Question

Solve the differential equation.

x y^2 y' = x + 1

Answer 1

$x y^2 y' = x + 1 \ \Rightarrow\ y^2 \frac{dy}{dx} = \frac{x+1}{x}\ \Rightarrow\ y^2 dy = \left(1 + \frac{1}{x}\right) dx\ \Rightarrow \\ \\ \int y^2 dy = \int \left(1 + \frac{1}{x}\right) dx\ \Rightarrow \frac{1}{3}y^3 = x + \ln|x| + C\ \Rightarrow \\ y^3 = 3x + 3\ln|x| + 3C\ \Rightarrow\ \\ \\y = \sqrt[3]{3x + 3\ln|x| + K}, \text{ where K = 3C }$