Question

Find the solution of y"+2y'+2y=delta(t-pi); y(0)=1, y'(0)=0.

Answer 1

delta(t-π) is the dirac delta function, correct?

Answer 2

Take the Laplace transform of both sides:
\[\left(s^2Y(s)-sy(0)-y'(0)\right)+2\left(sY(s)-y(0)\right)+2Y(s)=e^{-\pi s}\\
\left(s^2Y(s)+2sY(s)+2Y(s)\right)-s-2=e^{-\pi s}\\
Y(s)\left(s^2+2s+2\right)=e^{-\pi s}+s+2\\
Y(s)=\frac{e^{-\pi s}+s+2}{s^2+2s+2}\\
Y(s)=\frac{e^{-\pi s}}{s^2+2s+2}+\frac{s}{s^2+2s+2}+\frac{2}{s^2+2s+2}\]
It should be easy to find the inverse transforms.

Answer 3

Complete the square in each denominator:
\[s^2+2s+2=(s+1)^2+1\]
It might be easier to first write
\[Y(s)=\frac{e^{-\pi s}}{s^2+2s+2}+\frac{s+1}{s^2+2s+2}+\frac{1}{s^2+2s+2}\]
So you have
\[Y(s)=\frac{e^{-\pi s}}{(s+1)^2+1}+\frac{s+1}{(s+1)^2+1}+\frac{1}{(s+1)^2+1}\]

Answer 4

Thank you.