Hello!
This is just something I've asked myself, and not something I need help with for school.
Does the differential equation $\displaystyle f'(x)=f(x+1)$ have any nontrivial solutions?
i think that
$\displaystyle
k=ln \; k
$
there is no solution.
$\displaystyle
f(x)=sin(wx+a)
$
$\displaystyle
f'(x)=w \; cos(wx+a)
$
Solving
$\displaystyle
w \; cos(wx+a)=sin(w(x+1)+a)
$
gives if I am not mistaken
$\displaystyle
w=\pm 1
$
$\displaystyle
tg^2 \; a =-1
$
there is no solution.
If $\displaystyle \mathcal {L} \{f(t)}\}= \varphi(s)$, for the basic properties of the Laplace Tranform is...
$\displaystyle \displaystyle \mathcal {L} \{f^{'}(t)}\}= s\ \varphi(s) - f(0)$
$\displaystyle \displaystyle \mathcal {L} \{f(1+t)}\}= e^{s}\ \varphi(s) $ (1)
Now from the (1) it follows that a solution of the equation $\displaystyle f^{'} (t) = f(1+t)$ is [if any inverse LT exists]...
$\displaystyle \displaystyle f(t)= \mathcal{L}^{-1} \{\frac{c}{s-e^{s}}\}$ (2)
... being $\displaystyle c=f(0)$ a constant...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
If it is not requested that $\displaystyle f^{'}(t) = f(1+t)$ is valid for every t but for 'almost every t' we have several 'candidate functions'. One of the most simple is the periodic function with period T=1 defined as...
$\displaystyle f(t)= e^{t}$ , $\displaystyle 0<t<1$
Very suggestive is also the 'PSK modulated' periodic function with period T=4 defined as...
$\displaystyle f(t)=\left\{\begin{array}{ll}\sin 2 \pi t ,\,\,0 < t <1\\{}\\ \cos 2 \pi t ,\,\, 1 < t <2\\{}\\ -\sin 2 \pi t,\,\, 2<t<3 \\{}\\ -\cos 2 \pi t,\,\, 3<t<4\end{array}\right.$
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$