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November 1st, 2014, 01:58 PM   #1
szz
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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.
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November 1st, 2014, 03:14 PM   #2
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