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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:

$$Ae^{-2x}cos(x)+Be^{-2x}sin(x)$$

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Take problem a): Plug the initial values into the general solution to get 2 equations in 2 unknowns, A and B. Solve them for A and B. Proceed the same way for the other problems.

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