Solving Differential Equations: Understanding Variable Changes

[bassplayer142]

A little thing that may be stupid but I am confused about it. Say we take any equation like f=ma
If we take the derivative of both sides then we could either have

df=m da
or
df =a dm

Are both of these valid computations. If I am looking to change the integrating variable can I use this any way I want? And would this work with any equation relating 3 or more variables?

thanks in advanced

 

[Andy Resnick]

You are basically correct: the "rocket equation" comes from F = dp/dt = d(mv)dt = m dv/dt + v dm/dt.

For what you wrote, f = ma, df = m da/dt + a dm/dt.

 

[pam]

If f=abc, df=ab dc+ ac db + bc da, and so on for an number of variables.
You just differentiate one at a time.
Of course if any factor is a constant, then its differential is zero.

 

[pam]

You are basically correct: the "rocket equation" comes from F = dp/dt = d(mv)dt = m dv/dt + v dm/dt.

The "rocket equation" is a bit different, because the exit velocity of the gas enters instead of just v in the dm/dt term.

 

[bassplayer142]

thanks this clears some up. But as I said before could I take df=mda to subsitute df with da to integrate with respect to a, and in the same problem could I take df=adm to integrate with respect to m.

so you could take s=rTheta and make it ds=rdtheta?

I'm just seeing how flexible I can be when substituting vaiables to integrate or differentiate with.

Edit, I just realized that what you did there was the product rule which makes sense. Is what I just said above wrong then?>

 

[pam]

A derivative of all product of N variables will have N terms, each term being differentilated once. So d(r theta)=r dtheta+theta dr.