Question

Find the general solution if (1+y^2)+(x-e^tan-1y)dy/dx=0

Answer 1

Answer:

The Differential equation is,

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

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

Integrating factor

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

Multiplying both sides by integrating factor,

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

and now integrating with respect to y, we get

$xe^{tan^{-1}y}=\int {\frac{(e^{tan^{-1}y})^2}{1+y^2}} \, dx$

On, RHS, substitute,

$tan^{-1}y=z, \\\\ d z=\frac{1}{1+y^2}dy$

$xe^{tan^{-1}y}=\int {e^{2z}} \, dz\\\\ xe^{tan^{-1}y}=\frac{e^{2z} }{2}$

$xe^{tan^{-1}y}=\frac{e^{2tan^{-1}y}}{2}+K$, where K is any constant.

Answer 2

Step-by-step explanation: