# What is a solution to the differential equation y'-xy=0?

Jul 10, 2016

$y = C {e}^{{x}^{2}}$

#### Explanation:

this is separable

$y ' = x y$

$\frac{1}{y} \setminus y ' = x$

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

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

$\ln y = {x}^{2} + C$

$y = {e}^{{x}^{2} + C} = C {e}^{{x}^{2}}$

NB using C as a generic pointer for constant term.