Make the change of the independent variable setting $g(z)=f(e^z)$ then $g(z+a)=f(ke^z)$ where $k=e^a$. Now you have a linear difference equation. Difference equations have been well studied. A general method is some kind of Fourier transform, look for a solution in the form of exponential, then take a sum of those. See, for example, On equation $f(z+1)-f(z)=f'(z)$, or the book of Gelfond, Calculus of finite differences.

In your case, set $f(z)=az^2$. This will satisfy the equation if $\sum C_jk_j^2\neq 0$.

Make the change of the independent variable setting $g(z)=f(e^z)$ then $g(z+a)=f(ke^z)$ where $k=e^a$. Now you have a linear difference equation. Difference equations have been well studied. A general method is some kind of Fourier transform, look for a solution in the form of exponential, then take a sum of those. See, for example, On equation $f(z+1)-f(z)=f'(z)$, or the book of Gelfond, Calculus of finite differences.

In your case, set $f(z)=az^2$. This will satisfy the equation if $\sum C_jk_j^2\neq 0$.

Make the change of the independent variable setting $g(z)=f(e^z)$ then $g(z+a)=f(ke^z)$ where $k=e^a$. Now you have a linear difference equation. Difference equations have been well studied. A general method is some kind of Fourier transform, look for a solution in the form of exponential, then take a sum of those. See, for example, On equation $f(z+1)-f(z)=f'(z)$, or the book of Gelfand, Calculus of finite differences.

Make the change of the independent variable setting $g(z)=f(e^z)$ then $g(z+a)=f(ke^z)$ where $k=e^a$. Now you have a linear difference equation. Difference equations have been well studied. A general method is some kind of Fourier transform, look for a solution in the form of exponential, then take a sum of those. See, for example, On equation $f(z+1)-f(z)=f'(z)$, or the book of Gelfond, Calculus of finite differences.

In your case, set $f(z)=az^2$. This will satisfy the equation if $\sum C_jk_j^2\neq 0$.