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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 :

Option 1)
$\frac{1}{16}$

Option 2)
$\frac{1}{4}$

Option 3)
$1$

Option 4)
$\frac{1}{2}$

Answers (1)

S solutionqc

$\\(x^{2}+1)^{2}\frac{dy}{dx}+2x(x^{2}+1)y=1\\\; \; \; \\y(0)=0$

$\frac{dy}{dx}+\frac{2xy}{x^{2}+1}=\frac{1}{(x^{2}+1)^{2}}$

this is a linear differential equation with $I.F=e^{\int \frac{2x}{x^{2}+1}dx}=e^{\ln(x^{2}+1)}=x^{2}+1$

$\therefore Solution \; y.(x^{2}+1)=\int (x^{2}+1)\frac{1}{(x^{2}+1)^{2}}dx=\int \frac{dx}{x^{3}+1}$

$=\tan^{-1}\; x+c$

$y(0)=0\Rightarrow 0=\tan^{-1}(0)+1=0\Rightarrow c=0$

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

$y(x)=\frac{\tan^{-1}(x)}{x^{2}+1}$

$\sqrt{a} \; y(1)=\frac{\pi}{32}$

We have $y(1)=\frac{\tan^{-1}(1)}{2}=\frac{\frac{\pi}{4}}{2}=\frac{\pi}{8}$

$\frac{\sqrt{a}\; y(1)}{y(1)}=\frac{\frac{\pi}{32}}{\frac{\pi}{8}}\Rightarrow \sqrt{a}=\frac{8}{32}=\frac{1}{4}$

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

Option 1)

$\frac{1}{16}$

Option 2)

$\frac{1}{4}$

Option 3)

$1$

Option 4)

$\frac{1}{2}$

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