$$ 2y''+y'-y=e^{3t}; \text{ with } y(0)=2,\ y'(0)=0 $$
I got to this point:
$$ L(y)=\frac{1}{(s-3)^2}\cdot\frac{1}{(2s-1)(s+1)} $$
but now I'm not sure what to do with these polynomials. I know that the inverse Laplace of $\dfrac{1}{(s-3)^2}$ would be $\dfrac{te^{3t}}{1!}$, but because it is being multiplied by another polynomial I don't think it can be separated. Could someone guide me on the right track here?
On
You also needn't use the Laplace transform here. Your Ansatz could be $Ae^{3x}$ and you would have
$18Ae^{3x}+3Ae^{3x}-Ae^{3x}=e^{3x}$
$18A+3A-A=1$
which gives you the particular solution. The solution to the homogenous equation, as you know, will be a linear combination of $e^{r_1x}$ and $e^{r_2x}$. Then you can plug in the initial values. To me, this seems simpler than the transform.
In general, when you do Laplace transforms, you need to do partial fraction decomposition to separate all the terms: $$L(y)=\frac{1}{(s-3)^2}\cdot\frac{1}{2s-1}\cdot\frac{1}{s+1}$$ $$\frac{1}{(s-3)^2}\cdot\frac{1}{2s-1}\cdot\frac{1}{s+1}=\frac{A}{s-3}+\frac{B}{(s-3)^2}+\frac{C}{2s-1}+\frac{D}{s+1}$$ Clear the denominator: $$1=(s-3)(2s-1)(s+1)A+(2s-1)(s+1)B+(s-3)^2(s+1)C+(s-3)^2(2s-1)D$$ Solve for the coefficients by selecting values of $s$: $$s=3: ~ 1=0A+20B+0C+0D \rightarrow B=\frac{1}{20}$$ $$s=\frac{1}{2}: ~ 1=0A+0B+\frac{25}{4}\frac{3}{2}C+0D \rightarrow C=\frac{8}{75}$$ $$s=-1: ~ 1=0A+0B+0C-48D \rightarrow D=\frac{-1}{48}$$ Since we have a repeated root, pick a different value of $s$: $$s=0: ~ 1=3A-B+9C-9D \rightarrow A=\frac{-13}{400}$$ So your problem is now: $$L(y)=\frac{-13}{400}\frac{1}{s-3}+\frac{1}{20}\frac{1}{(s-3)^2}+\frac{8}{75}\frac{1}{2s-1}+\frac{-1}{48}\frac{1}{s+1}$$ This can be easily inverse transformed, thus providing your solution!