Laplace transformation of 1/(s-1)^4

Question

"Hi, I have a question about Laplace transformation.The question is:" $y ′′(t) − 2 y ′ (t) + y(t) = t e^ t , y(0) = y ′ (0) = 0$

"I know"

$Y(s)(s^2-2s+1)=1/(s-1)^2$

$Y(s)=1/(s-1)^4$

"and I know we can use Partial fraction decomposition, but I don't Know in this case when we have power 4. The answer is "

$Y(s)=(1/3!)(3!/(s-1)^4$

Answer 1

Recall the First Shifting Theorem for Laplace transform which states:

$$\mathcal{L}\{e^{at}f(t)\}(s) = \mathcal{L}\{f(t)\}(s-a).$$

In your case you have the last part of the equation $$\dfrac{1}{(s-1)^4} = \dfrac{1}{3!}\mathcal{L}\{t^3\}(s-1).$$

Proof of the theorem:

$$\mathcal{L}\{e^{at}f(t)\}(s) = \int^\infty_0e^{-st}e^{at}f(t)\,dt = \int^\infty_0 e^{-(s-a)t}f(t)\, dt = \mathcal{L}\{f(t)\}(s-a).$$

The inverse of $\mathcal L$ in the transform follows:

$$\mathcal{L}^{-1}\{{F(s-a)}\}(t) = e^{at}f(t).$$

In your case

$$Y(s) = \dfrac{1}{(s-1)^4} \Rightarrow y(t)= \frac{e^tt^3}{3!}.$$

Answer 2

If you had $Y(s) = 1/s^4$, then you know what to do, yes?

Remember also the exponential shift formula, $L[ e^{ct} y(t)] = Y(s-c)$ where $L[y] = Y$.

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Added:

$$L^{-1}\left[\frac{1}{s^4}\right] = L^{-1}\left[\frac{1}{3!}\frac{3!}{s^4}\right] = \frac{1}{3!} L^{-1}\left[\frac{3!}{s^4}\right] = \frac{1}{6} t^3$$

Now, by the Exponential Shift Formula,

$$L^{-1}\left[\frac{1}{(s-1)^4}\right] = e^t L^{-1}\left[\frac{1}{s^4}\right] = \frac{1}{6} e^t t^3$$