Why is $$ g'(t)-cg(t)=0 \iff g(t)=Ae^{ct} $$ and how do I know that? This is a part of the heat equation that I don't understand, I must have missed this part in some other course... What do you call this and where can I read about it?
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This is called a differential equation. See this answer to learn how to solve a more general kind of differential equation than the one presented. – Git Gud Oct 20 '14 at 09:08
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This is a first order ordinary linear differential equation. Let's solve it. Notice that:
$$g'(t) - cg(t) = 0 \iff e^{-ct}g'(t) - ce^{-ct}g(t) = 0 \iff (e^{-ct}g(t))' = 0$$
That means $e^{-ct}g(t)$ is a constant, say, $A$. Hence:
$$(e^{-ct}g(t))' = 0 \iff e^{-ct}g(t) = A \iff g(t) = Ae^{ct}.$$
Ivo Terek
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$$g'(t)=c g(t) \iff \frac{g'(t)}{g(t)}=c \stackrel{\int \mathrm{d}t}{\iff} \log |g(t)|=ct+\beta \iff g(t)=e^{ct}e^{\beta}\;\stackrel{e^{\beta}:=A}{=}\;A e^{ct}$$
UserX
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At the third step it'd be better, perhaps, to write $;\log |g(t)|=ct+\beta;$ – Timbuc Oct 20 '14 at 09:16
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$g'(t)-cg(t)=0$ can be written as (with $g(t)\neq 0$), $\dfrac{g'(t)}{g(t)}=c$
But if you know you derivatives well enough, you get $\dfrac{g'}{g}=(ln(g))'$, again, with all the restriction concerning positive values etc...
So, $ln(g(t))=ct+a$, or $g(t)=Ae^{ct}$, with $A=ln(a)$
Martigan
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You arrive at $Ae^{ct}$ by solving for $\mu\left(t\right)$, an integration factor that forces the left to look like the product rule. Thus in this case $$\frac{dg}{dt}+\left(-c\right)g=0,\:\: \mu\left(x\right)=\exp\left(\int -cdt\right)=e^{-ct}$$ which leaves us with $$e^{-ct}\cdot\frac{dg}{dt}+\left(-c\right)\cdot e^{-ct}g=0\Longleftrightarrow \frac{d}{dt}\left[e^{-ct}\cdot g\right]=0...$$
bjd2385
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