Finding $y=Ae^{kt}$ given $dy/dt$ at two points.

Question

How can I find the exponential growth equation, specifically $A$, the initial value and k when given

$dy/dt = 12.2$ at $t=1$

$dy/dt = 20.36$ at $t=8$

Answer 1

$$ y = Ae^{kt} \implies y' = kAe^{kt} $$ $$ \begin{align} y'\Big|_{t = 1} &= kAe^k = 12.2\tag{1} \\ y'\Big|_{t = 8} &= kAe^{8k} = 20.36\tag{2} \end{align} $$ $(2)/(1)$, $$\begin{align} e^{7k} &= \dfrac{20.36}{12.2} \\ 7k\ln e &= \ln\left(\dfrac{20.36}{12.2}\right) \\ k &\approx 0.0732 \end{align}$$ Plug $k$ in $(1)$ [or $(2)$] to find $A$.

Answer 2

We have $\frac{\mathtt{d}}{\mathtt{d}t}y=Ake^{kt}$. We know $Ake^{k}=v_1$ and $Ake^{8k}=v_2$. Thus potentiating and dividing: $$\frac{\left(Ak\right)^8 \cdot e^{8k}}{Ak \cdot e^{8k} }=\left(Ak\right)^7=\frac{v_1^8}{v_2}$$ We now know $Ak$. From $\frac{v_1}{Ak}=e^k$ we can calculate $k$ and finally $A$.

Specifically: $Ak = \left(\frac{v_1^8}{v_2}\right)^{\frac{1}{7}}=11.339$