# Laplace transform of $y''' - 3y'' + 3y' - y = (t^2)e^t$ where $y(0)=1$, $y'(0)=0$, $y'' = -2$

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

Hint

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?

Update

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

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

Regard

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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