Question

Solve the differential equation $(1+x^2)\frac{dy}{dx}+y=e^{{\tan }^{-1}x}$.

Answer 1

$(1+x^2)\frac{dy}{dx}+y=e^{ta{n}^{-1}x}$
$\frac{dy}{dx}+\frac{y}{1+x^2}=\frac{e^{ta{n}^{-1}x}}{1+x^2}$
Hence
$IF=e^{\int \frac{1}{1+x^2}dx}$
$=e^{ta{n}^{-1}x}$
Therefore
$e^{ta{n}^{-1}x}dy+\frac{e^{ta{n}^{-1}x}y}{1+x^2}dx=\frac{e^{2ta{n}^{-1}x}}{1+x^2}dx$
$d(e^{ta{n}^{-1}x}.y)=d\left(\frac{e^{2ta{n}^{-1}x}}{2}\right)$
Integrating both sides
$\int d(e^{ta{n}^{-1}x}.y)=\int d\left(\frac{e^{2ta{n}^{-1}x}}{2}\right)$
$e^{ta{n}^{-1}x}.y=0.5e^{2ta{n}^{-1}x}+c$
Or
$2y=e^{ta{n}^{-1}x}+2c{e}^{-ta{n}^{-1}x}$