Question

Let y = y(x) be the solution of the differential equation, $(x^2 + 1)^2 \frac{dy}{dx} + 2x (x^2 + 1) y = 1$ such that $y(0) = 0 $. If $\sqrt{a} y (1) = \frac{\pi}{32}$ , then the value of 'a' is

• A. $\frac{1}{2}$
• B. $\frac{1}{16}$
• C. $\frac{1}{4}$
• D. 1

Answer 1

Answer: (B)

$\frac{dy}{dx} + \left(\frac{2x}{x^{2} + 1}\right)y= \frac{1}{\left(x^{2} +1\right)^{2}} $
(Linear differential equation)
$\therefore I.F. = e^{\ell n\left(x^2+1\right)} = \left(x^{2} + 1\right) $
So, general solution is $y.\left(x^{2}+1\right)=\tan^{-1}x + c$
As $ y\left(0\right) =0 \Rightarrow c = 0$
$ \therefore y\left(x\right) = \frac{\tan^{-1}x}{x^{2} + 1} $
As, $ \sqrt{a} .y\left(1\right) = \frac{\pi}{32} $
$ \Rightarrow \sqrt{a} = \frac{1}{4} \Rightarrow a = \frac{1}{16} $