Question

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

x + y = tan^–1y   :   y² y′ + y² + 1 = 0

Answer 1

x + y = tan^-1y

1 = y’ = `1/(1 + y^2)` (y’)

⇒ (1 + y') (1 + y²) = y’

⇒ 1 + y² + y' + y²y' = y'

⇒  1 + y² + y²y' = 0

Hence, the given function x + y = tan^-1y is a solution to the given differential equation.