First order differential equation involving a square root

In summary: That would seem to require \frac{d}{dt} \left( \frac43 \pi \rho(t) R^3(t) \right) = 0 and hence \frac 1\rho \frac{d\rho}{dt} = - \frac{3}{R} \frac{dR}{dt} which can be wrangled into the ODE in the OP by taking \frac{d\rho}{dt} = 3 \rho\sqrt{\frac{GM}{R^3}}= 3 \rho \sqrt{\frac{4\pi G \rho}{3}}. I've no idea if that is physically
  • #1
Lilian Sa
18
2
Summary:: solution of first order derivatives

we had in the class a first order derivative equation:
##\frac{dR(t)}{dt}=-\sqrt{\frac{2GM(R)}{R}}##
in which R dependent of time.
and I don't understand why the solution to this equation is:
##R(t)={\frac{3}{2}}^{\frac{3}{2}}{2GM}{\frac{1}{3}}{(t_*-t)}^{\frac{2}{3}}##
thanks

[Moderator's note: moved from a technical forum.]
 
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  • #2
Are you saying you don't know how to plug that in and check it's a solution, or you don't know how to derive it without already knowing the solution?

Also, why is there an R in the numerator and denominator of the right hand side? Is there a typo?
 
  • #3
What is the function ##M(R)##? Note that this differential equation in general is separable.
 
  • #4
Office_Shredder said:
Are you saying you don't know how to plug that in and check it's a solution, or you don't know how to derive it without already knowing the solution?

Also, why is there an R in the numerator and denominator of the right hand side? Is there a typo?

let me be honest, both.
I just did not understand it.
In the numerator there is M as a function of R, and in the denominator there is the function R.
 
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  • #5
M is an arbitrary function of R? How does it relate to the M in the solution that you are given? Is that a constant?

I assumed it's supposed to be mass and this has something to do with orbital rotation given the general shape of it, which makes me surprised that M is a function of R to begin with.
 
  • #6
Office_Shredder said:
I assumed it's supposed to be mass and this has something to do with orbital rotation given the general shape of it, which makes me surprised that M is a function of R to begin with.
Same here. The only thing I can come up with is that the scenario represents a rocket whose mass steadily decreases as the fuel is used up. However, that's just a guess, as there was no additional information provided in the problem statement.
 
  • #7
[itex]M[/itex] being an aribtrary function of [itex]R[/itex] is not consistent with the proposed solution.

Rearranging the ODE gives [tex]
\frac12 \left(\frac{dR}{dt}\right)^2 = \frac{GM(R)}{R}[/tex] which is a gravitational potential (assuming that the tangential velocity vanishes and [itex]M[/itex] is constant).
 
  • #8
Mentor note: Fixed the LaTeX
It is a collapse of a non relativistic star under its own gravity.
##M(R)=4\pi \int\rho(t)r^2dr##
 
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  • #9
Lilian Sa said:
Mentor note: Fixed the LaTeX
It is a collapse of a non relativistic star under its own gravity.
##M(R)=4\pi \int\rho(t)r^2dr##
This doesn't make sense to me. M is a function of R, but R doesn't appear in the integral.
 
  • #10
Mark44 said:
This doesn't make sense to me. M is a function of R, but R doesn't appear in the integral.
replace r in R.
 
  • #11
Lilian Sa said:
replace r in R.
Is this what you mean?
##M(R)=4\pi \int\rho(t)R^2dR##
 
  • #12
Ok I am on it now, but why we have in the answer (t_*-t)^(2\3)
 
  • #13
Surely it's supposed to be ##M(R)= \int_0^R 4\pi \rho(r) r^2 dr##.

I'm still not sure what M in the proposed solution is supposed to be, since I guess it's not supposed to depend on R.

Edit to add: wait, ##\rho(r)## maybe is supposed to be ##\rho(t)## because as the star collapses its density increases. But if that's the case the mass is probably constant the whole time?
 
  • #14
Office_Shredder said:
Surely it's supposed to be ##M(R)= \int_0^R 4\pi \rho(r) r^2 dr##.

I'm still not sure what M in the proposed solution is supposed to be, since I guess it's not supposed to depend on R.

Edit to add: wait, ##\rho(r)## maybe is supposed to be ##\rho(t)## because as the star collapses its density increases. But if that's the case the mass is probably constant the whole time?
yes rho depends on time
 
  • #15
Office_Shredder said:
Surely it's supposed to be ##M(R)= \int_0^R 4\pi \rho(r) r^2 dr##.

I'm still not sure what M in the proposed solution is supposed to be, since I guess it's not supposed to depend on R.

Edit to add: wait, ##\rho(r)## maybe is supposed to be ##\rho(t)## because as the star collapses its density increases. But if that's the case the mass is probably constant the whole time?

That would seem to require [tex]\frac{d}{dt} \left( \frac43 \pi \rho(t) R^3(t) \right) = 0[/tex] and hence [tex]
\frac 1\rho \frac{d\rho}{dt} = - \frac{3}{R} \frac{dR}{dt}[/tex] which can be wrangled into the ODE in the OP by taking [tex]
\frac{d\rho}{dt} = 3 \rho\sqrt{\frac{GM}{R^3}}
= 3 \rho \sqrt{\frac{4\pi G \rho}{3}}.[/tex] I've no idea if that is physically justified.
 
  • #16
Lilian Sa said:
Mentor note: Fixed the LaTeX
It is a collapse of a non relativistic star under its own gravity.
##M(R)=4\pi \int\rho(t)r^2dr##
I would expect the equation to be
$$ \bigg(\frac{dR}{dt}\bigg)^2 = \frac{GM}{R} $$
if it where to describe the gravitational collapse of a non-relativistic star, as a fluid parcel located at distance ##R## is gravitational bound (the Viral theorem). But I might be wrong, I haven't really studied these type of systems since I went to gymnasium (college prep school, common in europe).

Nonetheless, I believe that the OP has made a typo and intended to write
$$ R(t) = \Big(\frac{3}{2}\Big)^\frac{2}{3}(2GM)^\frac{1}{3}(t_\ast - t)^\frac{2}{3}$$
as the solution. Which suggest that ## M(R) ## is taken to be constant. If so, the original ODE can be rewritten as
$$ \frac{d}{dt}R^\frac{3}{2} = - \frac{3}{2}\sqrt{2GM} $$
which can be integrated to give
$$ R(t) = \bigg(\frac{3}{2}\sqrt{2GM}(t_0 - t) + R_0^\frac{3}{2}\bigg)^\frac{2}{3} $$
and simplifies to
$$ R(t) = \Big(\frac{3}{2}\Big)^\frac{2}{3}(2GM)^\frac{1}{3}(t_0 - t)^\frac{2}{3},$$
assuming that we are given the "initial" condition ## R(t_0) = R_0 = 0 ##.
 
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1. What is a first order differential equation involving a square root?

A first order differential equation involving a square root is an equation that contains a square root function and its derivative. It is typically in the form of dy/dx = f(x,y), where f(x,y) is a function that involves a square root.

2. How do you solve a first order differential equation involving a square root?

To solve a first order differential equation involving a square root, we use the method of separation of variables. This involves isolating the variables on opposite sides of the equation and then integrating both sides to find the solution.

3. Can a first order differential equation involving a square root have multiple solutions?

Yes, a first order differential equation involving a square root can have multiple solutions. This is because the square root function has two possible solutions for a given input, positive and negative. Therefore, the equation can have two distinct solutions.

4. What are some real-world applications of first order differential equations involving square roots?

First order differential equations involving square roots have many real-world applications, such as in physics, engineering, and economics. For example, they can be used to model the motion of a falling object with air resistance, the growth of a population with limited resources, or the decay of a radioactive substance.

5. Are there any special techniques for solving first order differential equations involving square roots?

Yes, there are some special techniques that can be used to solve certain types of first order differential equations involving square roots. These include the substitution method, the Bernoulli method, and the exact method. It is important to carefully analyze the equation to determine the most appropriate method to use.

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