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Derivative of a Convolution 
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#1
Sep212, 01:17 AM

P: 122

Hi,
I want to verify that the form of a particular solution satisfies the following ODE: v' + (b/m)v = u/m with v_{part}= ∫e^{(b/m)(tr)} (u(r)/m) dr where the limits are from 0 to t So I tried to differentiate v with respect to t, in order to substitute it back into the equation. But, how do you do that when the integral is with respect to r? Is there a need to change variables? How can you do this? Cheers 


#2
Sep212, 07:16 PM

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v(t+δt)= ∫^{t+δt}f(t+δt, r).dr = ∫^{t}f(t+δt, r).dr + ∫_{t}^{t+δt}f(t+δt, r).dr So v' = ∫^{t}(d/dt)f(t, r).dr + f(t, t) 


#3
Nov812, 04:43 PM

P: 122

Sorry I'm not familiar with your method. I don't understand why you substitute "t+δt" for t. What approach are you using here? Could you elaborate or direct me to some further reading?
Cheers :) 


#4
Nov812, 06:50 PM

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Derivative of a Convolution



#5
Nov812, 06:54 PM

P: 122

Oh I see what you mean. Thanks for the clarification. I'm just not use to this notation :)



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