0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Paul Mar 9 '18 at 21:02
  • $\begingroup$ @Paul I just figured it out, dropped the 2 somewhere $\endgroup$ – TobyTobyo Mar 9 '18 at 21:03
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.