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!


1 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 $$\eqalign{x(0) &= 0\cr y(0) &= 0\cr}$$

  • $\begingroup$ Thanks for the help. Unfortunately I'm still a bit stuck. I can see that the solution probably needs to be of the form $Ae^{at}+Be^{-bt}$ and that $A=-B$ to satisfy the initial conditions, but can never get the functional differential equation to work out such that it is zero for all time. I'm not sure how writing the system of differential equations will help, since I am left with 3 things to solve for (I'm not sure how to reconcile $x(-t)$ with $x(t)$). $\endgroup$
    – Liz
    Sep 11, 2013 at 21:39

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