Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In my course notes, we are working on the stability of solutions, and in one example we start out with:

Consider the IVP on $(-1,\infty)$:

$x' = \frac{-x}{1 + t}$ with $x(t_{0}) = x_{0}$.

Integrating, we get $x(t) = x(t_{0})\frac{1 + t_{0}}{1 + t}$.

I can't produce this integration but the purpose of the example is to show that $x(t)$ is uniformly stable, and asymptotically stable, but not uniformly asymptotically stable.

But I can't verify the initial part and don't want to just skip over it.

Can someone help me with the details here?

Update: the solution has been pointed out to me and is in the answer below by Bill Cook (Thanks!).

share|cite|improve this question
up vote 3 down vote accepted

Separate variables and get $\int 1/x \,dx = \int -1/(1+t)\,dt$. Then $\ln|x|=-\ln|1+t|+C$

Exponentiate both sides and get $|x| = e^{-\ln|1+t|+C}$ and so $|x|=e^{\ln|(1+t)^{-1}|}e^C$

Relabel the constant drop absolute values and recover lost zero solution (due to division by $x$) and get $x=Ce^{\ln|(1+t)^{-1}|}=C(1+t)^{-1}$.

Finally plug in the IC $x_0 = x(t_0)=C(1+t_0)^{-1}$ so that $C=x_0(1+t_0)$ and there you go the solution is

$$ x(t) = x_0 \frac{1+t_0}{1+t} $$

share|cite|improve this answer
Wow thank you so much. I can't believe I forgot how to solve a separable first order ODE. :S Thanks very much though! – roo Oct 29 '11 at 2:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.