Answer
y = $C_{1}e^{\frac{-2+\sqrt {13}}{3}x}$ + $C_{2}e^{\frac{-2-\sqrt {13}}{3}x}$
Work Step by Step
Question: 3y"+4y'-3y=0
We know:
y = $e^{rx}$
y' = $re^{rx}$
y" = $r^{2}e^{rx}$
This results in $(3r^{2}+4r-3)e^{rx}$=0
The corresponding characteristic equation is:
$3r^{2}+4r-3=0$
Solving the characteristic equation with the abc-formula and complex numbers gives:
$\frac{-b\frac{+}{-}\sqrt {b^{2}-4*a*c}}{2*a}$ $\frac{-4\frac{+}{-}\sqrt {(4)^{2}-4*3*-3}}{2*3}$
$r=\frac{-4\frac{+}{-}\sqrt {52}}{6}$
This can also be written as:
$r=\frac{-2\frac{+}{-}\sqrt {13}}{3}$
Adding the constants gives the following general solution:
y = $C_{1}e^{\frac{-2+\sqrt {13}}{3}x}$ + $C_{2}e^{\frac{-2-\sqrt {13}}{3}x}$
