# B Separation of variables - rocket equation

1. Nov 27, 2016

### 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 :)

2. Nov 27, 2016

### Orodruin

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

3. Nov 27, 2016

### Januz Johansen

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

4. Nov 27, 2016

### Orodruin

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

5. Nov 27, 2016

### Januz Johansen

Thanks i see now :D

6. Nov 27, 2016

### Orodruin

Staff Emeritus
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 $m_i = m(v_i)$.

7. Nov 27, 2016

### 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

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8. Nov 27, 2016

### Orodruin

Staff Emeritus
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.

9. Nov 27, 2016

### 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

10. Nov 27, 2016

### Orodruin

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