# What is the particular solution of the differential equation  dy/dx = xy^(1/2)  with y(0)=0?

##### 1 Answer
Feb 4, 2018

$y = {x}^{4} / 16$

#### Explanation:

We have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = x {y}^{\frac{1}{2}}$ with $y \left(0\right) = 0$

This is a First Order Separable ODE, so we can can write:

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

Now, we separate the variables to get

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

Which consists of standard integrals, so we can integrate:

${y}^{\frac{1}{2}} / \left(\frac{1}{2}\right) = {x}^{2} / 2 + C$

Applying the initial condition, we have:

$0 = 0 + C \implies C = 0$

Thus, we have:

${y}^{\frac{1}{2}} / \left(\frac{1}{2}\right) = {x}^{2} / 2$
$\therefore {y}^{\frac{1}{2}} = {x}^{2} / 4$
$\therefore y = {x}^{4} / 16$