Question

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

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

Answer 1

Answer: (D)

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