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

How to you find the general solution of $4yy\prime -3{e}^{x}=0$?
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Leslie Braun
Changing the notation from Lagrange's notation to Leibniz's notation we have:

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

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

$\therefore \int 4ydy=\int 3{e}^{x}dx$

Integrating gives us:

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