Question

Solve the differential equation.

(y^2 + xy^2)y' = 1

Answer 1

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