Solve the initial value problem using Laplace Transforms.

$\displaystyle \ddot{y} - 6 \dot{y} + 9y = te^{3t}, y(0) = 1, \dot{y}(0) = 0 $

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- Aug 29th 2009, 09:28 PMMathTragicIVP with Laplace Transform
Solve the initial value problem using Laplace Transforms.

$\displaystyle \ddot{y} - 6 \dot{y} + 9y = te^{3t}, y(0) = 1, \dot{y}(0) = 0 $ - Aug 29th 2009, 11:32 PMChris L T521
Apply the Laplace transform to both sides of the equation to get

$\displaystyle s^2Y(s)-s y(0)-\dot{y}(0)-6\left(sY(s)-y(0)\right)+9Y(s)=\frac{1}{(s-3)^2}$

Apply the initial conditions and simplify to get

$\displaystyle s^2Y(s)-6sY(s)+9Y(s)-s+6=\frac{1}{(s-3)^2}\implies (s-3)^2Y(s)=\frac{1}{(s-3)^2}+s-6$ $\displaystyle \implies Y(s)=\frac{1}{(s-3)^4}+\frac{s}{(s-3)^2}-\frac{6}{(s-3)^2}$

Take the Inverse Laplace Transform of both sides. I leave it for you to continue from here.

__Spoiler__: - Aug 30th 2009, 09:15 PMwik_chick88
- Aug 30th 2009, 10:29 PMChris L T521
- Sep 2nd 2009, 09:26 PMwik_chick88
i get the answer to be:

$\displaystyle y(t) = e^{3t} (2t + 1)$

is this correct? - Sep 2nd 2009, 10:56 PMCaptainBlack