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What kind of differential equation is this?

  • Thread starter mathman44
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Homework Statement



It's been a long time since I've done DE's and now I'm stuck with a monster of this form:

y'(t) = a*g'(t) + b*g(t) + c*y(t)

where g(t) is a known function and a, b and c are constants. What kind of DE is this, and how can I solve for y(t) -- or better yet, what should I read to figure this out? The answer involves a convolution integral, that I know for sure.

Cheers.
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement



It's been a long time since I've done DE's and now I'm stuck with a monster of this form:

y'(t) = a*g'(t) + b*g(t) + c*y(t)

where g(t) is a known function and a, b and c are constants. What kind of DE is this, and how can I solve for y(t) -- or better yet, what should I read to figure this out? The answer involves a convolution integral, that I know for sure.

Cheers.
This is just a simple linear DE of the form y'(t) - c*y(t) = f(t), where f(t) is the KNOWN function f(t) = a*g'(t) + b*g(t). There is nothing at all monstrous about it.
 
  • #3
Dick
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Homework Statement



It's been a long time since I've done DE's and now I'm stuck with a monster of this form:

y'(t) = a*g'(t) + b*g(t) + c*y(t)

where g(t) is a known function and a, b and c are constants. What kind of DE is this, and how can I solve for y(t) -- or better yet, what should I read to figure this out? The answer involves a convolution integral, that I know for sure.

Cheers.
If g(t) is known, then if you put h(t)=a*g'(t)+b*g(t), it's y'(t)-c*y(t)=h(t). It's an inhomogeneous first order linear equation. Nothing to do with convolution I can see.
 
  • #4
Ray Vickson
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If g(t) is known, then if you put h(t)=a*g'(t)+b*g(t), it's y'(t)-c*y(t)=h(t). It's an inhomogeneous first order linear equation. Nothing to do with convolution I can see.
Well, the formula for the nonhomgeneous solution *does* involve something very like a convolution. In fact, if you do it by Laplace transforms you can see explicitly that part of the solution involves a product of Laplace transforms, hence involves convolution. However, the general solution may also involve some non-convolution parts, depending on initial conditions, etc.
 
  • #5
Dick
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Well, the formula for the nonhomgeneous solution *does* involve something very like a convolution. In fact, if you do it by Laplace transforms you can see explicitly that part of the solution involves a product of Laplace transforms, hence involves convolution. However, the general solution may also involve some non-convolution parts, depending on initial conditions, etc.
Ok, shows I don't know much about Laplace transform methods. I stand corrected.
 
  • #6
Ray Vickson
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Ok, shows I don't know much about Laplace transform methods. I stand corrected.
You don't need Laplace transforms; that is just one of the ways to see it. The general solution of
[tex] \frac{dx(t)}{dt} - a x(t) = f(t) [/tex]
is
[tex] x(t) = C e^{at} + \int_0^t e^{a(t-\tau)} f(\tau) \, d \tau.[/tex]
Here, the first term is the solution of the homogeneous equation and the second term is a 'particular' solution to the non-homogeneous equation. The second terms is a convolution of the two functions ##e^t## and ##f(t)##.
 

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