How to solve this differential equation involving a polynomial function? I am struggling with this question please help.. 
Suppose $f(x)$ is a polynomial function as well as continuous in $\mathbb{R} \to \mathbb{R}$. Given that $f(2x)=f'(x) f''(x)$, then find $f(3)$.
 A: Follow-up on my comment. We know that $f(x)=ax^3+bx^2+cx+d$, if neither $f'$ nor $f''$ is zero. Otherwise, you would have $f(x)=0$ for all $x$, which is also a solution.
Expand $f'(x)\cdot f''(x)$:
$$f'(x)\cdot f''(x)=(3ax^2+2bx+c)(6ax+2b)=18a^2x^3+18abx^2+2(3ac+2b^2)x+2bc$$
Equate this to $8ax^3+4bx^2+2cx+d$.


*

*Coefficient of $x^3$: $18a^2=8a$ hence $a=\frac{4}{9}$

*Coefficient of $x^2$: $18ab=4b$, hence a contradiction with the preceding unless $b=0$

*Coefficient of $x$: $c=3ac+2b^2=3ac$, hence $c=0$

*$d=2bc=0$


You have thus two solution to the equation $f(2x)=f'(x)\cdot f''(x)$, one is $f(x)=\frac49x^3$ and the other is $f(x)=0$.
Therefore, you can't know $f(3)$ for sure.
A: if f is a polynomial of the degree $n$ ($n>1$), then $f'$ and $f''$ will have the degrees $n-1$ and $n-2$. So, we'll have:
$$n=n-1+n-2\Rightarrow n=3$$. So our polynomial will be something like this:
$$f(x)=ax^3+bx^2+cx+d$$. By $f(2x)=f'(x)f''(x)$, we will have:
$$8ax^3+4bx^2+2cx+d=(3ax^2+2bx+c)*(6ax+2b) \Rightarrow 8ax^3+4bx^2+2cx+d= 18a^2x^3+(6ab+12ab)x^2+(6ac+4b^2)x+2bc \Rightarrow (8a-18a^2)x^3+(4b-18ab)x^2+(2c-6ac-4b^2)x+(d-2bc)=0 $$. Thus, we will have the following equations:
$$8a-18a^2=0 \Rightarrow a=\frac{8}{18}=\frac{4}{9}$$ Note that $a$ cannot be zero due to the degree of f.
$$4b-18ab=0\Rightarrow b=0$$
$$2c-6ac-4b^2=0\Rightarrow c=0$$
$$d-2bc=0\Rightarrow d=0$$. Thus,
$$f(x)=\frac{4}{9}x^3\Rightarrow f(3)=\frac{4}{9}*3^3=12$$
if $n<2$, then $f''=0$;hence, $f(2x)=0\Rightarrow f(3)=0$.
