Solving differential equation using Laplace Transform

[Edge Of Pain]

Homework Statement

solve the following differential equation using Laplace transforms:
$$y'' + 4y' + 4y = t^2 e^{-2t}, y_0 = 0, y'_0 = 0$$

y_0 and y'_0 are initial conditions.

Homework Equations

Using L to represent the Laplace transform, we have that

$$L(y) = Y$$
$$L(y') = pY - y_0$$
$$L(y'') = p^2 Y - py_0 - y'_0$$

The Attempt at a Solution

Taking the Laplace transform of the entire DE gives

$$p^2 Y - py_0 - y'_0 + 4pY - 4y_0 + 4Y = L(t^2 e^{-2t})$$

From Laplace transform tables (using M. Boas, page 469) we have

$$L(t^2 e^{-2t}) = 2/(p+2)^3$$
$$∴ p^2 Y - 0 -0 +4pY - 0 +4Y = 2/(p+3)^3$$
$$∴ (p^2 + 4p + 4)Y = 2/(p+2)^3$$
$$∴Y = 2/(p+2)^5$$

$$∴ y = L^{-1}(Y) = L^{-1}(2/(p+2)^5)$$

Then from tables (pg 469, M Boas, mathematical methods for the physics sciences), we use the following relation:
$$L(t^k e^{-at}, k>-1$$

becomes

$$k!/(p+a)^{k+1}$$

This all makes sense and I'm confident that this is correct so far

But somehow this means that the inverse Laplace transform of $$2/(p+2)^5$$ is $$2t^4 e^{-2t}/4!$$ (which is also the solution to the DE - since it's $$L^{-1}(Y) = y$$). I don't see how to make this jump and I don't see how that is correct.

 

[LCKurtz]

$s$ is more commonly used for the transform variable. Use $$
\mathcal L^{-1}F(p+a) = e^{-at}\mathcal L^{-1}F(p)$$where in your problem $F(p) = \frac 2 {p^5}$. You need to multiply numerator and denominator by $4!$ to make the formula fit.

 

[lurflurf]

that is just a particular case of
$$L\left(C\dfrac{t^k}{k!}e^{-a \, t}\right)=C(p+a)^{-(k+1)}$$
which we can get by combining
$$L\left(C\dfrac{t^k}{k!}\mathrm{f}(t)\right)=\dfrac{C}{k!}\left(-\dfrac{d}{dp}\right)^k L(\mathrm{f}(t))\\
L\left(e^{-a \, t}\right)=(p+a)^{-1}$$
into
$$L\left(C\dfrac{t^k}{k!}e^{-a \, t}\right)=\dfrac{C}{k!}\left(-\dfrac{d}{dp}\right)^k L(e^{-a \, t})=\dfrac{C}{k!}\left(-\dfrac{d}{dp}\right)^k (p+a)^{-1}=C(p+a)^{-(k+1)}$$