The first-order ODE

$$\frac{\dd y}{\dd t}=r\,y$$ where $r$ is a nonzero constant, can be rewritten in differential form as $$\frac{1}{y}\,\dd y=r\,\dd t$$ The left- and right-hand side can be worked out as $d(\ln|y|)$ and $d(r\, t)$, respectively. So we have the following equality between two differentials: $$\dd(\ln|y|))=\dd(r\, t)$$ The functions behind the ‘d's’ are thus equal to each other up to a constant: $$\ln(|y|)=r\,t+C$$ for some constant $C\tiny.$ Thus: $$|y| =e^{r\,t+C}=e^C\cdot e^{r\, t}=c\cdot e^{r\, t}$$ for some constant $c\gt 0$. Removal of the absolute value brackets leads to the explicit solution: $$y=c\cdot e^{r\, t}$$ for some constant $c$, which can being either positive or negative.

In other words, we have, in proven the following theorem.