Analytic function $f,$ such that $f(0) = 1$ and $f'(z) = zf(z),$ for all $z \in \mathbb{C}$

Question

I'm trying to find an example of an analytic function $f$ satisfying the IVP $$ f'(z) = z\,f(z), \quad f(0) = 1, $$ and for all $z \in \mathbb{C}$, but I'm somewhat at a loss of the best way to proceed.

Any hints or suggestions would be greatly appreciated.

Answer 1

Observe that $$ f'(z)=zf(z)\,\,\Rightarrow\,\,\mathrm{e}^{-z^2/2}\big(f'(z)-zf(z)\big)=0 \,\,\Rightarrow\,\,\big(\mathrm{e}^{-z^2/2}f(z)\big)'=0, $$ and thus $\mathrm{e}^{-z^2/2}f(z)$ is constant. In particular $$ \mathrm{e}^{-z^2/2}f(z)=\mathrm{e}^{-0^2/2}f(0)=1, $$ and thus $$ f(z)=\mathrm{e}^{z^2/2}. $$

Answer 2

Can you solve this ODE?: $y'=xy, y(0)=1\,.$