Need help solving Laplace Transform Question For context: I am just starting to learn LaPlace Transform and am stumped with this question.
Find the PS of $x''+x'-12x=sin(3t) + e^{-4t}+e^{3t}$
Where $x(0)=0, x'(0)=0$ using LaPlace Transform
My Approach:
Let X(s)=$\mathscr{L}[x(t)]$
Apply the LaPlace Transform to both sides of the equation
$\mathscr{L}[x''] + \mathscr{L}[x'] - 12\mathscr{L}[x]$ = $\mathscr{L}[sin(3t)] +\mathscr{L}[e^{-4t}] +\mathscr{L}[e^{3t}]$
to get $s^2X(s)-sx(0)-x'(0)+sX(s)-x(0)-12X(s) = \cfrac{3}{s^2+9}+\cfrac{1}{s+4}+\cfrac{1}{s-3}$
Plugging in the IC's $x(0) = 0$ and $x'(0)=0$ and then factoring $X(s)$ we get
$X(s)[s^2+s-12] = \cfrac{3}{s^2+9}+\cfrac{1}{s+4}+\cfrac{1}{s-3}$
You should then make the RHS have a common denominator so you get
$X(s)[s^2+s-12] = \cfrac{3(s+4)(s-3)}{(s^2+9)(s+4)(s-3)}+\cfrac{(s^2+9)(s-3)}{(s+4)(s^2+9)(s-3)}+\cfrac{(s+4)(s^2+9)}{(s+4)(s-3)(s^2+9)} = \cfrac{2s^3+4s^2+21s-27}{(s^2+9)(s-3)(s+4)}$
Dividing both sides by $[s^2+s-12]$ AKA $(s+4)(s-3)$ to get $X(s)$ by itself
We get
$X(s) = \cfrac{2s^3+4s^2+21s-27}{(s^2+9)(s-3)^2(s+4)^2}$
Now I Believe I'm supposed to use partial fraction decomposition here to make taking the Inverse LaPlace Transform easier so...
$\cfrac{2s^3+4s^2+21s-27}{(s^2+9)(s-3)^2(s+4)^2} = \cfrac{As+B}{s^2+9} + \cfrac{C}{s-3} + \cfrac{D}{(s-3)^2} + \cfrac{E}{(s+4)} + \cfrac{F}{(s+4)^2}$
Then multiplying both sides by the LHS denominator and simplifying we get
$2s^3+4s^2+21s-27 = (As+B)(s-3)^2(s-4)^2+C(s^2+9)(s-3)(s-4)^2+D(s^2+9)(s+4)^2+E(s+4)(s-3)^2(s^2+9)+F(s^2+9)(s-3)^2$
Here is where I get stuck. How do I solve for A,B,C,D,E,F? This seems ridiculously complex and seems like so much algebra that it makes me think that partial fraction decomposition isn't the right way. So what I've come to ask is partial fraction decomposition to right way to tackle this question? Is there an more efficient way besides partial fraction decomposition? If there's not and partial fraction decomposition is the correct way, How can I solve for A,B,C,D,E,F? Can someone help me determine the algebra, I'm not used to partial fractions being this long unless I've made an error somewhere...
Thank You For Any Help
 A: It's more simple to decompose fractions from this step:
$$X(s)[s^2+s-12] = \cfrac{3}{s^2+9}+\cfrac{1}{s+4}+\cfrac{1}{s-3}$$
$$X(s)(s+4)(s-3) = \cfrac{3}{s^2+9}+\cfrac{1}{s+4}+\cfrac{1}{s-3}$$
$$X(s)=\dfrac 1{(s+4)(s-3)}\left ( \cfrac{3}{s^2+9}+\cfrac{1}{s+4}+\cfrac{1}{s-3}\right)$$
And you have:
$$\dfrac 1{(s+4)(s-3)}=\dfrac 1 {7}\left (\dfrac 1{(s-3)}-\dfrac 1{(s+4)}\right)$$
Now what we have is this;
$$X(s)=\color {red}{\dfrac 37\left (\dfrac 1{(s-3)}-\dfrac 1{(s+4)}\right)\left ( \cfrac{1}{s^2+9}\right)}+\color {blue}{\dfrac 17\left (\dfrac 1{(s-3)^2}-\dfrac 1{(s+4)^2}\right)}$$
For the term in red you still need to decompose a little bit but for the term in blue you can already apply inverse Laplace Transform.
Note that:
$$\dfrac 1{(s-3)^2}=-\dfrac {d}{ds}\dfrac 1{(s-3)}$$
The inverse Laplace transform is therefore:
$$te^{3t}$$
And:
$$\dfrac 1{(s+4)^2}=-\dfrac {d}{ds}\dfrac 1{(s+4)}$$
The inverse Laplace transform is:
$$te^{-4t}$$

How to decompose this term ?
$$\dfrac 1{(s-3)}\cfrac{1}{(s^2+9)}$$
$$\dfrac {s+3}{(s^2-9)}\cfrac{1}{(s^2+9)}$$
$$(s+3)\dfrac {1}{(s^2-9)}\cfrac{1}{(s^2+9)}$$
$$(s+3)\left (\dfrac {A}{(s^2-9)}+\cfrac{B}{(s^2+9)}\right)$$
That's easy now ....
$$\dfrac {(s+3)}{18}\left (\dfrac {1}{(s^2-9)}-\cfrac{1}{(s^2+9)}\right)$$
$$\dfrac {1}{18}\left (\dfrac {1}{(s-3)}-\cfrac{s+3}{(s^2+9)}\right)$$
