I want to find the general solution to this equation and prove the solutions are unique please help

I know right, i'm so powerless

I mean I can find one solution,I don't know if its the general one

if you have a polynomial of degree N you'll have at most N real roots

thought it was N complex roots.

what you've got there is a difference differential equation

i'd start here
rand.org/content/dam/rand/pubs/reports/2006/R374.pdf

He said at most dumb dumb

The problem op posted is found in chapter 3.

Easy.

y' = y(x+1)
y' / y = x + 1

Now solve.

:^)

[math]f(x) = \Sigma_{n=0}^{\infty}\frac{f(n)}{n!}x^n [/math]

Fourier transform to get the recursive condition on its fourier image
[eqn]
\tilde f(k) = \frac{\tilde f(k-1)}{-ik} .
[/eqn]

I don't know if it's unique, but an 'obvious' solution is
[eqn]
\tilde f(k) = \frac{-ie^{i\pi k/2}}{\Gamma(k+1)} .
[/eqn]

Checking wolfram reminded me this is ambiguous up to an overall constant. This spells doom for the uniqueness hope you had, as the solution to the differential equation should be
[eqn]
y(x)=\int_{-\infty}^\infty \frac{dk}{2\pi}\tilde f(k) e^{-ikx}\,dk =\int_{-\infty}^\infty \frac{dk}{2\pi} \frac{-ie^{-ik(x-\pi/2)}}{\Gamma(k+1)}\,dk
[/eqn]
which is then itself ambiguous up to this overall constant.

It should be stated that I've made the friendliest convergence/existence assumptions possible throughout--under the hope the solution isn't 'too' pathological. The Gamma function in the denominator oscillates pretty wildly, but maybe there's a way to solve the integral.

I didn't get f'(x) = f(x+1) using the form of .

I have no idea if these are all the solutions, but let y(x) = e^(rx) for some r. Then from the differential equation, we have r*e^(rt) = e^(r(x+1)).

It follows that r must satisfy r = e^r, which can be solved by the Lambert W function. Then taking the real and imaginary parts of e^(rt) will give you a bunch of solutions.

Interesting. I guess the question is then if these solutions form a basis for smooth functions on the unit interval.

>I didn't get f'(x) = f(x+1)
you can show by induction that the n-th derivative of f in x is equal to f(x+n)

so if you write the taylor series for f in 0, you get the f(n)'s

the solutions are uniquely defined by their restriction to [0,1)

Plotted some of these. r=-W_k(-1), the k-th branch of Lambert's W function for k=0..4. Plots over interval [0,1]

clear all
k=4;
K=[0:k];
N=256;
x=linspace(0,1,N);
Y=zeros(length(K),N);
for j=1:length(K)
w=-lambertw(K(j),-1);
Y(j,:)=exp(w.*x);
end
figure(1);
hold off
plot(x,real(Y));
hold on
plot(x,imag(Y));
hold off

For large integer k,
[math]-W_k(-1)\approx \log\Big( 2\pi (k+\frac{1}{4} )\Big) -i\ 2\pi(k+\frac{1}{4}) [/math],
so the k-th solution is like a modulated sinusoid.

Separation of variables.
dy/dx = y(x+1)
dy/y = (x+1)dx
ln|y| = (x^2)/2 + x
y = c(e^((x^2)/2) + e^x)

y(x+1) is not y*(x+1), it is the function y evaluated at x+1.

notice this:
if Y(x) is the vector (y(x),y(x+1),y(x+2),...), then Y' = J*Y, where J is an infinite jordan matrix with all diagonal coefficients equal to 0. (the matrix only has ones above the diagonal).