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.