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Differential equation with Heaviside step function

  • #1
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Homework Statement



Find the solution to the differential equation using Laplace transforms.


Homework Equations



[tex]y'' - y' - 2y = 12 \theta{(t-\pi)}\sin{t}[/tex]

[tex]y(0) = 1[/tex]

[tex]y'(0) = -1[/tex]

The Attempt at a Solution



Using Laplace transforms...

[tex]\mathcal{L}(y'') = s^2F - s + 1[/tex]
[tex]\mathcal{L}(-y') = -sF-1[/tex]
[tex]\mathcal{L}(-2y) = -2F[/tex]

[tex]\mathcal{L}(12\theta{t-pi}\sin{t}) = \mathcal{L}(-12\theta{(t-\pi)}\sin{(t-\pi)}) = \frac{-12e^{-\pi s}}{s^2+1}[/tex]

[tex](s^2-s-2)F = \frac{-12e^{-\pi s}}{s^2+1}[/tex]

I rearrange and end up with partial fractions...

[tex]F = e^{-\pi s}[\frac{-12}{(s^2+1)(s-2)(s+1)}] = e^{-\pi s}(\frac{A}{s-2} \frac{B}{s+1} \frac{Cs+D}{s^2+1})[/tex]

From this point, I can rearrange everything and end up with 4 equations, 4 unknowns. I am thinking, however, that there is an easier way, simply by using one of the common transforms found in a Laplace transform table. Is there a better way to go about this problem without having to resort to partial fractions?
 
Last edited:

Answers and Replies

  • #2
vela
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You seem to have omitted the terms due to the initial conditions. Also, you should check the signs of those terms. I think you flipped one.

Do you know how to find the inverse Laplace transform using complex analysis? If not, partial fractions is your best bet.
 
  • #3
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where is the step function ? :)
 
  • #4
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Crap, I didn't surround the initial step function with parentheses. Also, I forgot the initial conditions. I'll fix this soon.
 
  • #5
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It's fixed. I don't see anywhere where I forgot a sign, and I do not know how to figure out the inverse using complex analysis. So, if partial fractions are the way to go, then that's what I'll do!
 
  • #6
vela
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You have a sign error in your transform for -y'. You also left out the plus signs between the terms of the partial fraction expansion.
 
  • #7
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You have a sign error in your transform for -y'. You also left out the plus signs between the terms of the partial fraction expansion.
For the second one, yikes, I wrote that on the paper, but left them out when transferring them over.


[tex]F = e^{-\pi s}[\frac{-12}{(s^2+1)(s-2)(s+1)}] = e^{-\pi s}(\frac{A}{s-2}+\frac{B}{s+1}+\frac{Cs+D}{s^2+1})[/tex]

As for the sign error in the transform...

[tex]\mathcal{L}(y') = s\mathcal{L}(y) - y'(0)[/tex]

So

[tex]\mathcal{L}(-y') = -s\mathcal{L}(y) + y'(0)[/tex]

So taking the initial condition that [tex]y'(0) = -1[/tex]...

[tex]\mathcal{L}(-y') = -s\mathcal{L}(y) - 1[/tex]

Three negatives = negative. Is this not right?

EDIT: wait, I see where I messed up. I got the original formula wrong, so I switched the 1 with a -1. I am fixing this now.
 
  • #8
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So, for the equations and unknowns, I am getting...

[tex]A - 2B + 0C - 2D = -12[/tex]
[tex]A + B - 2C - D = 0[/tex]
[tex]A - 2B - C + D = 0[/tex]
[tex]A + B + C + 0D = 0[/tex]

For the [tex]-12e^{\pi s}[/tex] term, and

[tex]A - 2B + 0C - 2D = -2[/tex]
[tex]A + B - 2C - D = 0[/tex]
[tex]A - 2B - C + D = 0[/tex]
[tex]A + B + C + 0D = 0[/tex]

For the [tex]2[/tex] term.
 

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