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

Question

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.

Answer 1

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