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Homework Help: Differential equations/initial-value problem

  1. Mar 30, 2005 #1
    Here's the initial-value problem I'm trying to solve:

    [tex]\sin (2x) \: dx + \cos (3y) \: dy = 0 \qquad y \left( \frac{\pi}{2} \right) = \frac{\pi}{3}[/tex]

    I can't see exactly where my mistake is, but I do know that my answer is not correct. My textbook gives:

    [tex]y=\frac{1}{3} \left[ \pi - \arcsin \left( 3 \cos ^2 x \right) \right]\mbox{.}[/tex]

    Here's what I've done:

    [tex]\sin (2x) \: dx + \cos (3y) \: dy = 0[/tex]

    [tex]\sin (2x) \: dx = - \cos (3y) \: dy [/tex]

    [tex]\int \sin (2x) \: dx = - \int \cos (3y) \: dy [/tex]

    [tex]-\frac{1}{2} \cos (2x) = -\cos (y) \sin (y) + \mathrm{C}[/tex]

    [tex]\cos (2x) = 2\cos (y) \sin (y) + \mathrm{C}[/tex]

    Applying the initial condition at this point gives:

    [tex]\mathrm{C} = -\frac{\sqrt{3}}{2} - 1[/tex]

    Thus, we have the following.

    [tex]\cos (2x) = 2\cos (y) \sin (y) -\frac{\sqrt{3}}{2} - 1[/tex]

    [tex]\cos (2x) = \sin (2y) -\frac{\sqrt{3}}{2} - 1[/tex]

    [tex]\sin (2y) = \cos (2x) +\frac{\sqrt{3}}{2} + 1[/tex]

    [tex]\sin (2y) = 2\cos ^2 (x) - 1 +\frac{\sqrt{3}}{2} + 1[/tex]

    [tex]\sin (2y) = 2\cos ^2 (x) +\frac{\sqrt{3}}{2}[/tex]

    [tex]y = \frac{1}{2} \arcsin \left[ 2\cos ^2 (x) +\frac{\sqrt{3}}{2} \right][/tex]

    As you can see, I don't get the same answer.

    Any help is highly appreciated.
     
  2. jcsd
  3. Mar 30, 2005 #2

    dextercioby

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    Are u claiming that

    [tex] -\int \cos 3y \ dy=-\cos y \sin y +C [/tex]

    If so,on what grounds?

    Daniel.
     
  4. Mar 30, 2005 #3

    HallsofIvy

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    If [tex]\integral sin 2x dx= -\frac{1}{2}cos(2x)[/tex]

    why isn't [tex]\integral cos 3x dx= \frac{1}{3}sin(3x)[/tex]?
     
  5. Mar 30, 2005 #4

    dextercioby

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    U mean with integral signs & integration constants,Halls,right?:uhh:

    Daniel.
     
  6. Mar 30, 2005 #5
    What I really did previously was [tex]-\int \cos (2y) \: dy = - \cos (y) \sin (y) + \mathrm{C}[/tex], which does not apply in this case. In fact, I missed one digit in the integrand. Thanks for pointing it out.

    I've just fixed that mistake, but not quite the problem. Here's what I have now:

    [tex]\sin (2x) \: dx + \cos (3y) \: dy = 0[/tex]

    [tex]\sin (2x) \: dx = - \cos (3y) \: dy [/tex]

    [tex]\int \sin (2x) \: dx = - \int \cos (3y) \: dy [/tex]

    [tex]-\frac{1}{2} \cos (2x) = -\frac{1}{3} \sin (3y) + \mathrm{C}[/tex]

    Applying the initial condition [tex]y \left( \frac{\pi}{2} \right) = \frac{\pi}{3}[/tex] gives [tex]\mathrm{C} = \frac{1}{2}[/tex].

    [tex]-\frac{1}{2} \cos (2x) = -\frac{1}{3} \sin (3y) + \frac{1}{2} [/tex]

    [tex]\sin (3y) = \frac{3 + 3\cos (2x)}{2}[/tex]

    [tex]\sin (3y) = \frac{3 + 3\left( 2\cos ^2 x - 1 \right)}{2}[/tex]

    [tex]\sin (3y) = 3\cos ^2 x[/tex]

    [tex]y = \frac{1}{3} \arcsin \left( 3\cos ^2 x \right)[/tex]

    I'm still confused!!
     
  7. Mar 30, 2005 #6

    dextercioby

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    The last formula

    [tex] y=\frac{1}{3}\arcsin\left(3\cos^{2}x\right) [/tex]

    does not verify the initial condition,becasue the range of "arcsin" is [itex] \left[-\frac{\pi}{2},+\frac{\pi}{2}\right] [/itex]...

    However,your penultimate formula is correct

    [tex] \sin 3y=3\cos^{2} x [/tex]

    I advise you to leave it in this implicit form or try to put in the one the problem is giving...

    Daniel.
     
  8. Mar 30, 2005 #7
    Okay, I just need to make some corrections in my last formula because of the problem with the range of arcsin. That would make it

    [tex]y = \frac{\pi}{3} - \frac{1}{3} \arcsin \left( 3\cos ^2 x \right)[/tex]

    Thank you
     
  9. Mar 30, 2005 #8

    dextercioby

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    It's good that u realized how to solve this trancendental equation

    [tex] \sin x = a [/tex]

    Daniel.
     
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