Question

Show that the solution of $(1+y^2)dx=\{ta{n}^{-1}y-x)dy$ is $x=c{e}^{-\arctan y}+\arctan y-1$

Answer 1

$\frac{dx}{dy}=\frac{ta{n}^{-1}y}{1+y^2}-\frac{x}{1+y^2}$
$\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{ta{n}^{-1}y}{1+y^2}$
The integrating factor is
$I.F=e^{\int \frac{1}{1+y^2}.dy}$
$=e^{ta{n}^{-1}y}$
Multiplying the entire equation by I.F
$e^{ta{n}^{-1}y}\frac{dx}{dy}+e^{ta{n}^{-1}y}\frac{x}{1+y^2}=\frac{ta{n}^{-1}y}{1+y^2}.e^{ta{n}^{-1}y}$
$\int d\left(e^{ta{n}^{-1}y}x\right)=\int \frac{ta{n}^{-1}y}{1+y^2}.e^{ta{n}^{-1}y}dy$
$e^{ta{n}^{-1}y}x=\int \frac{ta{n}^{-1}y}{1+y^2}.e^{ta{n}^{-1}y}$
Consider
$I=\int \frac{ta{n}^{-1}y}{1+y^2}.e^{ta{n}^{-1}y}$
Let
$ta{n}^{-1}y=t$
Then
$\frac{1}{1+y^2}dy=dt$
Hence
$I=\int t.e^t$
$=e^{t}t-e^t$
$=e^t(t-1)$
$=e^{ta{n}^{-1}y}(ta{n}^{-1}y-1)$
Hence
$e^{ta{n}^{-1}y}x=\int \frac{ta{n}^{-1}y}{1+y^2}.e^{ta{n}^{-1}y}$
$e^{ta{n}^{-1}y}x=e^{ta{n}^{-1}y}(ta{n}^{-1}y-1)+C$
$x=C{e}^{-ta{n}^{-1}y}+ta{n}^{-1}y-1$