Question

Find the general solution of the differential equation $\left(1+y^2\right)+\left(x-e^{{\text{tan}}^{-1}y}\right)\frac{dy}{dx}=0$

Answer 1

Given,

$(1+y^2)(x-e^{{\tan }^{-1}y})\frac{dy}{dx}=0$

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

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

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

$x\times I.F=\int Q\times I.Fdy$

$I.F=e^{\int \frac{1}{1+y^2}dy}=e^{{\tan }^{-1}y}$

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

substitute $\tan -1y=t$

$x{e}^t=\int e^t\times e^{t}dt$

$x{e}^t=\int e^{2t}dt$

$x{e}^t=\frac{e^{2t}}{2}+c$

$x=\frac{e^t}{2}+c{e}^{-t}$

$x=\frac{e^{\tan -1y}}{2}+c{e}^{-\tan -1y}$