Question

Solve the differential equation (1 + y²) tan^–1xdx + 2y(1 + x²)dy = 0.

Answer 1

Given differential equation is (1 + y²) tan^–1xdx + 2y(1 + x²)dy = 0

⇒ `2y(1 + x^2)"d"y = -(1 + y^2) . tan^-1x . "d"x`

⇒ `(2y)/(1 + y^2) "d"y = (tan^-1x)/(1 + x^2) . "d"x`

Integrating both sides, we get

`int (2y)/(1 + y^2) "d"y = -int (tan^-1x)/(1 + x^2) . "d"x`

⇒ `log|1 + y^2| = - 1/2(tan^-1x)^2 + "c"`

⇒ `1/2 (tan^-1x)^2 + log|1 + y^2|` = c

Which is the required solution.