OK so I use two properties of Laplace Transforms;
And I replace with
using my initial conditions;
Carry out the division;
So I guess this is the part I'm stuck on. It seems like the rightmost term is an expression of something like but I don't see any obvious way to handle the middle term. Have I made a mistake? Or do I not just see the obvious answer?
If you look at the transform table you'll see an entry for a pair where G(s) = 1/(s-a)n+1 which is what you've got here with a=-3, n=1
that transforms back to some g(t) that you'll see in the table.
Now, your F(s) is actually sG(s). Multiplying by s in the s domain corresponds with differentiation in the time domain.
so sG(s) <==> d/dt g(t)
so look up g(t) using the G(s) above and then take it's time derivative.
Are there any examples of differential equations problems where "Laplace transform method" is worth doing?
Here, the first thing I would do is see that y''+ 6y'+ 9y= 0 has characteristic equation and so has r= -3 as a double root. That tells us that the general solution can be written as . From that, .
All that remains is to find and so that and .
I have always disliked the Laplace transform method!
Your explanation makes this clearer too though. I just need to figure out what rule will let me correctly reference the fact that my last term is