The solution of (1+y^2)+(x-e^^-1y)dydx=0 is:

Question

The solution of $(1+y^2)+\left(x-e^{	an^{-1}y}ight)\dfrac{dy}{dx}=0$ is:

Accepted Answer

$\dfrac{dx}{dy}+\dfrac{x}{1+y^2}=\dfrac{e^{\tan^{-1}y}}{1+y^2}$; I.F.$=e^{\tan^{-1}y}$.

$$xe^{\tan^{-1}y} = \frac{e^{2\tan^{-1}y}}{2}+C \implies 2xe^{\tan^{-1}y}=e^{2\tan^{-1}y}+k.$$

Solutions · 1