\left ( \frac{\pi }{6}-1 \right )$
(C) $\frac{\pi }{3\sqrt{3}}$
(D) $\frac{\pi }{\sqrt{3}}$
My Approach :- Here, ordinary differential equation(ODE) is :-
$(1-t^{2})\;dx - tx\;dt = dt$
Divide the whole equation by $dt$
$(1-t^{2})\;\frac{dx}{dt} - tx = 1$
$\frac{dx}{dt} = \frac{1+tx}{1-t^{2}}$
$\frac{dx}{dt} +\; \left ( \frac{t}{t^{2}-1} \right )x= \frac{1}{1-t^{2}}$
Now, Integrating Factor (I.F.) will be $e^{\int \frac{t}{(t^{2}-1)}\;dt}$. Now,substituting $t^{2}-1 = z$.It will become $e^{\frac{1}{2}\int \frac{1}{z}\;dz} = \sqrt{z} = \sqrt{t^{2}-1}$.
Now, Solution of given ODE will be :-
$x\sqrt{t^{2}-1} = -\int \left ( \frac{1}{t^{2}-1} \right )\times \sqrt{t^{2}-1}\;dt + c$.
$x\sqrt{t^{2}-1} = -\int \left ( \frac{1}{\sqrt{t^{2}-1}} \right )\;dt + c$
$x\sqrt{t^{2}-1} = -ln|t+\sqrt{t^{2}-1}| + c$.
Now, using given condition $x(0)=1$
$1*i = -ln|i| + c$
Here, I am getting solution in terms of $’i’$ but it should be in in terms of $\pi$ and does not involve $'i'$. I am not getting where I am doing mistake. Please help.