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A classic example for this would be the exponential function, or sigmoid function, or even tanh function.

But is there a general form for all these functions, in such a way that the general form follows the same property? (ie the general form function's derivative can be written in terms of the function itself...)

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    $\begingroup$ I do not think there is a uniform way to explicitly solve a differential equation of the form $f'= \Phi(f)$. When $\Phi \equiv 1$ you have the exponential function, when $\Phi(t)= \sqrt{1-t^2}$ you have the $\sin$ function and so on, but in general I guess that the problem is hopeless. $\endgroup$ – Francesco Polizzi Aug 10 '17 at 14:26
  • $\begingroup$ Numerically it should be possible. But I suppose you mean algebraically or analytically. $\endgroup$ – mathreadler Aug 10 '17 at 14:51
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Essentially, you want to solve a differential equation of the form $$\frac{dy}{dx} = \Phi(y).$$ The usual separation of variables gives $$x = \int \frac{dy}{\Phi(y)},$$ so you obtain an explicit solution whenever you are able to calculate the indefinite integral $I(y) = \int \frac{dy}{\Phi(y)}$ and to invert (at least locally) the expression $x=I(y)$.

Of course, this is not always possible in terms of elementary functions. For instance, take $\Phi(y) = e^{y^2}$, so that $I(y)$ becomes the Gaussian integral $\int e^{-y^2} dy$.

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  • $\begingroup$ are you saying that basically any function that is invertible is the solution for it? $\endgroup$ – aditya sista Aug 10 '17 at 22:50

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