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Science & engineering math: integro-differential equation |
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| Mar18-12, 01:09 PM | #1 |
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Science & engineering math: integro-differential equation
1. The problem statement, all variables and given/known data
[itex]\int[/itex] y'(u)y(t-u)du = 24t3 The integral goes from t (top) to 0 (bottom) With y(0) = 0 2. Relevant equations I want to say it kind of looks like a convolution problem [itex]\int[/itex] f(u)g(t-u)du The integral goes from t (top) to 0 (bottom) 3. The attempt at a solution I have no idea... |
| Mar18-12, 09:33 PM | #2 |
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When you see a convolution in a homework problem, you should immediately think of a transform that reduces it to a multiplication. This transform can even handle the derivative easily...
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| Mar18-12, 10:00 PM | #3 |
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Well if it is convolution then it would just be
F(s)*G(s) But I was more concerned about whether or not it actually was convolution. Because its y' and y, and those are both completely different, then it would be convolution then? So I would need to take the Integral of y'(u) and y(t-u) then? And the integral of 24t3? |
| Mar18-12, 10:05 PM | #4 |
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Science & engineering math: integro-differential equation
it is a convolution between to functions, i.e., y'(t) and y(t), their Laplace transform are related, because one is the derivative of the other. So there is only one F(s) to solve for, the other is simply related to this one.
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| Mar19-12, 11:15 AM | #5 |
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Okay, understandable...but how do I start the problem? I'm still confused about how to start...
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| Mar19-12, 01:12 PM | #6 |
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Recognitions:
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RGV |
| Mar19-12, 11:56 PM | #7 |
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F(s)G(s)
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| Mar20-12, 07:49 AM | #8 |
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So "start" by taking the Laplace transform of both sides!
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| differential eq, integral |
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