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We then solve this equation: $$\begin{align*} \frac {\mathrm{d}N}{\mathrm{d}t}&=rN(t) \\ \frac {\mathrm{d}N}{N(t)}&=r\mathrm{d}t \\ ln\big(N(t)\big)&=rt+c \\ N(t)&=e^{rt}e^{c} \\ \therefore N(t)&=N_0e^{rt} \end{align*}$$