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The problem is: $y" + 9y = e^t$, with the initial conditions $y(0) = 0, y'(0) = 0$.

I'm stuck at the inverse Laplace transform part. Do I have to use partial fraction expansion or can I just split the equation $\frac{1}{(s-1)(s^2+9)}$?

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  • $\begingroup$ Please consider formatting mathematical formulae in LaTeX style. Also, please do not use uncommon English acronyms that might be unknown to the rest of the world (such as PFE). $\endgroup$
    – Alex M.
    Mar 25, 2015 at 20:22
  • $\begingroup$ What do you mean by splitting the equation? $\endgroup$
    – GFauxPas
    Mar 25, 2015 at 22:05
  • $\begingroup$ you need to break it up as $\frac{10}{(s-1)(s^2+9)} = \frac 1{s-1} - \frac{s}{s^2 + 9} - \frac 1{s^2 + 9}$ and look up the inverse transforms. $\endgroup$
    – abel
    Mar 30, 2015 at 1:56

1 Answer 1

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You do not have to use partial fractions. I have no idea though about your meaning of split the equation. I think most people would infer splitting the equation as partial fractions decomposition. If you don't want to use partial fractions, we can use the inverse Laplace transform definition. That is, $$ \mathcal{L}^{-1}\{F(s)\}(t) = \frac{1}{2i\pi}\int_{\gamma - i\infty}^{\gamma + i\infty}F(s)e^{st}ds = \sum_j\operatorname{Res}\{F(s);s_j\}\tag{1} $$ where $F(s) = \frac{1}{(s-1)(s^2 + 9)}$. The poles of $F(s)$ occur when $s = 1, \pm 3i$; thus, we only have simple poles. Using equation $(1)$, we have \begin{align} \frac{1}{2i\pi}\int_{\gamma - i\infty}^{\gamma + i\infty}\frac{e^{st}}{(s-1)(s^2+9)}ds & = \lim_{s\to 1}(s-1)\frac{e^{st}}{(s-1)(s^2+9)}+\lim_{s\to 3i}(s-3i)\frac{e^{st}}{(s-1)(s^2+9)}\\ &+\lim_{s\to -3i}(s+3i)\frac{e^{st}}{(s-1)(s^2+9)}\\ &= \lim_{s\to 1}\frac{e^{st}}{s^2+9}+\lim_{s\to 3i}\frac{e^{st}}{(s-1)(s+3i)}+\lim_{s\to -3i}\frac{e^{st}}{(s-1)(s-3i)}\\ &\tag{2} \end{align} After taking the limit in equation $(2)$ and simplifying, you should get $$ \mathcal{L}^{-1}\{F(s)\}(t) = \frac{1}{30}\bigl[3e^t - 3\cos(3t)-\sin(3t)\bigr] $$

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    $\begingroup$ I'm guessing the method is ... too complex. $\endgroup$
    – user170231
    Mar 30, 2015 at 1:37
  • $\begingroup$ Yes, it was a poor attempt at humor :) $\endgroup$
    – user170231
    Mar 30, 2015 at 1:41
  • $\begingroup$ @user170231 I enjoyed it. I just didn't know if it was accidental. $\endgroup$
    – dustin
    Mar 30, 2015 at 1:42
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    $\begingroup$ The OP asked if they need to use partial fractions to solve the problem. I have shown that partial fractions is not needed. If you are downvoting, it would be nice if you have the courtesy to explain. $\endgroup$
    – dustin
    Apr 2, 2015 at 0:41
  • $\begingroup$ I upvoted for precisely that reason. You offered a suggestion that doesn't involve partial fractions - exactly what was asked for, though probably not what was expected. Complex methods might be over the OP's head. (Not that I'm accusing OP of downvoting. At least one other user did too.) $\endgroup$
    – user170231
    Apr 2, 2015 at 5:05

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