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