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...)

  • 3
    $\begingroup$ 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

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 $\int e^{-y^2} dy$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.