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

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?

• you solved for a class of functions (up to a constant) all of which are smooth. You are correct Oct 16, 2016 at 0:16

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.

• @Misakov For your specific question: saying that some family of functions are the general solution of an ode is the same as saying that every solution of the ode is of that form. So if you have proved that that family of functions is the general solution then your proof is correct. If you haven't then you didn't get the general solution and so your proof would be incomplete.
– user378947
Oct 16, 2016 at 0:38
• Like your picture! That's what I would like to check. I'm not sure if all possible solutions have to be in that form -_-|| Oct 17, 2016 at 19:35
• @Misakov Thank you! It is from Steins;Gate ;) Well, my argument shows that all posible solutions need to be smooth, but my knowledge of general ode theory is a little rusty so I don't know how to prove that all possible solution is of that form.
– user378947
Oct 17, 2016 at 20:04

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 $j$.

• Hi, I've edited my post. Would you mind take a look? Thanks Oct 16, 2016 at 0:16