Solving Tricky ODEs: \dot{a}^2=\frac{C}{a}

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Homework Help Overview

The discussion revolves around solving the ordinary differential equation (ODE) given by \(\dot{a}^2=\frac{C}{a}\). Participants are exploring the relationship between the variables and the proposed solution \(a=\frac{1}{2}C(1-\cos{\eta})\), while grappling with the implications of the squared derivative.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss whether the goal is to find the solution or to verify a given solution. Some suggest splitting the ODE into two separate equations for analysis. Others express confusion regarding the interpretation of variables and the role of constants of integration.

Discussion Status

There is active engagement with various approaches to the problem, including attempts to derive the solution and verify its correctness. Some participants are questioning the assumptions regarding the independent variable and the interpretation of the derivative notation.

Contextual Notes

Participants note the potential confusion arising from the use of different variables (e.g., \(\eta\) versus \(t\)) and the lack of clarity in the problem statement regarding whether to find or verify the solution. There is also mention of integrating over specific intervals, which may affect the interpretation of the problem.

latentcorpse
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I'm having difficulty showing the solution of

[itex]\dot{a}^2=\frac{C}{a}[/itex] is [itex]a=\frac{1}{2}C ( 1 - \cos{\eta} )[/itex]
i think it's the fact that [itex]\dot{a}[/itex] is squared that's throwing me.
any advice?
 
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Do you have to find that solution or just confirm that it is a solution? If it's the latter, that's pretty easy. If it's the former, I would split it into two differential equations:
da/dt = sqrt(C) a-1/2 and da/dt = -sqrt(C) a-1/2 and go from there.
 
i get

[itex]a^{\frac{1}{2}} da = C^{\frac{1}{2}} dt \Rightarrow \frac{2}{3} a^{\frac{3}{2}} = C^{\frac{1}{2}} t \Rightarrow \frac{4}{9} a^3 = C t^2 \Rightarrow a= \left( \frac{9}{4} C t^2 \right)^{\frac{1}{3}}[/itex]

and the same from the other eqn as the negative dissappears at the squaring step.
 
In your second step above, you forgot the constant of integration. You should have (2/3)a3/2 = sqrt(C)*t + D
 
ok then

[itex]a^{\frac{3}{2}}=\frac{3}{3}C^{\frac{1}{2}}t+E[/itex]
[itex]a^3=\frac{9}{4}Ct^2+3EC^{\frac{1}{2}}t+E^2[/itex]

how do we get a cos out of all this? is it to do with expansions of cosine or something?
 
Two things:
1. a3/2 = (3/2) (sqrt(C)t + D) ==> a3 = (9/4) (sqrt(C)t + D)2
2. Your original problem has [tex]\dot{a}[/tex], which is usually interpreted to mean the derivative with respect to t. The right side of that equation has "eta." Is the derivative supposed to be with respect to eta? In any case, I did a quick check, and I'm not getting that a = C/2(1 - cos(eta)) is a solution of that equation. Can you verify that what you had in the OP is accurate?
 
my apologies. we should be trying to get [itex]a=(\frac{9C}{4})^{\frac{1}{3}} \tau^{\frac{2}{3}}[/itex] so the derivative is with respect to [itex]\tau[/itex] which makes sense because i should be dealing with proper time.

and we were integrating a from 0 to a and [itex]\tau[/itex] from 0 to [itex]\tau[/itex] so it works.

the one i am stuck on is finding [itex]a=\frac{1}{2}C(1-\cos{\eta})[/itex] is a solution of [itex]\dot{a}^2-\frac{C}{a}+1=0[/itex]
 
For the sake of easier typing, let's dispense with the Greek letters.

For the corrected problem, it's still not clear to me what you need to do. Your wording of find blah is a solution of blah blah is confusing.
Which one of these do we need to do?
a) Find the solution (which happens to be what you show).
b) Verify that a = C/2(1 - cos t) is a solution.

Just as before, if someone hands you a solution, it is pretty simple to verify whether it is a solution or not.

If the problem is to actually find the solution, you have da/dt = +/-sqrt(C/a - 1). Break into two cases, one for each sign, and separate and solve.



As I mentioned before, "a dot" usually indicates a derivative w.r.t. time, so is there some point in having the independent variable be eta?
 
i need to find the solution. but I am reading out of a book so it told me the answer and just skipped the working. i just want to "fill in the blanks" so to speak.

anyway they use eta in the book which is confusing to me also.
 

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