Is there a general solution to $\dot x = f(x)+g(t)$? Is there a general solution to $\dot x = f(x)+g(t)$? (for arbitrary continuously differentiable $f$ and $g$). 
You can just give me the name of this kind of ODE/solution method, so that I can find out about the solution myself.
 A: If $f$ is a linear function of $x$, (i.e. $f(x)=mx$), then the solution is $x=e^{-mt}+h(t)$ where $h$ depends on $g$. There are specific solutions to other cases of $f$ and $g$, some of them well studied.
As far as general solutions, they do not exist. Allowing arbitrary $f$ and $g$ means that you can use some really pathological functions. Imagine if $f$ was the Weierstrass function and $g$ was the Cantor function.
A: Your ''general solution'' is the integral formulation of Cauchy Problem.
Let $f(x)+g(t)=h(t,x)$ and $x:[t_{0},t]\rightarrow \mathbb{R}$ be a solution (if exist)
Then $x(t)-x(t_{0})=\int_{t_{0}}^{t}{h(s,x(s))\textrm{d}s}$. (Fund.Theorem of Calculus).
A: I don't think that you can solve this equation for arbitrary functions $f$ and $g$. In some cases, you might be able to integrate this equation by the method of eulerian multipliers (aka integrating factor).
Let us call $h(x(t),t)=-f(x)-g(t)$, then your ODE can be rewritten as:
$$h(x(t),t)+\dot{x}=0.$$
In some cases, one might come up with a eulerian multiplier $e(x(t),t)\neq 0$ for all $t$ and $x(t)$ such that by multiplying the ODE by $e(x(t),t)$ results in an exact differential equation
$$e(x(t),t)h(x(t),t)+e(x(t),t)\dot{x}=0$$
The function $e(x(t),t)$ is a eulerian multiplier if it is a solution to the following partial differential equation:
$$\dfrac{\partial }{\partial x} \left[eh\right] = \dfrac{\partial}{\partial t}e.$$
In most cases this partial differential equation is also difficult to solve, so one can use ansatz functions like $e=e(x)$, $e=e(t)$, $e=e(x+t)$, $e=e(xt)$ and so forth.
If a eulerian multiplier is found one can construct a potential to the ODE by integrating 
$$\dfrac{\partial}{\partial x}\Phi(x,t) = eh \qquad \text{and} \qquad \dfrac{\partial}{\partial t}\Phi(x,t) = e.$$
After having found the potential we know that $$\Phi(x,t)=\text{const.}$$ 
So the last problem is to solve this algebraic equation for $x(t)$.
A: Even for innocent-looking functions such as $f(x) = x^3$, $g(t) = t^3$, there seems to be no closed-form solution.  At least, none that Maple finds.
