Tougher Differential Equation

I did it similarly, but are there any other solutions??? Solutions which are not necessarily in this form..........How do we go about finding them???

I mean $f(x)=1$ and $3$ are solutions on their own.

@Alex Burgess Haha.. I didn't even notice the constant functions as solutions! Nice observation..

Ok, so one could consider a general Taylor series: $f(x)=\sum_{n=0}^\infty a_n x^n$ .

Then, looking at the orders of $O(x^n)$ in the equation, we get:

$(n+1)(2+2^n)a_{n+1} + 2^{n+2}a_n - \sum_{i=0}^n a_i (a_{n-i} + (n+1-i)a_{n+1-i}) - 3\delta_{n,0} = 0$

Then you can factor out $a_{n+1}$ : $a_{n+1} (n+1)(2 + 2^n - a_0) = g(a_0, a_1, ... a_n)$ for some function $g$ .

Hence the terms $a_n$ will be finite if $a_0 \neq (2 + 2^n) \forall n\in \mathbb{N}$ .

For large $n$ : $a_{n+1} (n+1)(2^n) \approx - 2^{n+2}a_n$ , so $a_{n+1} \approx -\frac{4}{n+1}a_0$ , and by the alternating series test, this converges.

For $a_0 = 3$ , we get $(0)a_1=(0)$ , so $a_1$ could take any value.

For $a_0 = 1$ , we get $a_n = 0 \forall n \in \mathbb{N}$ .

By f'(2x) should we not mean d/dx(f(2x))?

Nice guess. But if we want to find all possible solutions it is a headache. May be we can use existence uniqueness theorem or some other technique.