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 November 1st, 2014, 01:58 PM #1 Senior Member     Joined: Oct 2014 From: EU Posts: 224 Thanks: 26 Math Focus: Calculus ODE and Inverse Laplace Transform Hi all, I have a question related to an exercise regarding the resolution of ODE with the Inverse Laplace Transform. I have a worked example as such: "Use Laplace transforms to solve the differential equation" $\displaystyle 2 {\text{d}^2y \over \text{d}x^2} + 5 {\text{d}y \over \text{d}x} - 3y = 0$ "given when $\displaystyle y(0) = 4$ and $\displaystyle y'(0) = 9$" The the resolution starts: \displaystyle \begin{aligned} & 2 \mathcal{L} \left [ {\text{d}^2y \over \text{d}x^2} \right ] + 5 \mathcal{L} \left [ {\text{d}y \over \text{d}x} \right ] - 3 \mathcal{L} \left [ y \right ] = \\ & 2 ( s^2 \mathcal{L} [y] -sy(0) - y'(0) ) + 5(s\mathcal{L} [y] - y(0) ) -3 \mathcal{L} [y] = \\ & \cdots \end{aligned} And then the ODE is solved solving for $\displaystyle \mathcal{L} [y]$. If I am not wrong, given and ODE of the order $\displaystyle n$, we always apply the operator $\displaystyle s^n \mathcal{L} [y]$ where again, $\displaystyle n$ is the order of the differential equation, and then the same coefficient $\displaystyle n$ gets smaller as we evaluate de initial conditions.. So, for example, given a third order differential equation: $\displaystyle 6 {\text{d}^3y \over \text{d}x^3} = 0$ with the initial conditions: \displaystyle \begin{aligned} & y(0) = 2 \\ & y'(0) = 5 \\ & y''(0) = 3 \\ \end{aligned} I should proceed as follows: \displaystyle \begin{aligned} & 6 ( s^3 \mathcal{L} [y] - s^2y(0) - sy'(0) - y''(0) ) = \\ & 6 ( s^3 \mathcal{L} [y] - 2s^2 - 5s - 3 ) = \\ & {2 \over s} + {5 \over s^2} + {3 \over s^3} \end{aligned} Am I right ? Thank you in advance. Last edited by szz; November 1st, 2014 at 02:56 PM.
 November 1st, 2014, 03:14 PM #2 Math Team     Joined: May 2013 From: The Astral plane Posts: 1,807 Thanks: 722 Math Focus: Wibbly wobbly timey-wimey stuff. Looks good to me. -Dan Thanks from szz

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