Explanation of differential equation solution in survival analysis proof

I follow all the steps in the below derivation until the third to last line, "solving this differential equation for the survival analysis function shows that..."

Questions

I never took differential equations, though I did have a small amount of exposure 'on the job'. Could anyone explain what is happening in this step? Is this a PDE or ODE (my guess is ODE)? Is there a commonly known rule in that justifies this step in the derivaiton?

Edit: For the comment/request, here's the jump I'm unclear on:

$$\lambda(t) = \frac{-S'(t)}{S(t)}$$

Becomes

$$S(t) = -\int_{0}^{s} \lambda(s)ds$$

• Please type your question as text, do not just post a photograph or screenshot (see here). (I have problems reading the very small fonts in the image, and there are users of this site using screenreaders which do not work with images ... May 13 '21 at 2:12
• data.princeton.edu/wws509/notes/c7.pdf May 13 '21 at 2:16
• You don't have to know how to solve the differential equation: you only need to know how to check the solution (which is a matter of rote differentiation). Techniques of solving ODEs are the province of Mathematics.
– whuber
May 13 '21 at 13:04

1 Answer

The differential equation is a separable equation, so we can use some first year calculus to solve it.

Note that the equation for lambda can be written as

$$\lambda (t) = -\dfrac{d}{dt} \log(S(t))$$

Integrating both sides

$$\int _0^t \lambda (s) \, ds = -\log(S(t))$$

via algebra

$$S(t) = \exp \left( - \int_0^t \lambda (s) \, ds\right)$$

• I think I'm missing something super basic in rewriting $\lambda(t)$ as $-\frac{d}{dt}log(S(t))$. The best I came up with from the starting place $-\frac{S'(t)}{S(t)}$ is $-\frac{\frac{d}{dt}S(t)}{S(t)}$, but I'm not clear where the log comes in. May 13 '21 at 2:40
• Basic chain rule. $(\log(f(x))' = f'(x)/f(x)$. I just went backwards. May 13 '21 at 2:41
• Last question, how are the integral limits chosen/found? The author integrates from [0,s] the first time then [0,t] the second, which seems unusual. May 13 '21 at 15:38
• The lower limit is 0 because I assume $t$ stands for time and so negative time doesn't make sense (at least without some sort of reference time). The rationale for how the limits of the integral work is a topic for a first year calculus book-- and is too lengthy for explanation here. I will leave you to research that independently. May 13 '21 at 16:30
• ^Demetri, you might have noticed that the upper bounds on the integral switch. At the first the upper bound is s, then it later is t. Not sure if you caught this subtle detail, but it's at the root of my confusion related to me previous comment. May 14 '21 at 16:33