Let $k,A$ be constants. Then the differential equation

with inital condition $f(0)=A$ has unique solution $f(x)=Ae^{kx}.$

Proof. We verify that this $f$ works. Indeed $f(x)=Ae^{kx}$ has $f^{\prime}(x)=Ake^{kx}=kf(x),$ and $f(0)=Ae^{0}=A.$ Now let $g$ be any solution, and consider $h(x)=e^{-kx}g(x)$. Then $h(0)=e^{0}g(0)=A,$ and

for all $x$, so $h$ is a constant by Lemma 3.12. Hence $h(x)=h(0)=A$ for all $x$, and so $g(x)=Ae^{kx}$.