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Some Tricky ODE's

  1. Nov 5, 2009 #1
    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?
     
  2. jcsd
  3. Nov 5, 2009 #2

    Mark44

    Staff: Mentor

    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.
     
  4. Nov 6, 2009 #3
    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.
     
  5. Nov 6, 2009 #4

    Mark44

    Staff: Mentor

    In your second step above, you forgot the constant of integration. You should have (2/3)a3/2 = sqrt(C)*t + D
     
  6. Nov 6, 2009 #5
    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?
     
  7. Nov 6, 2009 #6

    Mark44

    Staff: Mentor

    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?
     
  8. Nov 6, 2009 #7
    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]
     
  9. Nov 6, 2009 #8

    Mark44

    Staff: Mentor

    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?
     
  10. Nov 6, 2009 #9
    i need to find the solution. but im 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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