Question

Find the general solution of the differential equation $(1+y^2)dx=({\tan }^{-1}y-x)dy$

Answer 1

$(1+y^2)dx=(ta{n}^{-1}y-x)dy$

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

$\frac{dx}{sy}+\frac{x}{1+y^2}=\frac{ta{n}^{-1}y}{1+y^2}$

$\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{ta{n}^{-1}y}{1+y^2}$ ________ (1)

Integrating factor = $e^{\int \frac{1}{1+y^2}dy}=e^{ta{n}^{-1}y}$

now, solution of equation (1) is

$x\times I.F=\int I.F\times \frac{ta{n}^{-1}y}{1+y^2}dy$

$x.e^{ta{n}^{-1}y}=\int e^{ta{n}^{-1}y}\frac{ta{n}^{-1}y}{1+y^2}dy$ _________ (2)

$x.e^{ta{n}^{-1}y}\Rightarrow$ let $ta{n}^{-1}y=p$

$\frac{1}{1+y^2}dy=dp$

So RHS in equation (2) is

$\int e^{ta{n}^{-1}y}.\frac{ta{n}^{-1}y}{1+y^2}dy=\int P.e^{p}dp=p{e}^p-p$

From equation (2)

$x{e}^{ta{n}^{-1}y}=ta{n}^{-1y}{e}^{ta{n}^{-1}y}-ta{n}^{-1}y+c$

$x{e}^{ta{n}^{-1}y}=ta{n}^{-1}y(e^{ta{n}^{-1}y}-1)+c$