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.
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:
Originally Posted by Hollysti
Lf''(s) = s^2 Lf(s) - s f'(0) - f(0)
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)