Solving ODE using Laplace Transforms Given function:
$$y''-2y'+5y=e^t$$
where $y(\pi)=2$ and $y'(\pi)=3$
I've got my inverse Laplace Transform as:
$\frac{1}{4}e^tcos(2t)+\frac{1}{4}e^t+y(0)e^tcos(2t)-1*e^tsin(2t)+y'(0)e^t\frac{1}{2}sin(2t)$
How should I use the initial conditions given from here?
 A: $$f(t)=\frac{1}{4}e^t\cos(2t)+\frac{1}{4}e^t+y(0)e^t\cos(2t)-e^t\sin(2t)+y'(0)e^t\frac{1}{2}\sin(2t)$$
Rewrite as: 
$$f(t)=\frac{1}{4}e^t\cos(2t)+\frac{1}{4}e^t+c_1e^t\cos(2t)-e^t\sin(2t)+c_2e^t\frac{1}{2}\sin(2t)$$
 Then apply first initial condition: 
$$f(\pi)=\frac{1}{2}e^{\pi}+c_1e^{\pi}=2$$
$$ \implies c_1=?$$
Differentiate $f(t)$ and apply second initial condition.

You can also rewrite $f(t)$ more simply as:
$$f(t)=\frac{1}{4}e^t+k_1e^t\cos(2t)+k_2e^t\sin(2t)$$
And:
$$f'(t)=\frac{1}{4}e^t+(k_1+2k_2)e^t\cos(2t)+(k_2-2k_1)e^t\sin(2t)$$

Edit2:
You find:
$$f(t)=\frac{1}{4}e^t\cos(2t)+\frac{1}{4}e^t+\frac{2-\frac{1}{2}e^t}{e^t}*e^t\cos(2t)-1*e^t\sin(2t)+\frac{1+2e^t}{e^t}*e^t*\frac{1}{2}\sin(2t)$$
It should be :
$$f(t)=\frac{1}{4}e^t\cos(2t)+\frac{1}{4}e^t+\frac{2-\frac{1}{2}e^{\pi}}{e^{\pi}}*e^t\cos(2t)-1*e^t\sin(2t)+\frac{1+2e^{\pi}}{e^{\pi}}*e^t*\frac{1}{2}\sin(2t)$$
You see the difference ? The coefficients $c_1,c_2$ have $\pi$ and the functions keep their $t$ variable. Now if you check $f({\pi})=2$ it looks correct to me. Simplify:
$$\boxed {f(t)=\frac{1}{4}e^t+\left (\frac 2 {e^{\pi}}-\dfrac 14 \right )e^t\cos(2t)+\frac{e^t}{2e^{\pi}}\sin(2t)}$$
