I'm stuck with a differential equation of the following form: \begin{equation} a f'(x) + b f(x) + \exp(c f(x)) = w \end{equation} where $a,b,c,w$ are constants.
Is there a closed form solution to this type of differential equation? If yes, how can it be solved, is there any good reference for a specific solution technique?
Thanks, Chris
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You can linearize the equation with $c \lll 1$. Then $e^{cf} \to 1 + cf$, and the integral becomes
$$ x(f) = \int \frac{a}{(w-1) - (b+c)f}df = -\frac{a}{b+c}\big[\ln \big|(w-1)-(b+c)f\big|- \ln k\big] $$
Then $$ f(x) = \frac{(w-1) - k\exp\big(-\frac{b+c}{a}x\big)}{b+c} $$
This is the best I can get to a "closed form"
This is to explain the comments from Yuriy and Claude Leibovici : $$a \frac{df}{dx} + b f + \exp(c f) = w$$ $$\frac{dx}{df}=\frac{a}{- b f - \exp(c f) + w}$$ $$x=a\int \frac{df}{- b f - \exp(c f) + w}+\text{constant}$$ This is the solution implicitly expressed on the form of a function defined by an integral.
As far as I know, there is no closed form for this kind of integral.
A-fortiori, the inverse function $f(x)$ cannot be expressed with a finite number of standard functions.
Approximate solution can be expected on the form of finite series or more likely in practice thanks to numerical calculus.