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Any ideas?

I got:-

$$s^3 - 2s^2 + 3s - 4/(s(s^2 + 3) + 1))$$

but I got it wrong, obviously, because it does not simplify into any inverse laplaces.

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Please, anyone know what I can do? –  user57190 Jan 14 '13 at 1:16
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2 Answers


Taking the Laplace Transform of your equation, we arrive at:

$[s^{3}y(s) - s^{2}Y(0) - sY'(0) - Y''(0)] -3[s^{2}y(s) - sY(0) -Y'(0)] + 3[sy(s)-Y(0)] -y(s) = \frac{2}{(s -1)^{3}}$

Collect like terms, do some algebra, do partial fraction expansion and this results in $$y(s) = \frac{1}{s-1} - \frac{1}{(s-1)^{2}} -\frac{1}{(s-1)^{3}} + \frac{2}{(s-1)^{6}}$$

Then, do the inverse Laplace Transform to arrive at $y(t)$.

Does that all make sense and you can it from there?


$y(t) = e^{t} -te^{t} - \frac{t^{2}e^{t}}{2} + \frac{t^{5}e^{t}}{60}$

Test it against the DEQ, and IC's and verify that this is the solution.


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That's exactly what I did and got L{y} = s^2 - 3s + 1/(s^3 - 3s^2 + 3s - 1). But this still does not easily simplify... –  user57190 Jan 14 '13 at 1:28
waiiit a minute! Yes it does! Is it e^-t - te^t - t^2e^t? –  user57190 Jan 14 '13 at 1:31
No, that is not correct. Solve and show y(s) and lets see where you went astray. Y(t) should have four terms when you are done. Smartphones are hard to type on! Regards –  Amzoti Jan 14 '13 at 1:40
I'm lost, no worries, thanks for your help so far! –  user57190 Jan 14 '13 at 1:47
OK, I've simplified it and got L{y}(s^3 - 3s^2 + 3s - 1) - s^2 + 3s + 1 = 2/((s-1)^3) Is this right so far? –  user57190 Jan 14 '13 at 1:55
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Not sure what you did, but you need to take the initial values into account when Laplace transforming the higher derivatives in the diff eq'n.

For example,

$$ \int_0^{\infty} dt \: f'(t) e^{-s t} = -f'(0) + s \hat{f}(s)$$

where $\hat{f}(s)$ is the LT of $f$. Similar expressions may be derived for the higher derivatives.

Sme algebra is involved, but I get

$$ \hat{f}(s) = \frac{2}{(s-1)^6} -\frac{1}{(s-1)^3}-\frac{1}{(s-1)^2} + \frac{1}{s-1}$$

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That's strange, I just re-did it, but got 1/(s-1)-1/(s-1)^2-1/(s-1)^3 –  user57190 Jan 14 '13 at 1:31
Can you summarize what you did? I still like my result. –  Ron Gordon Jan 14 '13 at 1:36
@user57190: That is missing a sixth order term. Regards –  Amzoti Jan 14 '13 at 1:42
How did you get there? –  user57190 Jan 14 '13 at 1:46
@rlgordonma: Something looks wrong, if you simplify my y(s), there should be four terms. Regards –  Amzoti Jan 14 '13 at 1:48
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