# What is a general solution to the differential equation y'=sqrtxe^(4y)?

Jul 19, 2016

$y = - \frac{1}{4} \ln \left(4 \left(\frac{2}{3} {x}^{\frac{3}{2}} + C\right)\right)$

#### Explanation:

This is a separable differential equation.

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{\frac{1}{2}} {e}^{4 y}$

$\frac{1}{{e}^{4 y}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{\frac{1}{2}}$

${e}^{- 4 y} \mathrm{dy} = {x}^{\frac{1}{2}} \mathrm{dx}$

$\int {e}^{- 4 y} \mathrm{dy} = \int {x}^{\frac{1}{2}} \mathrm{dx}$

$- \frac{1}{4} {e}^{- 4 y} = \frac{2}{3} {x}^{\frac{3}{2}} + C$

$- 4 y = \ln \left(- 4 \left(\frac{2}{3} {x}^{\frac{3}{2}} + C\right)\right)$

$y = - \frac{1}{4} \ln \left(4 \left(C - \frac{2}{3} {x}^{\frac{3}{2}}\right)\right)$