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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!).

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1 Answer 1

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

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

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Wow thank you so much. I can't believe I forgot how to solve a separable first order ODE. :S Thanks very much though! –  Kyle Schlitt Oct 29 '11 at 2:07

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