When is $$f(x) = f'(x)\int{f(x)}dx$$? I just though of the problem but couldn't solve it myself.

  • $\begingroup$ $f(x)=0$ could work.. :-) $\endgroup$ – Simply Beautiful Art Mar 30 '17 at 22:11
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    $\begingroup$ It might be better to write $g(x)=\int f(x)dx$; then this equation becomes $g'(x)=g''(x)g(x)$. $\endgroup$ – Steven Stadnicki Mar 30 '17 at 22:25
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    $\begingroup$ Keep in mind that $\int f(x) \, dx$ is a family of functions. Are we permitted to choose any constant? Maybe it would be better to write it as $f'(x) = f''(x) f(x)$? (I see that Steven has beaten me to the punch!) $\endgroup$ – Brian Tung Mar 30 '17 at 22:32
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    $\begingroup$ There appear to be no nice solutions: Maple gives a general implicit solution in terms of an elliptic integral function. On the other hand, if we replace the left-hand side with $f(x)^2$ the resulting equation is relatively easy to solve. $\endgroup$ – Travis Willse Mar 30 '17 at 22:51
  • $\begingroup$ I was thinking of using the constant 0 but feel free if you have a general solution! $\endgroup$ – mtheorylord Mar 30 '17 at 22:55

We can write your equation as follows substituting $y+C_1=\int f(x)$ $$y'=(y+C_1)y''$$ $$y''=y'/(y+C_1)$$ Now integrating each side gives $$y' = \ln({y+C_1})+C_2$$ $$\frac{1}{\ln{(y+C_1)}+C_2}y'=1$$ And this is as far as we will go. $$\int \frac{1}{\ln{(y+C_1)}+C_2}dy=x$$ So, the crux of this matter comes to finding $\int \frac{1}{\ln{y}}dy$ for which there is no closed form. From here we would take the inverse of this non closed form function, and then take the derivative of that...

So it didn't work out as nicely as one would hope, but nice thinking. I'd have to agree that this is a cool function, one which is equal to the product of it's derivative and integral, despite the fact that it is not going to have a nice closed form.


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