1. ## Laplace

Is this going in the right direction with this problem?

$y''-4y'+13y=e^{3t}; y(0)=0; y'(0)=1$

$s^2Y(s)-1-4sY(s)+13Y(s)=\frac{1}{s-3}$

$Y(s)[s^2-4s+13]=\frac{1}{s-3}+1$

$Y(s)=\frac{1}{(s-3)[(s-2)^2+11]}+\frac{1}{(s-2)^2+11}$

Before I continue down a futile path, should this work or is there a more efficient way?

I don't need an answer just a yes this way is fine or no try this, thanks.

2. Eh, I didn't check the Laplace transform so that will probably work, but here's another way: solve the general equation by looking for a solution to the homogenous and adding a particular solution. Since the funciton on the right is exponential, the simplest guess for a particular solution could be $\alpha e^{\beta t}$. If you plug this and equate coefficients, you should get $\beta = 3$ (coulda guessed that actually) and $\alpha(\beta^2-4\beta+13) = 1$ or $\alpha=1/10$, so a particular solution is $\frac1{10}e^{3t}$. The homogenous solution is $e^{2x}(c_1\sin(3x) + c_2 \cos (3x))$ so add these together and you should have a general solution.

The homogenous solution is $e^{2x}(c_1\sin(3x) + c_2 \cos (3x))$ so add these together and you should have a general solution.
I don't understand how this part came about.

4. Oh, I'm sorry. Its the solution to the homogenous equation $y''-4y'+13y=0$ (which you can always find by taking a linear combination of solutions of the form $e^{\lambda t}$). The solution of an inhomogenous equation is the sum of any particular solution to that equation and the solution to the corosponding homogenous equation. Do you known how to solve the homogenous equation? (It involves the characteristic polynomial...)

5. So I continued on the Laplace method and it ended up similar but different.
From this point, I did this:

$Y(s)=\frac{1}{(s-3)[(s-2)^2+11]}+\frac{1}{(s-2)^2+11}$
After partial fractions
$Y(t)=\frac{1}{10}L^{-1}\bigg[\frac{1}{s-3}\bigg]+\frac{1}{10}L^{-1}\bigg[\frac{s}{(s-2)^2+11}\bigg]-\frac{1}{10}L^{-1}\bigg[\frac{1}{(s-2)^2+11}\bigg]$

$Y(t)=\frac{1}{10}e^{3t}+\frac{1}{5}e^{2t}cos(\sqrt {11}t)-\frac{1}{10}e^{2t}sin(\sqrt{11}t)$