Question

Solve: (1 + y²)dx = (tan^–1y – x)dy.

Answer 1

We have,

So, the solution is given by

Putting the value of I in (1), we get

\(xe^{\tan^{-1}y} = e^{\tan^{-1}y} (\tan^{-1}y - 1)+C\)

Answer 2

The given differential equation is

(1 + y²)dx = (tan^–1y – x)dy

= ∫te^tdt + c, t = tan^–1y

= (t – 1)e^t + c  

= (tan^{– 1}y – 1)e^{tan^–1 y} + c 

which is the required solution.