Question

Solve : $(1+y^2)dx=({\tan }^{-1}y-x)dy$

Answer 1

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

$\frac{dx}{dy}=\frac{{\tan }^{-1}y}{1+y^2}-\frac{x}{1+y^2}$

$\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{{\tan }^{-1}y}{1+y^2}$

Hence

$IF=e^{\int \frac{1}{1+y^2}.dy}$

$=e^{{\tan }^{-1}y}$

Hence the above differential equation changes to

$e^{{\tan }^{-1}y}.\frac{dx}{dy}+\frac{x{e}^{{\tan }^{-1}y}}{1+y^2}=\frac{e^{{\tan }^{-1}y}{\tan }^{-1}y}{1+y^2}$

$e^{{\tan }^{-1}y}.dx+\frac{x{e}^{{\tan }^{-1}y}}{1+y^2}dy=\frac{e^{{\tan }^{-1}y}{\tan }^{-1}y}{1+y^2}dy$

$d(e^{{\tan }^{-1}y}.x)=d\left(e^{{\tan }^{-1}y}\right)$

Integrating both sides give us

$e^{{\tan }^{-1}y}.x=e^{{\tan }^{-1}y}+C$