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Homework Help: Solving A Differential Equation

  1. Nov 12, 2006 #1
    Given the two equations:

    [tex]y(1) = \pi[/tex]

    [tex]\frac{dy}{dx} = \frac{6x^{2}}{2y + \cos{y}}[/tex]

    Solve for x:

    [tex]\begin{align*}
    \int(2y+\cos{y})dy &= \int(6x^{2})dx\\
    y^{2} + \sin{y} &= 2x^{3}\\
    \end{align*}[/tex]

    But in order to set a value to the function y, I need some way to exclude y and then plug in 1 for x. How do I exclude y when it's in two separate terms?

    Or is there an easier way to do this?
     
    Last edited: Nov 12, 2006
  2. jcsd
  3. Nov 12, 2006 #2
    When x = 1, y = pi. Does that make it any simpler? Don't forget the constant of integration
     
  4. Nov 12, 2006 #3
    Oh right!! I forgot the C. That's why it wasn't working. I'm silly without my morning coffee... Haha. Thanks.
     
  5. Nov 12, 2006 #4
    I just need to make sure I did this correctly.

    [tex]\begin{align*}
    (\pi)^{2} + \sin{(\pi)} &= 2(1)^{3} + C\\
    C &= \pi^{2} - 2\\
    \\
    2x^{3} + \pi^{2} - 2 &= y^{2} + \sin{y}\\
    2x^{3} &= y^{2} + \sin{y} - \pi^{2} + 2\\
    x &= (\frac{y^{2} + \sin{y} - \pi^{2}}{2} + 1)^{\frac{1}{3}}
    \end{align*}[/tex]
     
    Last edited: Nov 12, 2006
  6. Nov 12, 2006 #5
    I guess the step before the last would do just fine.
     
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