Master Non-Seperable ODEs for Physics: A Step-by-Step Guide | Crep Inc.

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In summary, the conversation discusses the attempt to solve a physics problem involving a rocket with thrust, drag, and acceleration due to gravity. The problem involves a second-order equation and a generalized function of the first derivative, making it more difficult to solve. The conversation explores potential methods of solving the problem, including using an integrating factor or making a transformation. The assumption of the rocket being incredibly massive compared to its fuel is also discussed. The conversation concludes with the suggestion of using numerical methods to solve the problem.
  • #1
crepincdotcom
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Hey,

I've been teaching myself some DEs that I can use for physics and whatnot. I am comfertable with seperable equations, but I can't figure out how to solve this problem.

Let's assume we have some rocket with thrust F(t) and drag r(v), plus acceleration due to gravity, g=9.8 m/s/s.

Overall acceleration: (m is the mass, assume constant)

[tex]a(t)=\frac{F(t)}{m} - \frac{r(v)}{m} - g[/tex]

[tex]\frac{dv}{dt}=\frac{F(t)}{m} - \frac{r(v)}{m} - g[/tex]

Now as you can see, we can't move dt over to the other side, because there are multiple terms there. Can we simply distribute it across them, and get:

[tex]\int{dv}=\int{\frac{F(t)dt}{m}} - \int{\frac{r(v)dt}{m}} - \int{gdt}[/tex]

Also, we neet to relate v to t in the r(v) term, but we don't have a v(t)...

Thanks,

-Jack Carrozzo
jack _{at}_ crepinc.com
http://www.crepinc.com/
 
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  • #2
You effectively have a second-order equation, with a generalized function of the first derivative. You will need to be more specific about r(v) if you want to make any progress. If r(v) is linear, you're in luck - otherwise it may be quite tough to solve.
 
  • #3
[tex]r(v)=(\frac{1}{2}rAc)v^2[/tex]

or simply

[tex]r(v)=kv^2[/tex]

which is not linear, per se.

How would I go about this?

Thanks,

-Jack Carrozzo
jack _{at}_ crepinc.com
http://www.crepinc.com/
 
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  • #4
I'd wager that you could probably use an integrating factor type equation or something along those lines.
 
  • #6
Sorry - my bad. What you actually have (if r(v) is proportional to v^2) is a Riccati equation of the form

[tex]v^{\prime} + v^2 + f(t) = 0 [/tex]

in v (give or take a constant or two). This is not generally soluble for any f(t). One thing you can do is make the transformation

[tex]v = \frac{u^{\prime}}{u}[/tex]

and sub, giving you the Hill equation

[tex]u^{\prime \prime} + f(t) u = 0 [/tex]

which only helps if you actually know a solution and can therefore work backwards.

There are solutions of this Riccati equation for specific forms of f(t) - perhaps if you could give us some clue what the thrust term might be, we could explore it further.

sorry for the first (slightly misleading) post.

edit: as a guide to what sort of Riccati equations are soluble, you can check out

http://eqworld.ipmnet.ru/en/solutions/ode/ode-toc1.htm

as they have some special cases listed.
 
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  • #7
Thanks, I need to let that sink in for a bit before I can try to apply it.

Matthew Rodman said:
perhaps if you could give us some clue what the thrust term might be, we could explore it further.

The thrust equation is for the most part a type of step function: x Newtons for 0.5s, y Newtons for 2 sec, and z Newtons for 4 secs. That's a bit hard to express in an equation unless we use peice-wise.

Which come to think of it may work... if we solve the problem with T constant from t=0 to 0.5, then 0.5 to 2.5 and 6.5, each time carrying v0, perhaps we could solve it.

What do you think?

-Jack Carrozzo
jack _{at}_ crepinc.com
http://www.crepinc.com/
 
  • #8
By the way - the equation is slightly doubtful - for a rocket you need to include the loss of mass due to fuel expenditure. Therefore m = m(t). You'll have to think about how this fits in - i.e., is the rate of fuel loss proportional to time?

Basically, you have to resort to a momentum equation, e.g.

[tex] \frac{d (m(t)v)}{dt} = F(t) - r(v) - m(t)g [/tex]
 
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  • #9
I was using the assumption that the mass lost compared to the total mass of the vehicle was negligable in order to make calculations easier, but if I were to include it, it would be some function of the form

[tex]m(t)=m_{v} - km_{p0}t[/tex]

or the mass of the vehicle plus the initial mass of the propellant decreasing linearly with time.

-Jack Carrozzo
jack _{at}_ crepinc.com
http://www.crepinc.com/
 
  • #10
^actually you'll have

dp/dt=dm/dt v +dv/dt m

so you can plug in the terms that give you the force.

F(t)-r(v)=v dm/dt + m dv/dt

so

F(t)- r(v)- v dm/dt = m dv/dt

so

F(t)/m(t) -r(v)/m(t) -v/m(t) dm/dt =a

as I recall rocketry problems are usually solved by a series of applications of Newtons laws and conservation of energy, ie no plug and play differential equations.
 
  • #11
out of curiosityin this problem is it allowed for us to assume that the rocket is incredibly massive in comparison to its fuel, thus assuming that mass is a constant?

also this appears to form an integral equation withintegral r(v) dt= v + q(t)

where q(v) is all of the other t terms grouped together.I think an equation of this form can usually be solved with a laplace transform correct?
 
  • #12
truthfully I've never heard of laplace transforms, so I'll take your word for it.

I've only ever seen the problem addressed as a numerical method wherein each part is solved for some small dt. I wanted to do it with diff EQs if possible.

Thanks guys

-Jack Carrozzo
jack _{at}_ crepinc.com
http://www.crepinc.com/
 

1. What is a non-separable ODE?

A non-separable ODE (Ordinary Differential Equation) is a type of differential equation where the variables cannot be separated and solved independently. This means that the equation cannot be written in the form of y = f(x) and requires more advanced techniques to solve.

2. How do I solve a non-separable ODE?

The most common method for solving a non-separable ODE is using numerical methods such as Euler's method or Runge-Kutta methods. These methods involve approximating the solution through a series of steps rather than finding an exact solution.

3. Can a non-separable ODE have an analytic solution?

Yes, some non-separable ODEs can have an analytic solution, but they are rare and usually involve special cases or assumptions. In most cases, an analytic solution is not possible, and numerical methods must be used.

4. Are there any real-world applications for non-separable ODEs?

Yes, non-separable ODEs have many real-world applications in fields such as physics, engineering, and economics. They are commonly used to model systems that involve rates of change, such as population growth, chemical reactions, and electrical circuits.

5. Is there any software available to help solve non-separable ODEs?

Yes, there are many software packages that can help solve non-separable ODEs, such as MATLAB, Mathematica, and Maple. These programs have built-in functions and methods specifically designed for solving differential equations, including non-separable ones.

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