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  4. If (1+x2) (dy/dx)+y=tan-1x, y(0)1, then y((π/4))=

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Q. If $\left(1+x^{2}\right) \frac{dy}{dx}+y=tan^{-1}x, y\left(0\right)1$, then $y\left(\frac{\pi}{4}\right)=$

Differential Equations

Solution:

$\frac{dy}{dx}+\frac{y}{1+x^{2}}=\frac{tan^{-1}\,x}{1+x^{2}}$
It is linear differential equation with
$I.F.=\text{exp} \int \frac{dx}{1+x^{2}}=e^{tan^{-1}\,x}$
$\therefore $ Solution is $ye^{tan^{-1}\,x}=\int e^{tan^{-1}\,x}\cdot\frac{tan^{-1}\,x}{1+x^{2}}dx$
$ye^{tan^{-1}\,}x=\left(tan^{-1}\,x-1\right)e^{tan^{-1}\,x}+c$
$\Rightarrow y=tan^{-1}\,x-1+ce^{-tan^{-1}\,x}$
$x=0$,
$y=1$
$\Rightarrow c=2$
$\Rightarrow y\left(x\right)=tan^{-1}\,x-1+2e^{-tan^{-1}\,x}$
$\therefore y\left( \frac{\pi}{4}\right)=\frac{2}{e}$.