Thread: 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.

You do the last step

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.

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