Find the solution of each of the following initial value problems:
$a) y''-5y'+6y=0 \space \space \space \space \space \space y(1)=e^2 \space \space y'(1)=3e^2$
$b) y''-6y'+9y=0 \space \space \space \space \space \space y(0)=0 \space \space y'(0)=5$
$c) y''+4y'+5y=0 \space \space \space \space \space \space y(0)=1 \space \space y'(0)=0$
I can easily find the general solution for each $a)$ , $b)$ and $c)$, but I'm not entirely sure what to do, or how I use the initial value.
What I have so far, for part $a)$:
$ y''-5y'+6y=0$ $r^2-5r+6=0$ $(r-3)(r-2)=0$ So, the general solution is: $y=Ae^{2x}+Be^{3x}$
For part $b)$:
$ y''-6y'+9y=0$ $r^2-6r+9=0$ $(r-3)^2=0$ So, the general solution is: $y=Ae^{3x}+Bxe^{3x}$
For part $c)$:
$r^2+4r+5=0$ $\frac{-4 \pm \sqrt{16-4*1*5}}{2}$ $-2 \pm i$ So, the general solution is:
$Ae^{-2x}cos(x)+Be^{-2x}sin(x)$