# How do you find the particular solution to yy'-e^x=0 that satisfies y(0)=4?

Jan 13, 2017

$y \left(x\right) = \sqrt{2 {e}^{x} + 15}$

#### Explanation:

The differential equation:

$y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{x} = 0$

is separable, so we can solve it by separating he variables and integrating:

$y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x}$

$y \mathrm{dy} = {e}^{x} \mathrm{dx}$

$\int y \mathrm{dy} = \int {e}^{x} \mathrm{dx}$

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

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

for $x = 0$ we have:

$y {\left(0\right)}^{2} = {e}^{0} + C$

$16 = 1 + C \implies C = 15$

So the particular solution we are seeking is:

$y \left(x\right) = \sqrt{2 {e}^{x} + 15}$