Question

Solve the IVP Problem.
\[x''+4x'+4x = 1 + \delta (t-2); x(0)=x'(0)=0 \]

What I have gotten the following which from the book is wrong, can someone help me?

\[X(s^2+4s+4) = 1/s + e^{-2t}\]

Answer 1

looks right to me

Answer 2

solving with laplace atm

Answer 3

Does it? The book is saying that that should turn into
\[ x(t)=1/4(1-e^{-2t}-1/2te^{-2t}+u_2(t)(t-2)e^{-2(t-2)} \]

Where as wolfram is giving me something different.
http://www.wolframalpha.com/input/?i=inverse+laplace+transform+%281%2Fs+%2B+e%5E%28-2t%29%29%2F%28s%5E2%2B4s%2B4%29

I have not done it by hand yet though.

Answer 4

actually i realized what's going on. all you did was find the laplace transform.

now you need to manipulate it, by first isolating F(s) = X to

\[X = \frac{ 1 + se^{-2s} }{ s(s^2 + 2)^2 }\]

now find inverse transform

x(t), which is the solution.

Answer 5

let me know if you'd like help finding inverse transform

Answer 6

Alright give me a moment and ill try it by hand.

Answer 7

might actually be simpler to simplify like this:

\[X = \frac{ 1 }{ s(s^2 + 2)^2 } + \frac{ e^{-2s} }{ (s^2 + 2)^2 }\]

my bad

Answer 8

Alright I got it and I found out what wolfram alpha was doing that make it look different. Thanks a lot!

Answer 9

glad i could help :)