Question

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

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

Answer 1

The correct option is A $\frac{1}{16}$

Given,
$(x^2+1{)}^2\frac{dy}{dz}+2x(x^2+1)y=1$
$\frac{dy}{dx}+\left(\frac{2x}{x^2+1}\right)y=\frac{1}{(x^2+1{)}^2}$

Integrating Factor(I.F.)$=e^{\int \frac{2x}{1+x^2}dx}=(1+x^2)$
Solution of differential equation:-
$y(1+x^2)=\int \frac{1}{(1+x^2{)}^2}\times (1+x^2)dx$
$y(1+x^2)={\tan }^{-1}\left(x\right)+c$

Given that
$y\left(0\right)=0$
$c=0$
$y(1+x^2)={\tan }^{-1}x$
Put $x=1$
$y\times 2=\frac{\pi }{4}$
$y=\frac{\pi }{8}$
$\frac{1}{4}\cdot y=\frac{\pi }{32}$
$\Rightarrow \sqrt{a}=\frac{1}{4}$
$a=\frac{1}{16}$