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, $\sqrt{a}.y\left(1\right)=\frac{\pi }{32}$
$\Rightarrow \sqrt{a}=\frac{1}{4}\Rightarrow a=\frac{1}{16}$