# What is a solution to the differential equation dy/dx=x^2(1+y) with y=3 when x=0?

Aug 22, 2016

$y = 4 {e}^{{x}^{3} / 3} - 1$

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2} \left(1 + y\right)$

This is separable
$\frac{1}{1 + y} \frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2}$

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

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

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

$1 + y = {e}^{{x}^{3} / 3 + C} = {e}^{{x}^{3} / 3} {e}^{C} = C {e}^{{x}^{3} / 3}$

$y = C {e}^{{x}^{3} / 3} - 1$

Applying the IV:

$3 = C {e}^{0} - 1 = C - 1 \implies C = 4$

$y = 4 {e}^{{x}^{3} / 3} - 1$