Solve Laplace Transform DE: y''+4y'+4y=delta_3(t)

[daxfeliz]

Homework Statement

Solve the differential equation:

y'' + 4y' + 4y = delta_3 (t)

y(0) = 0 and y'(0) = 1

Homework Equations

1) lapace [delta_a] = e^-as
2) laplace inverse [ 1 / (s+a)^n] = [(t^n-1) / (n-1)! ] *e^-at

The Attempt at a Solution

so I began by taking the laplace transform of both sides to get :

s^2(laplace[y]) -s*0 -1 + 4s(laplace[y]) + 4*laplace[y] = laplace[delta_3]

when further simplied, and calculated with equation 1,

(s^2 +4s + 4)*laplace[y] = 1+ e^-3s

and thus

y = laplace inverse[ (1+ e^-3s) / (s^2 +4s +4) ]

which can be separated by linearity to get

y = laplace inverse[1/(s^2 + 4s +4)] + laplace inverse [ (e^-3s) /( s^2 +4s +4) ]

I can simplify the first half by equation 2 (which honestly I'm not sure if it's true because I found it online and it's not in my textbook in any example nor have I ever seen it) so that,

y = t*e^-2t laplace inverse [ (e^-3s) /((s+2)^2) ]

Any help would be GREATLY appreciated :)

 

[vela]

Your book probably has

\mathcal{L}[t^n] = \frac{n!}{s^{n+1}}

\mathcal{L}[e^{-at}f(t)] = F(s+a)

Combine them and you have your equation 2.

Your book should also have an entry relating e^-asF(s) to the f(t). Use that to deal with the second term.