Differential Equation of Two Functions

[Gallagher]

Homework Statement

f '(x)-cf(x)+g(x)=0

Homework Equations

f(X)=1

f(x)=e^-c(X-x) + ∫_x^X e^-c(r-x) * g(r)dr

X is a constant.

The Attempt at a Solution

Ok, so the part of this problem that is confusing me is the integral within the function, f(x). It is an exponential function times an unknown function, g, which also is in the differential equation so my first thought was to use integration by parts. Before I get to that though, let me put down how far I've gotten into solving this quandary. Plugging f(x) into the DE I've got:

f '(x)= ce^-c(X-x) + e^-c(X-x)g(X)-g(x)

Therefore:

f '(x)-cf(x)+g(x)=0 ==>

ce^-c(X-x) + e^-c(X-x)g(X)-g(x)-c(e^-c(X-x) + ∫_x^X e^-c(r-x) * g(r)dr)+g(x)=0

The +-g(x) and +-ce^-c(X-x) cancel to leave us with:

e^-c(X-x)g(X) - c∫_x^X e^-c(r-x) * g(r)dr=0

Or equivalently;

e^-c(X-x)g(X) = c∫_x^X e^-c(r-x) * g(r)dr

Now back to my idea of integrating by parts. Again, it's just an unknown function and an exponential so it seems easy but I'm getting lost. I set:

u=g(r) v'=e^-c(r-x) such that

u=g(r) v=-(e^-c(r-x))/c
u'=g'(r) v'=e^-c(r-x)

Then my uv-∫u'v would mean

e^-c(X-x)g(X) = c([-g(r)(e^-c(r-x))/c]|_x^X + ∫_x^X g'(r)(e^-c(r-x))/c dr

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Getting here at least means that I have g'(r) in my integral instead of g(r) but, from this point I don't know what to do. I'm not even sure my math is right up to this point. I feel like if I try to integrate by parts again then I will just get caught in an infinite loop. Was this even the correct method in the first place? Any help or guidance would be greatly appreciated.

 

[tiny-tim]

Hi Gallagher!

f(x)=e^-c(X-x) + ∫_x^X e^-c(r-x) * g(r)dr

X is a constant.
…

f '(x)= ce^-c(X-x) + e^-c(X-x)g(X)-g(x)

nooo …

X' = 0, so that middle term shouldn't be there, should it? ​

 

[Gallagher]

Hi Tiny Tim!

I don't think I understand your advice. As X is a constant, not a function, I don't think you could take X'. What I think you're referring to is the initial condition of f(X)=1. As such you would think that f'(X) would = 0 but, as I will show, that is not the case and f'(X)=c as f(x) is an exponential function.

To clarify, I found f'(x) by first factoring the -c in the exponential:

f(x)=e^-c(X-x) + ∫_x^X e^-c(r-x) * g(r)dr
f(x)=e^(-cX+cx) + ∫_x^X e^-c(r-x) * g(r)dr

Differentiating brings the positive c down and leaves you to evaluatie the derivative of the integral which should just be the integrand over the interval (Right?).

f'(x)=ce^(-cX+cx) + [e^-c(r-x) * g(r)dr]_x^X

Leading to:

f'(x)=ce^(-cX+cx) + e^-c(X-x)g(X) - e^-c(x-x)g(x)

And then as e^-c(x-x)=e⁰=1

f'(x)=ce^(-cX+cx) + e^-c(X-x)g(X) - g(x)

Plugging in X in gives:

f'(X)=ce^-c(X-X) + e^-c(X-X)g(X) - g(X)
f'(X)=ce⁰+g(X)e⁰-g(X)=c

Im not really sure where I was going with that. It's all well and good but it doesn't get me any closer to solving the DE. This is where I am:

f(x)=e^-c(X-x) + ∫_x^X e^-c(r-x) * g(r)dr

f'(x)=ce^(-cX+cx) + e^-c(X-x)g(X) - g(x)

Solve: f'(x)-cf(x)+g(x)=0

Equivalently: (ce^(-cX+cx) + e^-c(X-x)g(X) - g(x)) - c(e^-c(X-x) + ∫_x^X e^-c(r-x) * g(r)dr) + g(x) = 0

I feel like I'm close to a solution here but may be completely off. If you Tiny Tim or anyone else could ponder this a bit and perhaps offer some guidance I would be forever grateful.

 

[tiny-tim]

Hi Gallagher!

f'(x)=ce^(-cX+cx) + [e^-c(r-x) * g(r)dr]_x^X

no, that last term should be X'[e^-c(r-x) * g(r) * g'(r)]_X - x'[e^-c(r-x) * g(r) * g'(r)]_x …

(and X' = 0 and x' = 1)

to put it more simply, you differentiate each limit, and ignore the ∫

I forgot to mention that in this case the integrand (as well as the limits) depends on x, so you also need to add an ∫ with the integrand differentiated …

see http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign ​