# Solving DE- which approach and more importantly - why

1. Apr 18, 2013

### CJDW

My end goal is to solve for $G(m)$ in terms of the other functions, but first I have to solve the DE :

$\frac{d}{dm}[F(m) G(m)] = (\frac{d}{dm}F(m)) D(m)$.

What I've done is to say (using integration by parts)

$F(m) G(m) = F(m)D(m) - \int F(m) (\frac{d}{dm}D(m)) dm$.

This is one method I tried. Another approach would be a separation of variables, which I tried as...

$\frac{d}{dm}[F(m) G(m)] = \frac{dF}{dm} G + F \frac{dG}{dm} = \frac{dF}{dm} D$

$\ln |F| = \int \frac{dG}{D - G} dm$ and I believe that the RHS cannot be evaluated without specifying functions $D$ and $G$ (if the RHS can be evaluated without specification - please let me know...and please show me the way).

Help on understanding the correct method would be extremely appreciated.

2. Apr 18, 2013

### Simon Bridge

If you just need an expression for G in terms of F and D, then the first one looks more like it gets you there.

I noticed that $$\renewcommand{\dd}[2]{\frac{d #1}{d #2}} \renewcommand{\dm}[1]{\frac{d #1}{dm}} G\dm{F}+ F\dm{G} = D\dm{F}$$ rearranges to $$\dm{G}= \frac{D-G}{F}\dm{F}$$ ... integrate both sides wrt m ... but it's early and I havn't had my coffee yet...

3. Apr 19, 2013

### CJDW

Thanks for your suggestion, but I think I'll stick with the integration by parts method, since I have convenient expressions for $F(m) (\frac{d}{dm}D(m))$ that I will need to integrate.

Once again, thanks for the help.

4. Apr 20, 2013

### JJacquelin

Hi !
it's a separable EDO :

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5. Apr 20, 2013

### Simon Bridge

Yeah - that's where I was going ;)