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

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Answer 1 · 2016-11-09T01:38:43

$y = sqrt(3/2e^x + C)$

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

$4ydy/dx-3e^x = 0$
$:. 4ydy/dx = 3e^x$

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

$:. int 4y dy = int3e^x dx$

Integrating gives us:

$2y^2 = 3e^x + C_1$
$:. y^2 = 3/2e^x + C$ (by writing $C_1=1/2C$ )
$:. y = sqrt(3/2e^x + C)$

How to you find the general solution of 4yy'-3e^x=04yy'-3e^x=0?