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Solving differential equation using Laplace Transform
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[QUOTE="Edge Of Pain, post: 4962479, member: 506799"] [h2]Homework Statement [/h2] 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. [h2]Homework Equations[/h2] 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] [h2]The Attempt at a Solution[/h2] 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. [/QUOTE]
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Solving differential equation using Laplace Transform
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