I might be losing my mind this morning (I am, for sure), but I can't these two techniques to give me the same answer to a basic differential equations problem.
The problem is $y''-8y'+27y=0$ with the initial conditions $y(0)=-8$ and $y'(0)=5$. Using Laplace transforms, I get the solution $y=\frac{69}{\sqrt{11}}e^{4t}\sin(\sqrt{11}t)-8e^{4t}\cos(\sqrt{11}t)$, which is wrong.
If I use Undetermined Coefficients, I get $y=\frac{37}{\sqrt{11}}e^{4t}\sin(\sqrt{11}t)-8e^{4t}\cos(\sqrt{11}t)$, which is right.
The problem is that I can't find my mistake with the LT technique. The process is straight forward enough:
$\mathcal{L}(y''-8y'+27y)=\mathcal{L}(s^2-8s+27)-s(-8)-5+8(-8)=\mathcal{L}(s^2-8s+27)+8s-69=0$
$\mathcal{L}(y)=\frac{69-8s}{s^2-8s+16+11}=\frac{69-8s}{(s-4)^2+11}$
Then I split up the fraction, scale by $\frac{69}{\sqrt{11}}$ for the $\sin$ term and I'm done. The $\cos$ term is right, but that 69 is the bugaboo for some reason. Please point out the idiotic thing I missed!
$\mathcal{L}(y)$ is correctly computed. Your error is in computing the inverse Laplace transform: $$ \frac{69-8\,s}{(s-4)^2+11}=\frac{37}{(s-4)^2+11}-\frac{8(s-4)}{(s-4)^2+11}. $$