# What is a general solution to the differential equation (y^2+xy^2)y'=1?

Jul 11, 2016

${y}^{3} / 3 = \ln \left(1 + x\right) + C$

#### Explanation:

separate it

$\left({y}^{2} + x {y}^{2}\right) y ' = 1$

${y}^{2} \left(1 + x\right) y ' = 1$

${y}^{2} \setminus y ' = \frac{1}{1 + x}$

$\int {y}^{2} \setminus y ' \setminus \mathrm{dx} = \int \setminus \frac{1}{1 + x} \setminus \mathrm{dx}$

$\int {y}^{2} \setminus \mathrm{dy} = \int \setminus \frac{1}{1 + x} \setminus \mathrm{dx}$

${y}^{3} / 3 = \ln \left(1 + x\right) + C$

personally i'd leave it there but you can "tidy" ie clutter it more as

$y = \sqrt[3]{3 \ln \left(1 + x\right) + C}$