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## Homework Statement

solve the following differential equation using Laplace transforms:

[tex] y'' + 4y' + 4y = t^2 e^{-2t}, y_0 = 0, y'_0 = 0 [/tex]

y_0 and y'_0 are initial conditions.

## Homework Equations

Using L to represent the Laplace transform, we have that

[tex]

L(y) = Y

[/tex]

[tex]

L(y') = pY - y_0

[/tex]

[tex]

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

[/tex]

## The Attempt at a Solution

Taking the Laplace transform of the entire DE gives

[tex]

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

[/tex]

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

[tex]

L(t^2 e^{-2t}) = 2/(p+2)^3

[/tex]

[tex]

∴ p^2 Y - 0 -0 +4pY - 0 +4Y = 2/(p+3)^3

[/tex]

[tex]

∴ (p^2 + 4p + 4)Y = 2/(p+2)^3

[/tex]

[tex]

∴Y = 2/(p+2)^5

[/tex]

[tex]

∴ y = L^{-1}(Y) = L^{-1}(2/(p+2)^5)

[/tex]

Then from tables (pg 469, M Boas, mathematical methods for the physics sciences), we use the following relation:

[tex]

L(t^k e^{-at}, k>-1

[/tex]

becomes

[tex]

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

[/tex]

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

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