Laplace transforms

• May 13th 2007, 09:39 AM
Hollysti
Laplace transforms
Solve the initial value problem by laplace transforms:
y" - 4y' - 5y = 2 + e^(-t); y(0) = y'(0) = 0.

I have the first couple steps done, but I keep getting stuck, so if someone would help me with this problem I would appreciate it. It is on the review for my final and I think there will be one like it on the test, so if you could go into detailed explaination and show every step, maybe I will be able to do the problem on the final. Thank you.
• May 13th 2007, 12:16 PM
ThePerfectHacker
You do the last step
• May 14th 2007, 11:15 PM
Hollysti
Would you be able to tell me how you got the first step of the problem? I would have been able to work the problem, I think, but I didn't know where to start. Also, are you sure that the partial derivatives towards the bottom are correct, because I am getting different answers for them. Thank you.
• May 15th 2007, 12:08 AM
CaptainBlack
Quote:

Originally Posted by Hollysti
Would you be able to tell me how you got the first step of the problem? I would have been able to work the problem, I think, but I didn't know where to start. Also, are you sure that the partial derivatives towards the bottom are correct, because I am getting different answers for them. Thank you.

You are supposed to know that:

Lf''(s) = s^2 Lf(s) - s f'(0) - f(0)

and that:

Lf'(s) = s Lf(s) - f(0)

Lf(s) = 1/s when f(t) = 1 for all t

Lf(s) = 1/(s-a) when f(t)=e^(at)

Then you assemble these into:

s^2 Lf(s) - s f'(0) - f(0) -4[s Lf(s) - f(0)] -5 Lf(s) = 2/s + 1/(s+1)

Putting Lf = F, and using the initial conditions this becomes:

s^2 F(s) - 4 s F(s) - 5 F(s) = 2/s + 1/(s+1)

RonL