Differential equations:
x''+2x'+2x=2*delta(t-pi), x(0)=x'(0)=0
I am not sure how to do this problem... I found X(s) to be 2*e^(-pi*s)/(s^2+2s+2), but how do you do the laplace transform of 1/(s^2+2s+2)? Is there a laplace transform for functions with complex roots?

Question

Differential equations:
x''+2x'+2x=2*delta(t-pi),     x(0)=x'(0)=0
I am not sure how to do this problem... I found X(s) to be 2*e^(-pi*s)/(s^2+2s+2), but how do you do the laplace transform of 1/(s^2+2s+2)? Is there a laplace transform for functions with complex roots?

anonymous · Answer

1/(s^2+2s+2) is on the s domain, so its already a transformed function. ¿did you mean the inverse transform for this function?

anonymous · Answer

sorry I clicked on this kid's question and it was this
basically when I did it, I came out with the wrong answer and was looking for some help

anonymous · Answer

would you know how to do this?

anonymous · Answer

Find the inverse transform for 1/(s^2+2s+2)?

anonymous · Answer

well eventually you would take the inverse laplace transform of : (2e^(-pi*s))/(s^2+2s+2)

anonymous · Answer

so I got that far, and how did the inverse turned out to be wrong

anonymous · Answer

First, ¿are you sure about the laplace transform of delta function?