Question

Solve the differential equation. (y^2 + xy^2)y' = 1

Answer 1

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