Solution to the differential equation $\frac{dg(0,w)}{dw}=-\lambda g(0,w)$

Question

Some background:

Let $g(x,w)$ denote the probability of $x$ changes in each interval of length $w$. Furthermore, let the symbol $o(h)$ represent any function such that $\lim_{h \to 0}[o(h)/h] = 0$. The Poisson postulates are the following:

$1. \ g(1,h) =\lambda h + o(h),$ where $\lambda$ is a positive constant and $h > 0 $.
$2. \ \sum_{x=2}^{\infty} g(x,h) = o(h).$
$3. \ $The numbers of changes in nonoverlapping intervals are independent.

We take $g(0,0) = 1$

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I understand what the postulates mean and all, but on the way to showing that $$g(x,w)=\frac{(\lambda w)^xe^{-\lambda w}}{x!}, \ \ \ \ x=1,2,3, \ldots$$

my text states that the solution to the differential equation $$\frac{dg(0,w)}{dw}=-\lambda g(0,w)$$ is $$g(0,w) = ce^{-\lambda w}$$

I'm not sure how they got there. My knowledge of solving differential equations is about 6 years behind me. I tried integrating both sides, but how would I go from here? $$g(0,w) = \int-\lambda g(0,w) \ dw$$

Answer 1

Let's write - for brevity - $h(w) := g(0,w)$. Then $h'(w) = \frac{dh}{dw} = \frac{dg(0,w)}{dw}$. The idea is to separate the variable before you integrate. So, everything containing an $h$ to one side, everything with an $w$ (no occurence of $w$ here) to the other side. That is $$ h'(w) = -\lambda h(w) $$ Dividing by $h(w)$ gives $$ \frac{h'(w)}{h(w)} = -\lambda $$ Now we integrate: $$ \int \frac{h'(w)}{h(w)} \, dw = \int \frac 1{\eta}\, d\eta = \log \eta + C = \log h(w) + C $$ where we substituted $\eta = h(w)$, i. e. $d\eta = h'(w) \, dw$. The right hand side gives: $$ -\int \lambda \, dw = -\lambda w + C $$ Equating both sides gives us: $$ \log h(w) = -\lambda w + C \iff h(w) = \exp(-\lambda w + C) = c\exp(-\lambda w) $$

Answer 2

Many ways to solve differential equations of this type. One way by noting the equation is separable: \begin{align} \frac{dg}{dw} & = -\lambda g \\ \int \frac{dg}{g} & = \int - \lambda \ dw \\ \ln g & = -\lambda w + K \\ g & = e^K e^{-\lambda w} := Ce^{-\lambda w} \end{align}