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Differential equatn involving cos^2 and sin^2 in denominator

  1. Nov 29, 2011 #1

    sharks

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    The problem statement, all variables and given/known data
    (9cos^2(x) + 4sin^2 (x)).dy/dx = 6

    The attempt at a solution
    Rearranging:

    (1/6).dy = 1/(9cos^2(x) + 4sin^2 (x))

    (1/6).dy = 1/((3cos(x))^2 + (2sin(x))^2)

    L.H.S becomes y/6 after integrating.

    The R.H.S. (denominator):
    (3cos(x))^2 + (2sin(x))^2
    I take out cos^2(x)
    This leaves me with:
    (9 + 4tan^2(x))
    Then i use the standard integral formula for arctan
    And i get: (1/3)arctan (2tanx/3)

    Then i'm stuck...
     
  2. jcsd
  3. Nov 29, 2011 #2
    What are you trying to find?
     
  4. Nov 29, 2011 #3

    dextercioby

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    It's almost perfect. The LHS is y/6 + C. The RHS is 1/6 tan^(-1)((2 tan(x))/3).

    What's y(x) then ?
     
  5. Nov 30, 2011 #4

    sharks

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    Solving the ODE to get y in terms of x.


    dextercioby, for the R.H.S, after i worked it out, i get 1/2 tan^(-1)((2 tan(x))/3)
    I used substitution t=tanx to integrate R.H.S.

    Based on the formula: integral of 1/(a^2 + x^2) is 1/a tan^(-1)(x/a) + C

    So, i don't understand how you got 1/6 instead of 1/2

    I got the 1/2 by: 1/3 x 3/2 (obtained by dividing by the differentiation of 2t/3)
     
    Last edited: Nov 30, 2011
  6. Dec 2, 2011 #5

    sharks

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    Any help?
     
  7. Dec 2, 2011 #6

    dextercioby

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    Hmm...

    [tex] ... = \frac{1}{9}\int \frac{dx}{\left(\frac{2}{3}\sin x\right)^2 + \cos^2 x} = ...= \frac{3}{18} \int \frac{dp}{p^2 +1} = ... [/tex]

    Fill in the blanks.
     
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