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First order differentials: separating variables

  1. Nov 13, 2006 #1
    "Find the general solution to the differential equation by separating variables:
    3tany - dy/dx(secx) = 0"

    This is what I set up:

    3tany dx = secx dy
    1/secx dx = 1/3tany dy
    cosx dx = 1/3tany dy
    [int] cosx dx = [int] 1/3tany dy
    sinx = (1/3)ln|sinx|

    I'm stuck as to what to do next, how to solve the equation.. did I mess up in the beginning or what am I doing wrong?
     
  2. jcsd
  3. Nov 13, 2006 #2
    you have to seperate the same variables. So:

    [tex] 3\tan y \; dx = \sec x \; dy [/tex]

    Multiply by [tex] \frac{1}{dy\cdot dx} [/tex]. Then you get:

    [tex] \frac{dx}{\sec x} = \frac{dy}{3\tan y} [/tex]

    [tex] \int \cos x \; dx = \frac{1}{3}\int \cot y + C [/tex]

    [tex] \sin x = \frac{1}{3}\ln|\sin y| [/tex]

    [tex] 3\sin x = \ln|\sin y| [/tex]

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

    [tex] y = \arcsin(e^{3\sin x}) + C [/tex]
     
    Last edited: Nov 13, 2006
  4. Nov 13, 2006 #3
    It looks like the y's became x's between these two steps :)
     
  5. Nov 14, 2006 #4

    HallsofIvy

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    Every thing exactly right up to here.

    No, sin x= (1/3) ln |cos(y)|+ C. You accidently wrote x instead of y and forgot the constant of integeration.

    [/quote]I'm stuck as to what to do next, how to solve the equation.. did I mess up in the beginning or what am I doing wrong?[/QUOTE]
    What exactly do you mean by "solve the equation". In general you cannot solve for y. What you have, with the corrections, is the general solution.
     
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