# Suppose $f : \mathbb{R} \to \mathbb{R}$ is twice differentiable and that $f''(x)+f(x)=0$ for all $x$.

Prove that if $f(0)=0$ and $f'(0)=0$, then $f(x)=0$ for all $x$.

Hint: The idea is to multiply both sides of the equation $f''(x)+ f(x) = 0$ by something that makes the left-hand side of the equation into the derivative of something.

I'm not sure how to proceed and don't really understand the hint.

$$f''(x)+ f(x) = 0$$

$$f'(x) f''(x) +f'(x)f(x) =0$$

$$(1/2)(f^2 + f'^2 )' =0$$

$$f^2 + f'^2=C$$

Since $$(f^2 + f'^2)(0)=0$$

We get $C=0$, that is $f(x)=0$

• could you explain the last step, I'm really confused. – MANONMARS45 May 12 '18 at 8:11
• we know $f(0)=f'(0)=0$, thus$f^2(0)+f'^2(0)=0$. That gives us $C=0$ from which we get $f^2+f'^2=0$, so$f=f'=0$ for all $x$. – Mohammad Riazi-Kermani May 12 '18 at 9:15

Multiply by f'. It becomes $((f^2)'+(f'^2)')/2$.