This is from reading a book on probabilities.
How is (3) developed?
The text reads:
From
$$(+Δ)−()=−()()Δ\tag{1}$$
We take limits
$$\frac{()}{}=−()()\tag{2}$$
Thus we have
$$()=^{−\int_0^t\lambda(\tau)(\tau)}\tag{3}$$
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Are you familiar with the method of integrating factors for solving first-order ODEs? (Because that's what has been used.) – Hans Lundmark Mar 26 '23 at 10:40
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(Also, the conclusion is only true if $V(0)=1$. Otherwise you need a factor $V(0)$ on the right-hand side.) – Hans Lundmark Mar 26 '23 at 14:56
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@Hans Lundmark, Why is that? – Steve Mar 26 '23 at 16:56
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Well, what do you get if you let $t=0$ in equation (3)? – Hans Lundmark Mar 26 '23 at 17:01
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@Hans Lundmark, If t=0, is it like this: $V(0) = e^{-(\lambda(0)-\lambda(0))}=1$? – Steve Mar 26 '23 at 17:12
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Yes, if you replace $\lambda(0)$ with $\Lambda(0)$, where $\Lambda$ is an antiderivative of $\lambda$. Or just use that $\int_0^0$ is zero no matter what you are integrating. – Hans Lundmark Mar 26 '23 at 19:36
1 Answers
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You can either use an integrating factor (since this is a first order linear ODE), as other comments suggested, or you could notice that $$(\ln y)'=\frac{y'}{y}$$So in this example, $$\frac{V'(t)}{V(t)}=-\lambda(t)\Rightarrow \ln\left|V(t)\right|=-\int \lambda(t)\, dt+c$$ and depending on your constraints and your initial conditions, $$V(t)=\exp\left(-\int \lambda(t)\, dt\right)$$
kon gram
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