Ordinary Differential Equation - Comparing 2 Solutions

1. Apr 27, 2014

AntSC

1. The problem statement, all variables and given/known data

$${g}'\left ( s \right )+\mu g\left ( s \right )={f}'\left ( -s \right )+\mu f\left ( -s \right )$$
Integrate up to get
$$g\left ( s \right )=-f\left ( -s \right )+2\mu e^{-\mu s}\int_{-s}^{\infty }e^{-\mu {s}'}f\left ( {s}' \right )d{s}'$$

2. Relevant equations

As above

3. The attempt at a solution

I've seen a few attempts at this, one that arrives at the correct solution, which is the method i used and had no trouble with. However, another attempt i've seen, which doesn't yield the correct solution is causing me much headache as i can't work out where it's going wrong. Some pointing out here would be really helpful. Thanks

Using the following integrating factor on the equation above
$$e^{\int \mu ds}=e^{\mu s}$$
$$e^{\mu s}{g}'\left ( s \right )+e^{\mu s}\mu g\left ( s \right )=e^{\mu s}\left [ {f}'\left ( -s \right )+\mu f\left ( -s \right ) \right ]$$
At this point i wrote
$$\frac{d}{ds}\left [ e^{\mu s}g\left ( s \right ) \right ]=e^{\mu s}\left [ {f}'\left ( -s \right )+\mu f\left ( -s \right ) \right ]$$
$$\left. e^{\mu {s}'}g\left ( {s}' \right ) \right|_{-\infty }^{s}=\int_{-\infty }^{s}e^{\mu {s}'}\left [ {f}'\left ( {-s}' \right )+\mu f\left ( {-s}' \right ) \right ]d{s}'$$
Then to skip a few steps i did by parts integration to get the following (can include missing steps if needed)
$$e^{\mu s}g\left ( s \right )=-e^{\mu s}f\left ( -s \right )+ 2\mu \int_{-\infty }^{s}f\left ( {-s}' \right )e^{\mu {s}'}d{s}'$$
$$g\left ( s \right )=-f\left ( -s \right )+ 2\mu e^{-\mu s}\int_{-\infty }^{s}f\left ( {-s}' \right )e^{\mu {s}'}d{s}'$$
Swapping limits and changing 's' for '-s' gives
$$g\left ( s \right )=-f\left ( -s \right )+ 2\mu e^{-\mu s}\int_{-s}^{\infty }f\left ( {s}' \right )e^{-\mu {s}'}d{s}'$$
As required

The other route wrote took the following path
$$\frac{d}{ds}\left [ e^{\mu s}g\left ( s \right ) \right ]e^{-\mu s}=\frac{d}{ds}\left [ f\left ( -s \right )e^{-\mu s} \right ]e^{\mu s}$$
$$\frac{d}{ds}\left [ e^{\mu s}g\left ( s \right ) \right ]=-\frac{d}{ds}\left [ f\left ( -s \right )e^{-\mu s} \right ]e^{2\mu s}$$
$$\left. e^{\mu {s}'}g\left ( {s}' \right ) \right|_{-\infty }^{s}=\int_{-\infty }^{s}-e^{2\mu {s}'}\frac{d}{ds}\left [ f\left ( {-s}' \right )e^{-\mu {s}'} \right ]d{s}'$$
$$e^{\mu s}g\left ( s \right )=\left. -e^{2\mu {s}'}f\left ( {-s}' \right )e^{-\mu {s}'}\right|_{-\infty }^{s}+ \int_{-\infty }^{s}2\mu e^{2\mu s}f\left ( {-s}' \right )e^{-\mu {s}'}d{s}'$$
$$e^{\mu s}g\left ( s \right )=-e^{\mu s}f\left ( -s \right )+ 2\mu e^{2\mu s}\int_{-\infty }^{s}f\left ( {-s}' \right )e^{-\mu {s}'}d{s}'$$
$$g\left ( s \right )=-f\left ( -s \right )+ 2\mu e^{\mu s}\int_{-\infty }^{s}f\left ( {-s}' \right )e^{-\mu {s}'}d{s}'$$
Swapping limits and changing 's' for '-s' gives
$$g\left ( s \right )=-f\left ( -s \right )+ 2\mu e^{\mu s}\int_{-s}^{\infty }f\left ( {s}' \right )e^{\mu {s}'}d{s}'$$
This is giving a sign swap in the final solutioin that is having my head spin. Can't see the route of this. Can anyone spot the mistake?

2. Apr 27, 2014

HallsofIvy

Staff Emeritus
Swapping $\int_{-s}^\infty ds'$ for $\int_{-\infty}^s ds'$ is equivalent to making the substitution $u= -s'$. Then $du= -ds'$. When $s'= -\infty$, $u= \infty$ and when $s'= s$, $u= -s$.

So $$\int_{-s}^\infty f(-s')e^{-\mu s'}ds'= \int_u^{-\infty}f(u)e^{\mu u}(-du)= \int_{-\infty}^u f(u)e^{\mu u}du$$