Graph ODEs in Mathematica: y''+2y'+2y & y''+3y'+2y

[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.

 

[James R]

I don't understand what you want to graph?

Don't you need the solution first?

 

[saltydog]

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.

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.

 

[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.

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.

 

[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 :)

 

[saltydog]

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 :)

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

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.