# Laplace Transform with the Dirac Delta Function

We just went over dirac delta functions today, and the examples seemed quite a bit easier than some of the assigned homeworks. I went back to the textbook, and the problems we did in class were simply their examples, so it didn't help.

My problem is: $$y''+2y'+5y=2\delta(t-\pi), y(0)=2, y'(0)=4$$

I start solving: $$(s^2Y-2s-4)+2(sY-2)+5(Y)=2e^{-\pi s}$$ $$Y(s^2+2s+5)-2s-8=2e^{-\pi s}$$ $$Y(s^2+2s+5)=2e^{-\pi s}+2s+8$$ $$Y=\frac{2e^{-\pi s}+2s+8}{s^2+2s+5}$$

Then I get stuck at the next step(s): $$Y=\frac{2e^{-\pi s}+2s+8}{(s+1)^2+2^2}$$ $$Y=e^{-\pi s}\frac{2}{(s+1)^2+2^2}+2\frac{s}{(s+1)^2+2^2}+4\frac{2}{(s+1)^2+2^2}$$

What do I do now? I'm trying to figure out how to turn this into something of like $u(t-a)sin(t-a)$, but I'm not sure of where to go, since my denominators of all of them contain a $(s-c)^2$ and $a^2$ term, and I'm not sure what to do with that $(s-c)$

EDIT: I rearranged the above equation into $$Y=e^{-\pi s}\frac{2}{(s+1)^2+2^2}+2\frac{(s+1)}{(s+1)^2+2^2}+3\frac{2}{(s+1)^2+2^2}$$

This got me to $$e^{-(t-\pi)}\sin{(t-\pi)}u(t-\pi)+2e^{-t}\cos(t)+3e^{-t}\sin(t)$$

But the program says this is incorrect. Any ideas?

• Look at the Laplace transforms of $e^{-at}sin(bt)$ and $e^{-at}cos(bt)$. Your second and third terms together is a linear combination of these. The first term is a time shift of the transform of a $e^{-at}sin(bt)$ term. – Paul Mar 9 '18 at 21:02
• @Paul I just figured it out, dropped the 2 somewhere – TobyTobyo Mar 9 '18 at 21:03

I rearranged the above equation into $$Y=e^{-\pi s}\frac{2}{(s+1)^2+2^2}+2\frac{(s+1)}{(s+1)^2+2^2}+3\frac{2}{(s+1)^2+2^2}$$
This simplifies to $$e^{-(t-\pi)}\sin(2(t-\pi))u(t-\pi)+2e^{-t}\cos(2t)+3e^{-t}\sin(2t)$$
Not $$e^{-(t-\pi)}\sin{(t-\pi)}u(t-\pi)+2e^{-t}\cos(t)+3e^{-t}\sin(t)$$