Differential Equations for Noncausal Functions

Question

I am struggling to find the solution to the differential equation:

$\dot x(t)+x(-t)=0; x(0)=0$

Basically, I can't figure out how the reflection in $t$ affects the solution, and from there how to include the initial condition. I know the function that solves this must cross the origin with no slope (so you can't just shift over a trig function), and I would assume it is related to the exponential function, since the derivative will need to cancel out the reflection of the original function. I would really appreciate it if someone could explain how I can approach this problem (preferably without giving the solution). Thanks!

Answer 1

It's not a differential equation, it's a functional differential equation.
Hint: Write a system of differential equations for $x(t)$ and $y(t)$ where $y(t) = x(-t)$.

Further hint: it's very easy to guess a solution. Checking whether this is the only solution takes a bit of work.

EDIT: Apparently you didn't understand my hint. If $y(t) = x(-t)$, you have $\dot{x}(t) = -y(t)$ and the chain rule says $\dot{y}(t) = -\dot{x}(-t) = y(-t) = x(t)$. So your system is $$ \eqalign{\dot{x}(t) &= -y(t)\cr \dot{y}(t) &= x(t)\cr}$$ It's easy to solve, but you don't really need to know the general solution, just the fact that it is a homogeneous system, because the initial condition is $$\eqalign{x(0) &= 0\cr y(0) &= 0\cr}$$