I think I have this one all figured out, I just need someone to tell me if I got it or not...
so I have:
$$ \begin{cases} 9y''+12y'+4y = e^{-2t}\\ \\ y(0)=y'(0)=1 \end{cases} $$
so using the usual method, where $L[f'']=s^{2}F(s)-sf(0)-f'(0)$
and $L[f']=sF(s)-sf(0)$
I managed to get to
$(9s^{2}+12s+4)F(s)-30=\frac{1}{s+2}$
and thus
$F(s) = \frac{31}{(s+2)(3s+2)^{2}}$
using partial fractions, I eventually get down to
$\frac{31}{(s+2)(3s+2)^{2}} = \frac{31/16}{s+2}-\frac{93/48}{3s+2}+\frac{93/4}{(3s+2)^{2}}$
and therefore:
$F(s)=\frac{31}{16}e^{-2t}-\frac{93}{48}e^{-2/3t}+\frac{93}{4}te^{-2t}$
Is this alright?? If no, where did I go wrong?? I usually have suspicions that I got something wrong when the constants are very large, like I have right here
thanks very much, in advance
-GD
This line is not correct $L[f']=sF(s)-sf(0)$. It should be $L[f']=sF(s)-f(0)$
Then here you cant have 31 at the numerator with
$$(9s^{2}+12s+4)F(s)-30=\frac{1}{s+2}$$
Should be
$$F(s)=\frac{30s+61}{(s+2)(3s+2)^2}=\frac{1}{(s+2)(3s+2)^2}+\frac{30}{(3s+2)^2}$$
Another mistake you did is that it is not 31 but rather $9s+21$
So you should have this instead:
$$(9s^{2}+12s+4)F(s) \color{red}{-9s-21}=\frac{1}{s+2}$$ $$F(s)=\frac{1}{(s+2)(3s+2)^2}+\frac{3(3s+7)}{(3s+2)^2}$$ $$.......$$ Edit
$$\frac{3(3s+7)}{(3s+2)^2}=\frac{3(3s+2)}{(3s+2)^2}+\frac{15}{(3s+2)^2}=\frac{1}{(s+\frac 2 3)}+\frac 5 3\frac{1}{(s+\frac 2 3)^2}$$
$$\implies e^{-\frac {2t} 3}+\frac 5 3 te^{-\frac {2t} 3}$$
For the first fraction try to find A B C such that: $$\frac{1}{(s+2)(3s+2)^2}=\frac{A}{(s+2)}+\frac{B}{(3s+2)}+\frac{C}{(3s+2)^2}$$