Solving differential equation using Laplace Transform

  • #1

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.
 

Answers and Replies

  • #2
LCKurtz
Science Advisor
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##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.
 
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  • #3
lurflurf
Homework Helper
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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
$$L\left(C\dfrac{t^k}{k!}e^{-a \, t}\right)=\dfrac{C}{k!}\left(-\dfrac{d}{dp}\right)^k L(e^{-a \, t})=\dfrac{C}{k!}\left(-\dfrac{d}{dp}\right)^k (p+a)^{-1}=C(p+a)^{-(k+1)}$$
 
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