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Solving a system of 2 diffrential equations with 3 unknowns.

  1. Jun 2, 2009 #1
    1. The problem statement, all variables and given/known data
    r''-(theta')^2=-g reads: second derivative of r minus the first derivative of theta squared equals negative g

    2(r')theta'+(r)theta''=0 reads: 2 times the first derivative of r times the first derivative of theta plus r times the second derivative of theta.

    both r and theta are a function of time. Oh and the derivatives of each variable are with respect to time.

    2. Relevant equations

    3. The attempt at a solution
    I really have no clue as to how I'm going to solve this system of differential equations. I'm trying to solve for r'
  2. jcsd
  3. Jun 3, 2009 #2


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    Hi torquerotates! :smile:

    (have a theta: θ and try using the X2 tag just above the Reply box :wink:)

    (and it's ok using ' to mean derivative … you needn't explain it! :wink:)
    Hint: trick is to multiply the second one by r …

    that gives you 2rr'θ' + r2θ'' = 0,

    which you can integrate to give … ? :smile:
  4. Jun 3, 2009 #3
    Im thinking about it but i just cant seem to solve it. At first I was thinking chain rule. But that clearly doesn't work.
  5. Jun 3, 2009 #4


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    To make use of Tim's hint recognize the product rule.
  6. Jun 3, 2009 #5
    Oh I see. So if i integrate, I get (r^2 )theta'=C where C is a constant. Is this the right thinking?
  7. Jun 3, 2009 #6
    Ok so solving for theta' I get theta'=C/(r^2)

    And plugging into the other equation, I get r''-(C^2)/(r^4)=-g

    But I have no clue solve this one as well. Maybe if r^4 was in the numerator, I would get somewhere.
  8. Jun 3, 2009 #7


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    Yup! :biggrin:

    (but what happened to that θ i gave you? :confused:)
    Hint: this time, multiply by r'. :wink:
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