1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

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

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

    Daniel.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Differential equations/initial-value problem
  1. Initial value problem (Replies: 3)

  2. Initial value problem (Replies: 8)

Loading...