# Let y=y(x) be the solution of the differential equation,

Question:

Let $y=y(x)$ be the solution of the differential equation, $\left(x^{2}+1\right)^{2} \frac{d y}{d x}+2 x\left(x^{2}+1\right) y=1$ such that $y(0)=0$. If

$\sqrt{a} y(1)=\frac{\pi}{32}$, then the value of ' $a$ ' is :

1. (1) $\frac{1}{4}$

2. (2) $\frac{1}{2}$

3. (3) 1

4. (4) $\frac{1}{16}$

Correct Option: , 4

Solution:

$\left(1+x^{2}\right)^{2} \frac{d y}{d x}+2 x\left(1+x^{2}\right) y=1$

$\Rightarrow \frac{d y}{d x}+\left(\frac{2 x}{1+x^{2}}\right) y=\frac{1}{\left(1+x^{2}\right)^{2}}$

Since, the above differential equation is a linear differential equation

$\therefore$ I.F $=e^{\int \frac{2 x}{1+x^{2}} d x}=e^{\log \left(1+x^{2}\right)}=1+x^{2}$

Then, the solution of the differential equation

$\Rightarrow y\left(1+x^{2}\right)=\int \frac{d x}{1+x^{2}}+c$

$\Rightarrow y\left(1+x^{2}\right)=\tan ^{-1} x+c$.....(1)

If $x=0$ then $y=0 \quad$ (given)

$\Rightarrow 0=0+c$

$\Rightarrow c=0$

Then, equation (1) becomes,

$\Rightarrow y\left(1+x^{2}\right)=\tan ^{-1} x$

Now put $x=1$ in above equation, then

$2 y=\frac{\pi}{4}$

$\Rightarrow 2\left(\frac{\pi}{32 \sqrt{a}}\right)=\frac{\pi}{4} \quad\left[\sqrt{a} y(1)=\frac{\pi}{32}\right]$

$\Rightarrow \sqrt{a}=\frac{1}{4}$

$\Rightarrow a=\frac{1}{16}$