# Solving ODEs of the form $x'(t)=F(x(t))+f(t)$

In some physical problem the following differential equation appears

$\dot{x}=F(x)+f(t)$,

where the dot denotes derivative with respect to $t$. $x$ is evidently a function of $t$. I'm wondering what the theory of differential equations knows about how to solve these kind of equation. Generic solution for arbitrarily nonlinear functions $F$ and and $f$ is desirable without making any assumptions on theses functions.

The solution can easily be found for $F(x)=x$ and for arbitrary $f(t)$. In principle, one could Taylor expand $F(x)$ in the vicinity of the origin, and do perturbation theory in terms of higher corrections, but that's quite trivial and is \textit{not} the subject of my question.

UPD. By making the substitution $x = u'/u$ one can reduced the above equation to a Riccati type equation provided that $F(x)$ is a quadratic polynomial in $x$.

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What IS the subject of your question? Closed form solutions? Hopeless without restricting F or f in some way. Numerical solutions? Pick your textbook. –  Michael Renardy Apr 5 '11 at 22:29
Does this DE appear in the above generality for your physical problem, or do $F$ and $f$ live in a fairly specific family of functions? –  Ryan Budney Apr 5 '11 at 23:01
Yes, indeed, $F(x)$ needs to be generic. Last thing I can give up on is $f(t)$ -- let's say it's harmonic, but it does not really make the problem easier. –  Peter Apr 5 '11 at 23:13
Surely, if such a general theory does not yet exist, it would be worthwhile to create it: I think this is interesting. –  Zen Harper Apr 6 '11 at 8:38
I would be surprised if solutions exist for ARBITRARY choices of $F$ and $f$. You probably need some weak hypothesis on the functions $F$ and $f$, for example that they are Lipschitz continuous. –  Tom LaGatta Apr 6 '11 at 21:07

Right, Thanks, I recently realized myself that for some special cases it can be reduced to Riccati equation by substituting $x=u'/u$ for example. –  Peter Apr 6 '11 at 20:38
There is no general closed-form solution method for this differential equation. Even in some rather simple special cases (e.g., I believe, $F(x) = x^3$, $f(t) = t$, which is an Abel d.e.), there is no known closed-form solution.
Yes, indeed, only $F=x + const$ is the case I know the solution for. But since the $x$ and $t$ dependence is additively separated, I thought one may figure it out somehow... –  Peter Apr 5 '11 at 23:15
I imagine, however, that the additive separation of $x$ and $t$ dependencies may simplify any analysis of the qualitative behaviour of the solutions, not that I can offer any concrete advice in that regard. This is an entirely different kettle of fish of course, but it is frequently what is really desired. A closed form solution, if sufficiently complicated, may even get in the way of your qualitative understanding. –  Harald Hanche-Olsen Apr 6 '11 at 6:53