# How to you find the general solution of 4yy'-3e^x=0?

Nov 9, 2016

$y = \sqrt{\frac{3}{2} {e}^{x} + C}$

#### Explanation:

Changing the notation from Lagrange's notation to Leibniz's notation we have:

$4 y \frac{\mathrm{dy}}{\mathrm{dx}} - 3 {e}^{x} = 0$
$\therefore 4 y \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {e}^{x}$

This is a First Order sepaable DE, and "seperating the variables" give is

$\therefore \int 4 y \mathrm{dy} = \int 3 {e}^{x} \mathrm{dx}$

Integrating gives us:

$2 {y}^{2} = 3 {e}^{x} + {C}_{1}$
$\therefore {y}^{2} = \frac{3}{2} {e}^{x} + C$ (by writing ${C}_{1} = \frac{1}{2} C$)
$\therefore y = \sqrt{\frac{3}{2} {e}^{x} + C}$