# Mathematica Question

1. Dec 1, 2005

### amcavoy

Could someone tell me how to graph the following ODE's in Mathematica?:

$$y''+2y'+2y=\delta\left(t-\pi\right)$$

$$y''+3y'+2y=\delta\left(t-5\right)+u_{10}\left(t\right)$$

Thanks.

2. Dec 1, 2005

### James R

I don't understand what you want to graph?

Don't you need the solution first?

3. Dec 2, 2005

### saltydog

Can you solve them using Laplace Transforms first? That is:

$$\mathcal{L}\left\{\delta(t-a)\right\}=e^{-as}$$

and:

$$\mathcal{L}^{-1}\left\{f(s)\right\}=e^{-at}\mathcal{L}^{-1}\left\{f(s-a)\right\}$$

Edit: I tried using NDSolve in Mathematica. Having a problem with it as the first equation usually has the initial conditions both set to zero with the unit pulse at pi.

Last edited: Dec 2, 2005
4. Dec 2, 2005

### saltydog

Well, I'm impressed. At Mathematica anyway. At first I thought it couldn't handle:

$$\text{InverseLaplaceTransform}\left[\frac{e^{-\pi s}}{s^2+2s+2}\right]$$

returning a complex expression but then I used:

$$\text{Simplify[ComplexExpand[InverseLaplaceTransform}\left[\frac{e^{-\pi s}}{s^2+2s+2}\right]]]$$

and it returned the correct value. Amcavoy, you gettin' all of this? Really would recommend to you to do all the work by hand first. Right? Just use the relations I gave above and you can invert that last expression by hand.

Set the initial conditions for the first one to 0. As far as the second one, that u10 just ain't happening for me.

Edit:
Oh great, suppose you need another relation:

$$\mathcal{L}^{-1}\left\{e^{-as}F(s)\right\}=f(t-a)u(t-a)$$

where u(t) is the unit step function.

Last edited: Dec 2, 2005
5. Dec 2, 2005

### amcavoy

Yeah I did solve them (I had some more too), but they looked too messy for me to "sketch" by hand so I wanted to see what they looked like in Mathematica. Thanks for the help :)

6. Dec 2, 2005

### saltydog

Well, this is the first one in Mathematica for y(0)=0, y'(0)=0:

Code (Text):
f[t_] :=
If[t ≤  Pi,
Return[0];
,
Return[Exp[-(t - Pi)] Sin[t - Pi]];
];
Plot[f[x], {x, 0, 4 Pi}, PlotRange -> {{0, 4 Pi}, {-0.4, 0.4}}]
Edit: Jesus, don't even need the 'Module' part.

Last edited: Dec 2, 2005