Original question :
Find $f(x)$, $x\geq 0$, such that $$f'(x) = \frac{\sqrt{f(2x)-1}}{\sqrt{2}}$$
My work :
$\begin{align} f'(x) &= \frac{\sqrt{f(2x)-1}}{\sqrt{2}}\\ f(2x) &= 2(f'(x))^2 + 1\tag{1.1}\\ f(x) &= 2\left(f'\left(\frac x2\right)\right)^2 + 1 \end{align}$
Suppose i want $f'(x)=x$ (a random guess). Then $f(x)=\frac{x^2}{2} + C$ and $$f(2x) = 2x^2 + C$$ So, we have $C=1$ from $(1.1)$
Thus, $f(x)=\frac{x^2}{2}+1$. But my friend said there's another solution involving exponential function idk maybe a hyperbolic? Since there's a similar property of hyperbolic function like this. But again, how do we know if there are no other solutions? How do you solve this?
Please help. Thanks
Your friend is correct $f(x) = (e^x + e^{-x})/2$ is also a solution. As for the second part of you're question I'm not $100\%$ sure about the details but I think existence and uniqueness theorem for ODEs suggest that for many initial conditions $f(0) = y_0$ there will exist solutions. This makes we think that this problem might have infinitely many solutions however I don't know how many of these solutions will be globally defined.
EDIT: I thought about this more and what I said above doesn't make sense. However I think still think that more solutions should exist because we can substitute $f(x) = \sum_n c_n x^n$ and then solve for $c_n$ in the resulting equation.