Derivative of a Convolution: Solving an ODE with an Integral Solution

[Shaybay92]

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)(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

 

[haruspex]

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)(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 + ∫_t^t+δtf(t+δt, r).dr
So v' = ∫^t(d/dt)f(t, r).dr + f(t, t)

 

[Shaybay92]

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 :)

 

[haruspex]

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 :)

The equation you posted for v(t) is generic - i.e. it's true for all t. So it's true both for a given t and for a later time t+δt. So you can write a second equation substituting t+δt for t consistently. Taking the difference, diving by δt, then letting δt tend to zero gives you v'. That is the standard process of differentiation.

 

[Shaybay92]

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