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Solving a system in polar coordinates

  1. Mar 3, 2009 #1
    Hey Everybody.

    for the system:

    [tex] r' = r(1-r) [/tex]
    [tex] \theta' = 1 [/tex]

    with

    [tex]r(0) = x; \theta(0) = 0 [/tex];

    the answer is

    [tex]r(t) = \frac{xe^{t}}{1-x+xe^{t}} [/tex]

    [tex]\theta(t) = t [/tex]

    This answer was given in class as part of a process, and I can't remember how that answer is calculated. Can someone help me?
     
  2. jcsd
  3. Mar 3, 2009 #2

    HallsofIvy

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    Pretty basic. You have two completely separate equations- just solve each one separately. [itex]\theta'= 1[/itex] so, integrating, [itex]\theta= t+ C[/itex]. Since [itex]\theta(0)= 0+ C= 0[/itex], C= 0 and [itex]\theta(t)= t[/itex]. [itex]r'= r(1- x)[/itex] is only slightly harder:[itex]dr/dt= r(1-r)[/itex] can be rewritten as
    [tex]\frac{dr}{r(1- r)}= dt[/tex]
    and that can be integrated by "partial fractions".
     
  4. Mar 4, 2009 #3
    [tex] \int\frac{dr}{r(1-r)} \ = \ \int dt [/tex]

    where [tex] \int\frac{dr}{r(1-r)} \ = \ \int\left(\frac{1}{r} \ + \ \frac{1}{1-r} \right)dr \ = \ ln(r) \ - \ ln(1-r) + C_{1} [/tex]

    so [tex] ln(r) \ - \ ln(1-r) + C_{1} = t + C_{2} [/tex]

    [tex] 2r -1 + C_{3} = e^{t} + C_{4} [/tex]

    [tex] r = \frac{e^{t} + C + 1}{2} [/tex]

    [tex] r(0) = \frac{e^{0} + C + 1}{2} = x [/tex]

    [tex] C = x-1 [/tex];

    [tex] r = \frac{e^{t} + x}{2} [/tex];

    I'm still not getting the right answer.
     
    Last edited: Mar 4, 2009
  5. Mar 4, 2009 #4

    HallsofIvy

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    No. ln(r/(1-r))= t+ C' (I've combined your C1 and C2).

    Now r/(1-r)= Cet. (C is eC')
    So r= Cet- rCet, (1- Cet)r= Cet, and
    [tex]r(t)= \frac{Ce^{t}}{1- Ce^{t}}[/tex]
    Since r(0)= x,
    [tex]r(0)= \frac{C}{1- C}= x[/tex]
    so C= (1- C)x, C= x- Cx, C+ Cx= C(1+ x)= x so
    [tex]C= \frac{x}{1+x}[/tex]

     
  6. Mar 4, 2009 #5
    Thanks so much, I get it now.
     
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