Solve the functional equation $\,\,f(2x)=2x f^\prime(x)$

Question

If $f(x)$ is a real analytic functions on $\mathbb R$, and $$2xf'(x)=f(2x),$$ then find $f(x)$.

My idea: express $f$ as: $$f(x)=\sum_{n=0}^{\infty}a_{n}x^n.$$

Thank you

Answer 1

Answer. The function $\,f$ is of the form $\,f(x)=ax+bx^2,\,\,$ for some constants $a$ and $b$.

Explanation. Assume that $\,\,\displaystyle f(x)=\sum_{n=0}^\infty a_nx^n$, where $\lvert x\rvert<r$, for some $r\in(0,\infty]$. Then $$ 2xf'(x)=\sum_{n=1}^\infty 2na_nx^n, $$ while $$ f(2x)=\sum_{n=0}^\infty a_n2^nx^n, $$ and $2xf'(x)=f(2x)$ implies that $$ 2na_n=2^na_n, \quad\text{for all}\,\,n\in\mathbb N, $$ and hence $a_0=0$, $a_1,a_2$ can be anything since $$ 2\cdot a_1=2^1\cdot a_1 \quad\text{and}\quad 2\cdot 2\cdot a_2=2^2a_2, $$ and $$ a_n=0, \quad\text{for all}\,\,n>2. $$

Indeed $f(x)=ax+bx^2$ satisfies the (delay differential) equation.