Equation involving integrals I saw on a shirt: $\int f(x) = \frac{1}{f'(x)} $

Question

The equation is $$ \int f(x) = \frac{1}{f'(x)} $$ I saw it on a shirt so I thought it is probably a joke, but I couldn't get it. So I tried to solve it. On the first look I thought this is never true. Then I thought of differentiating the whole thing.

$$ f(x) = - \frac{f''(x)}{f'(x)^2} $$

This is what I got. I don't see any easy solution. This looks like a complicated differential equation. The only diff. eq. that I know how to solve are the most basic ones that appear in newtonian mechanics.

So, where is the joke if there is one. Or is this equation special in any other way.

Answer 1

There is a solution. Replacing $y=f(x)$ and playing with the derivatives, you should end with $$x''=x' y$$ Then, reduction of order $x'=p$ gives $$p'=p y \implies p=c_1 e^{\frac{y^2}{2}}$$ One more integration to get $$x=\sqrt{\frac{\pi }{2}} c_1 \text{erfi}\left(\frac{y}{\sqrt{2}}\right)+c_2$$ I do not think that we could get the inverse of it.

Answer 2

We have $f'f=-f''/f'$ so $f^2/2+C=-\ln f'$, i.e. $f'=A\exp -f^2/2$ and $x=A^{-1}\int\exp (f^2/2) df$.