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First order Ordinary Differential Equation

  1. Sep 14, 2009 #1

    PAR

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    1. The problem statement, all variables and given/known data
    Solve the differential equation:

    [tex]y' = 8sin(4yt) ; y(1) = 4[/tex]



    2. Relevant equations



    3. The attempt at a solution

    The integrating doesn't apply because I can't get the equation into:
    y' + p*y = f(x) form

    Also, I have tried separating variables, but I can't get the y inside of the sin(4yt) outside of the sin(). Basically, I need some help to get started.

    Thanks in advance
     
    Last edited: Sep 15, 2009
  2. jcsd
  3. Sep 15, 2009 #2

    Avodyne

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    Science Advisor

    Try changing variables: y(t)=z(t)/t.
     
  4. Sep 15, 2009 #3

    PAR

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    Correction! The differential equation should be y' = 8sin(4yt) instead of y'=8y[tex]^{3}[/tex]sin(4yt)

    I've tried substitution where y(t) = z(t)/t

    This gave me:

    [tex]y' = z'/t - z/t^{2} = 8sin(4z)[/tex]

    The problem now is that I still have a function of z on the right hand side of the equation, and I don't know how to combine it with the left hand side to give me an a nice integrating factor or separation of variables. I was thinking I could divide the whole thing by sin(4z) anyway, and then integrate with respect to t. The z' would be integrable and so would the 8, but the resulting -z/(t[tex]^{2}[/tex]sin(4z)) I don't know how to integrate with respect to t. Great help so far, but could still use more :).
     
  5. Sep 15, 2009 #4
    [tex]\frac{dy}{dt} = 8\sin (4yt)[/tex]

    [tex] y=\frac{z}{t} \Rightarrow 8 \sin (4yt) = 8 \sin (4z) = \frac{dy}{dt} [/tex]


    [tex] y=\frac{z}{t} [/tex]

    [tex] \frac{dy}{dt}=-\frac{dz}{t^2 dt} [/tex]

    [tex] \frac{dz}{dt} = -8t^2\sin (4z) [/tex]

    [tex] \int \frac{dz}{\sin (4z)} = \int -8t^2 dt [/tex]

    Maybe that will work?
     
  6. Sep 15, 2009 #5

    PAR

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    Don't you have to apply the product rule when you differentiate y?

    Since y = z/t = zt[tex]^{-1}[/tex] and z is a function of t by definition shouldn't

    y' = z'/t - z/t[tex]^{-2}[/tex] instead of y' = [tex]-\frac{dz}{t^2 dt} [/tex]
     
  7. Sep 15, 2009 #6
    Oh yeah god, use the product rule :$.
     
  8. Sep 15, 2009 #7

    PAR

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    Thank you Gregg and Avodyne for the help, but I don't have a solution to this problem. Is there a different way to solve first order ODE other than separation of variables and the integration factor? Please, I need some help.
     
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