# Solving an unusual ODE

I'm looking to find a solution $f(x)$ to the following differential equation:

$$\frac{d(f^2(x))}{dx} = \left(\frac{df}{dx}\right)^2$$

but I can't seem to figure out where to start, since this isn't a typical sort of ODE. Anyone able to point me in the right direction?

Hint: Using the chain rule, this becomes $$2f(x)f'(x)=f'(x)^2$$ Therefore, $$f'(x)(2f(x)-f'(x))=0$$

Hint: First of all $$\frac{d(f^2)}{dx} = 2f\frac{df}{dx}$$ so your differential equation can be rearranged to give $$\frac{df}{dx}\left(\frac{df}{dx} - 2f\right) = 0.$$

Hint:

Starting from the ODE,

$$\frac{d(f^2(x))}{dx} = \left(\frac{df}{dx}\right)^2,$$

use the chain rule on the RHS to get:

$$2ff'=f'^2.$$

Dividing through on both sides by $f'$,

$$2f=f'.$$

• And POOF division by zero. $f(x) = 1$ is a solution to the ODE, but not your "simplified form"... Please fix that. ($\pm 0$) – AlexR May 27 '14 at 8:33
• @AlexR When I write hints, I don't feel obligated to explicitly spell out details like that. I expect a college level student advanced enough in their mathematics education to be taking an ODE course to know how to not divide by zero. The hint was just meant as a prod to help him get past what he was stuck on. At least, that's my philosophy here. To each his own. :) – David H May 27 '14 at 8:52

$$\frac{df(x)^2}{dx}=2f(x)\frac{df(x)}{dx}$$ So the equation becomes: $$2f(x)\frac{df(x)}{dx}=\frac{df(x)}{dx}^2$$ It's now easy to find the solutions of this ODE