Answer
y = $C_{1}$ + $C_{2}e^{\frac{4}{3}x}$
Work Step by Step
Question: 3y"=4y'
We write: $3y"- 4y'=0$
We know:
y = $e^{rx}$
y' = $re^{rx}$
y" = $r^{2}e^{rx}$
This results in $(3r^{2}-4r)e^{rx}$=0
The corresponding characteristic equation is:
$3r^{2}-4r=0$
Solving the equation gives:
$r(3r-4)=0$
$r=0$ or $r=\frac{4}{3}$
Adding the constants gives the following general solution:
y = $C_{1}e^{0}$ + $C_{2}e^{\frac{4}{3}x}$
$e^{0} = 1$ so
y = $C_{1}$ + $C_{2}e^{\frac{4}{3}x}$
