Separation of variables - rocket equation

[Januz Johansen]

hello there
Im trying to do a derivation of tsiolkovsky's rocket equation, but i got stuck at the step when i have to use separation of variables (marked with red in the pic), i used maple to solve it, so i could get on with it, but i want to understand what is happening to solve this, so can anyone explain how to solve this step with separation of variables?
Thanks :)

 

[Orodruin]

It is exactly what they have done. What step in particular do you have problems with?

 

[Januz Johansen]

It is exactly what they have done. What step in particular do you have problems with?

ok thanks so i have done some right ;)

im having trouble explaining what is happening, or i think i do.
I can explain the first steps, just isolate the variables on each side of the equation.
But what rules are used/how is this integrated (the bordered step)
Thanks :D

 

[Orodruin]

You integrate both sides between the same points. It is essentially making an integration and then making a change of variables.

 

[Januz Johansen]

You integrate both sides between the same points. It is essentially making an integration and then making a change of variables.

Thanks i see now :D

 

[Orodruin]

Oh, and the integration boundaries on the LHS should be $v_f$ to $v_i$. In this particular example it does not matter for the result because the integrand is constant. You then make a change of variables to $m(v)$ and use $m_f = m(v_f)$ and $m_i = m(v_i)$.

 

[Januz Johansen]

Hello
Do you mean like this?

im not 100% sure what you mean with the change of variables
Again thank you for helping

 

[Orodruin]

Yes. Consider the integral
$$
\int_{v_f}^{v_i} dv.
$$
Now, you know that $m$ is a function of $v$ so change variables to $m$. The integral changes to
$$
\int_{m(v_f)}^{m(v_i)} \frac{dv}{dm} dm.
$$
Insert the known differential equation and perform the new integral.

 

[Januz Johansen]

so i have it like so:

or do i get -1/u*m(vi)-m(vf)?
thank you for your patience and help

 

[Orodruin]

No, you have the wrong integration boundaries in the first integral. They are what I said in my post.