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Dif Eq

  1. Sep 21, 2005 #1
    How do I go about solving

    x^2y' - cos (2y) = 1

    This is unlike anything I've seen so far. Or so I think.
     
    Last edited by a moderator: Sep 21, 2005
  2. jcsd
  3. Sep 21, 2005 #2

    saltydog

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    It's separable right?
     
  4. Sep 21, 2005 #3

    lurflurf

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    Yes
    Trig is your friend.
    as a matter of personal taste I enjoy the use of the identity
    1-cos(2y)=2(sin(y/2))^2
     
  5. Sep 21, 2005 #4
    Ok, I got it, thanks for the help.

    I got y = arctan ( c = 2/x)
     
    Last edited by a moderator: Sep 21, 2005
  6. Sep 21, 2005 #5
    18 down, 2 more to go:

    y' = (4x + 2y - 1)^(1/2)

    and

    (1-xy)y' + y ^2 + 3xy^3 = 0

    The last one kinda looks like Riccati, but I can't get it to work.


    edit: I got the second one. One more ODE remains... I don't know how to approach that one.
     
    Last edited by a moderator: Sep 21, 2005
  7. Sep 22, 2005 #6

    saltydog

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    Regarding:

    [tex]\frac{dy}{dx}=\sqrt{4x+2y-1}[/tex]

    How about squaring it for starters. Then let:

    [tex]p=\frac{dy}{dx}[/tex]

    Differentiate it to get an ODE in p. Then separate variables.
     
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