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Solve sin(2x) dx + cos(3y) dy = 0, where y(pi/2) = pi/3

  1. Oct 2, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve the equation [itex]\sin(2x) dx + cos(3y) dy = 0[/itex], where [itex]y(\pi/2) = \pi/3[/itex]

    2. Relevant equations

    N/A

    3. The attempt at a solution

    I understand the process that gets from the original equation to [itex]y = \frac{\arcsin(\frac{3}{2} (\cos(2x) + 1))}{3}[/itex]

    However, I don't understand why the answer provided is that but with a small change: [itex]y = \frac{\pi - \arcsin(\frac{3}{2} (\cos(2x) + 1))}{3}[/itex]

    I believe it has something to do with the properties of [itex]\sin[/itex], but I just can't see it.

    Thank you

    EDIT: Removed repeating problem, equations, attempt formats
     
  2. jcsd
  3. Oct 2, 2012 #2
    When you integrate you obtain a constant term. For example,

    [tex] \int{x}dx = \frac{x^2}{2} + C [/tex]

    Use the boundary condition to find it.
     
  4. Oct 2, 2012 #3
    When you say "boundary condition," are you referring to how [itex]y(\pi/2) = \pi/3[/itex]?

    If so, I already found the constant [itex]C[/itex] from the integration, namely [itex]C=\frac{1}{2}[/itex]:

    Integrating, I got [itex]\frac{1}{3} \sin(3y) = \frac{1}{2} \cos(2x) + C[/itex]

    Substituting, I find [itex]0 = -\frac{1}{2} + C \Longrightarrow C = \frac{1}{2}[/itex]

    In solving, I got the equation in the OP

    EDIT: Thanks for the reply
     
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