If $f''(x)+2f'(x)+3f(x)=0$, then $f$ is infinitely differentiable

Question

I came across this problem: which of the following statements are true regarding differentiability.

Is the following statement true?

If $f$ is twice continuously differentiable in $(a,b)$ and if for all $x\in(a,b)$ , $$f''(x)+2f'(x)+3f(x)=0$$, then $f$ is infinitely differentiable in $(a,b)$.

I understand the argument using induction. However, I am wondering if the following argument makes sense or not?

My argument: Solve the differential equation $y''+2y'+3y=0$, we get the general solution $$y=C_1 e^{-x}\sin{\sqrt{2}x}+C_2e^{-x}\cos{\sqrt{2}x}$$ , which is infinitely differentiable.

My friend thinks that my argument is not correct, since I cannot guarantee that all possible $f$ has to be in form of the general solution. I am confused. Is my reasoning correct?

Answer 1

You don't need to solve the differential equation. Just write it like $$ f''=-2f'-3f.$$ The RHS is a continuously differentiable function (by hypothesis) so $f''$ must be continuously differentiable also; i.e. $f$ must be three times continuously differentiable. Differenciate the equation to get $$f'''=-2f''-3f'$$ and the same reasoning applies now to $f'''$, allowing you to conclude that $f$ must be four times continuously differentiable. By induction then it follows that $f$ must be infinitely differentiable.

Answer 2

If $f$ is a function that is $k$ times differentiable and if $f^{(k)}$ can be expressed as a linear combination of $f^{(j)}$ for $j<k$ then $f$ is infinitely differentiable. Indeed, if $$f^{(k)} = \sum a_i f^{(i)}$$ then $f^{(k)}$ is differentiable, because all terms in the right side of the equation are, and thus --- working inductively --- you can show that $f^{(j)}$ exists for every $j$.