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here is the question:

$$ {\rm y}''\left(t\right) + 2\,{\rm y}'\left(t\right) + 5\,{\rm y}\left(t\right) = 0; \qquad\qquad {\rm y}\left(0\right) = 2\,,\quad {\rm y}'\left(0\right) = -1. $$

  • $\mathcal{L} (y''(t)) = s^2y(s) -s y(0) -y'(0)$
  • $\mathcal{L} (+2y'(t)) = 2(sy(s) -y(0))$
  • $\mathcal{L} (5y(t)) = 5y(s)$

I find that

$y(t)=\dfrac{2s+3}{s^{2}+2s+5}$

It is irreducible, so I write the transform as a function of $\varepsilon = s + 1$.

$y(t)=\dfrac{2\varepsilon+1}{\varepsilon^2+4}$

I apply fraction by parts then use laplace transform table and find the result:

$y(t)=e^{-t} (2\cos{2t}+\sin{2t})$, but the result has $\frac{1}{2}$ before $\sin{2t}$. What am I missing here? Where does the $\frac{1}{2}$ come from?

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  • $\begingroup$ Are you sure you wrote the problem correctly? $\endgroup$
    – Amzoti
    Apr 22, 2014 at 20:33
  • $\begingroup$ The result you gave would require $y(0)=2$ and $y'(0)=-1$, if I'm not mistaken. $\endgroup$
    – Matt L.
    Apr 22, 2014 at 20:38
  • $\begingroup$ Oops, @MattL. is right. That 0 is extra. I'm editing it. What about my question now? $\endgroup$
    – nope
    Apr 22, 2014 at 20:46
  • $\begingroup$ Check my answer. The result is correct. $\endgroup$
    – Matt L.
    Apr 22, 2014 at 20:59

2 Answers 2

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If the DE is written correctly, we should have arrived at:

$$y(s)=\dfrac{2s+3}{s^{2}+2s+5}$$

This yields:

$$y(t) = \dfrac{1}{2} e^{-t} (\sin(2 t)+4 \cos(2 t))$$

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  • $\begingroup$ I'm sorry, there was an error. I updated initial values. $\endgroup$
    – nope
    Apr 22, 2014 at 20:47
  • $\begingroup$ Oh, it was a stupid error I see. I took 1/s^2+4 as sint but I need 2 in dividend part in order to inverse transform it. Thank you! $\endgroup$
    – nope
    Apr 22, 2014 at 21:08
  • $\begingroup$ You are quite welcome - regards! $\endgroup$
    – Amzoti
    Apr 22, 2014 at 21:08
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With the correct initial values you get

$$y(s)=\frac{2s+3}{s^2+2s+5}=2\frac{s+1}{(s+1)^2+4}+\frac{1}{2}\frac{2}{(s+1)^2+4}$$

which corresponds to

$$y(t)=2e^{-t}\cos(2t)+\frac{1}{2}e^{-t}\sin(2t)$$

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  • $\begingroup$ Thank you! It was a stupid error. I needed 2 in dividend part of 1/s^2+4 in order to inverse transform it. $\endgroup$
    – nope
    Apr 22, 2014 at 21:08

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