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Math Help - Slope Field with Solution Passing Through Point

  1. #1
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    Slope Field with Solution Passing Through Point

    Hi,

    This is for Calc II homework..

    The figure below shows the slope field for the equation y' = sin (x) cos (y).

    (If you want I can post the picture, but I don't think its needed)

    And the question is

    What is the equation of the solution passing through (0(2n+1)2), where n is any integer?
    y=

    Now, I know that the slope is 0 along any of these points but I just don't know what to do with the rest of the information.

    Thanks for any help
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  2. #2
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    Quote Originally Posted by pcannons View Post
    Hi,

    This is for Calc II homework..

    The figure below shows the slope field for the equation y' = sin (x) cos (y).

    (If you want I can post the picture, but I don't think its needed)

    And the question is

    What is the equation of the solution passing through (0(2n+1)2), where n is any integer?
    y=

    Now, I know that the slope is 0 along any of these points but I just don't know what to do with the rest of the information.

    Thanks for any help
    You need to integrate y'= sin(x)cos(x) (which should be easy). You will, of course, have a "constant of integration". Use the fact that y(0)= \frac{(2n+1)\pi}{2} to find the value of that constant.
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  3. #3
    MHF Contributor Calculus26's Avatar
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    y' = sin (x) cos (y).


    has equiilibrium solutions of the form y= constant where cos(y) = 0

    which is (2n+1)pi/2

    the solution curves are y = (2n+1)pi/2 which are horizontal lines

    In general for dy/dx = f(x)g(y) equillibrium solutions are precisely the zeroes of g(y)
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