Integrating complicated functions of time

[ck99]

Homework Statement

m' = G O^1/2 m^(-1/2)

where m' = dm/dt. Find m by integrating this expression wrt to time (indefinite).

Homework Equations

We have G = m'/m

and O = constant/G^2

and m is a function of time m(t)

The Attempt at a Solution

I know that integral of m'/m = ln m, so that means I can integrate the G part. I have tried to integrate O by taking the constant outside the integral, which leaves an integral of [1/(G^2)]^(1/2) or just [G^(-2)]^(1/2) or [(m'/m)^(-2)]^(1/2) I am not sure what form is easiest to work with, and have no idea how to start with this part! Could I use the substitution rule? I can't see how...

The last part is an integral of m^(-1/2) which is 2m^(1/2).

I also have no idea how to combine the three different parts together to give the final equation for the integral.

 

[phyzguy]

First, put it all in terms of m and m'. If you know G in terms of m and m', and O in terms of G (which you know in terms of m and m'), then you can eliminate G and O, and get it in the form m' = f(m), where f is some function only of m (and perhaps some constants). Then you can integrate to find m in terms of t.

 

[ck99]

Thanks phyzguy. I have been having a go at that, a lot of the m terms cancel out, so I am left with

a' = a^(-0.5) {times a constant}

Again a is some function of time. How do I integrate this wrt time t, without knowing how a and t are related? I think I am looking for the opposite of the "total differential", where you can just put a'(t) to indicate the derivative of some arbitrary function a of t. How do I do this in integration?

 

[phyzguy]

If you have a' = a^(-n), then you can write:
$$\frac{da}{dt} = a^{-n}$$
$$a^n da = dt$$
$$\int a^n da = \int dt = t + C$$

Hopefully you can do the a integral, then solve for a as a function of t.