I'm trying to prove that one is the same as the other : $$\ln y = -kt+c$$ $$y=ce^{-kt}$$
Where c is undefined and k is defined constant. I got as far as: $$y=e^{-kt+c}$$ So by what rule would c be multiplied by e? Could someone explain please? Thank you.
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$\begingroup$They are not the same, but usually, when one is solving a differential equation (I can't imagine another instance where these two would be "equivalent" in some sense), one writes:
$$\ln y = -kt + c \\ y = e^{-kt + c} \\ y = e^c e^{-kt}$$
Now $e^c$ is just a constant, so we may have given $c$ the name $a$ and then we would let $c = e^a$, so that $y =c e^{-kt}$
$\endgroup$ 1 $\begingroup$That's because they aren't! Usually, in the context of differential equations, the $A$ in $$ y=Ae^{-kt} $$ Really is a more concise stand in for some exponentiated constant, in your case $e^c$, i.e. $$ \ln(y)=-kt+c\Rightarrow y=e^{-kt+c}=e^ce^{-kt}\Rightarrow y=Ae^{-kt} $$
You can view the $A$ as absorbing constants you are adding on.
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