How Do I Solve This Differential Equation?

[StevensElectronicAccount]

How do I solve the equation f'(t+h)+f(t)=0 where h is a constant?
I know that one solution for f'(t+1)+f(t)=0 is f(t) = k*m^t where m is some weird number such that m^m = e that natural growth number.
This question has been confusing me because it combines the difficulty of differential equations with the difficulty of equations like f(t+h)-f(t)=0.
I know the equation f(t+h)-f(t)=0 solves for periodic functions.
The equation f'(t+h)+f(t)=0 probably solves for something even weirder.
Edit: Just remembered the word. This has the difficulty of a recurrence relation multiplied with the difficulty of a differential equation.

[sherlock ohms]

h=0 works algebraically. K.I.S.S.

[StevensElectronicAccount]

You seem to be referencing the second equation I listed f(t+h)-f(t)=0.
I was giving that equation as an similar example I already know all about.

I was asking how do I solve the equation f'(t+h)+f(t)=0 where h is a constant for possible functions f(t)?

[Sch3mat1c]

To get the two points to sum to zero, you'll be looking at, for example, the half period of an *odd* periodic function, for instance sin(t) at h = (2N + 1) * pi (for all integer N). There are numerous functions which satisfy this, for example square waves (which are simply built of components of odd harmonics, i.e. all values of N, with decreasing amplitudes).

I'm not sure if it's necessary for the function to be periodic, though if it is, I'm pretty sure it must be a period of 2h.

If you do the good old chain rule, you get:
df(t) / dt + df(t + h) / d(t + h) * d(t + h) / dt = 0
and of course we know that d(t+h) = dt, since h is a constant, so the differential equation of the original equation tells us nothing we did not already know. In other words, this will help us constrain our choice of functions (because we know the derivative has to follow the same rule), but it won't help us choose or create such functions, because it's just as general.

Tim

[migueloreles]

hi,

i think you can solve this by fourier analysis. don't know if you're familiar with it, but simply put, you solve it for a generic function as
f(t)=A exp(i w t)
and then your final function will be a sum (actually an integral) over your solutions.
the w's are the frequencies, so you're basically decomposing a function in its spectrum. since the differentiating operator is linear you can find several solutions, add them together and you still have a solution.
so in this case
f'(t) = i w A exp(i w t)
f'(t+h) = A i w exp(i w t) exp( i w h) = iw exp(iwh) f(t)
and you want f'(t+h)=-f(t)
so
iw exp(iwh) f(t) = -f(t)
which means
i w exp(i w h)=-1
now you have a way of finding which "h" is satisfied for each w:
h(w)=log (i/w)/(iw)
now you can build any function from a sum of these functions or an integral (a Fourier transform).
hope this helped

miguel

[StevensElectronicAccount]

Thank you very much.
I was trying to solve this via taylor series.
Using a fourier series seems like it would make a lot more sense as this function has some kind of periodic stuff but I don't know how fourier series work yet so I couldn't solve it that way.