## Derivative of a Convolution

Hi,

I want to verify that the form of a particular solution satisfies the following ODE:

v' + (b/m)v = u/m

with

vpart= ∫e-(b/m)(t-r) (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
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 Quote by Shaybay92 Hi, I want to verify that the form of a particular solution satisfies the following ODE: v' + (b/m)v = u/m with vpart= ∫e-(b/m)(t-r) (u(r)/m) dr where the limits are from 0 to t. So I tried to differentiate v with respect to t,... How can you do this?
v(t) = ∫tf(t, r).dr
v(t+δt)= ∫t+δtf(t+δt, r).dr
= ∫tf(t+δt, r).dr + ∫tt+δtf(t+δt, r).dr
So v' = ∫t(d/dt)f(t, r).dr + f(t, t)
 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 :)

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