Let $X$ be a r.v. such that $E[X]=0,E[X^2]=\sigma^2,$ suppose that the characteristic function $\varphi$ of $X,$ verifies the following : $$\varphi\varphi''-(\varphi')^2=-\sigma^2\varphi^2.$$
Prove that $$\forall x \in \mathbb{R},\varphi(x)=e^{-\frac{1}{2}x^2\sigma^2}.$$
Since $\varphi(0)=1,\varphi'(0)=0,$ and on $[-\eta,\eta],\eta>0,$ (since $\varphi$ is continuous at $0$), we have $$\frac{\varphi''}{\varphi}-(\frac{\varphi'}{\varphi})^2=-\sigma^2$$ Solving this equation of $[-\eta,\eta]$ gave $\varphi(x)=e^{-\frac{1}{2}x^2\sigma^2}.$
How to have the result on $\mathbb{R}?$ Is it possible to prove, under the above conditions, that $\forall x \in \mathbb{R},\varphi(x) \neq0?$