How Do You Integrate dy/dt with Respect to y in Ballistic Equations?

[gendoikari87]

[URL]http://latex.codecogs.com/gif.latex?(d^2Y)/(dTdY)=%20-c(dY/dT)-9.8(dT/dY)[/URL]

basically how do you integrate dy/dt with respect to y, I know dy/dt integrated with respect to t is simply Y, but the other I have no idea.

background: C is a constant that is a function of air pressure and is from the drag equation.

 

[Curl]

the equation you posted doesn't make sense to me. but to answer your second question
"how do you integrate dy/dt with respect to y"
you should know that dy/dt = 1/(dt/dy) so you can integrate with respect to y if you can write t=t(y).

 

[gendoikari87]

actually I re wrote the thing in an easier form but it gets messy if you just use separation of variables

dv/dt=-fV^2-9.8

I can't remember how to do this using ODE, any help?

 

[BruceW]

I'm having trouble understanding your first problem.
\frac{d^2Y}{dTdY}=-c \frac{dY}{dT} -9.8 \frac{dT}{dY}
The bit on the left-hand side could be rearranged:
\frac{d}{dT} ( \frac{dY}{dY} )
which is equal to zero, right? So then you'd have:
(\frac{dY}{dT})^2 = \frac{-9.8}{c}
Does this look right? Are you trying to model an actual physical process, or is it just a maths problem?

And your second equation:
\frac{dv}{dt}=-f V^2 -9.8
Is f a constant or a function? And is v the same thing as V? If so, then the equation is nonlinear in v, which doesn't have a general method to solve, although maybe its possible...

 

[rcgldr]

\frac{dv}{dt}=-c v^2 -9.8

\frac{dv}{v^2 + 9.8 / c}= -c \ dt

This is similar to the formula used for free fall, wiki article:

wiki_free_fall.htm