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DE Initial Value Problem

  1. Oct 17, 2011 #1
    I'm given the following DE and initial conditions:
    [tex]y''=2yy'[/tex]
    [tex]y(0)=0, y'(0)=1[/tex]

    I started by doing a reduction of order like so:
    [tex]w=y', w'=y'', \int w=y=\frac{w^{2}}{2}+c[/tex]

    which then gave me this:
    [tex]w'=2w(\frac{w^{2}}{2}+c)[/tex]
    [tex]w'=w^{3}+2wc[/tex]

    Now I'm stuck on where to go from here. I can't use any of the usual techniques like variation of parameters. There is a trick in my notes here for dealing with equations missing the dependent variable, however I don't fully grasp how it works.

    Here is the trick that's given:
    For [tex]yy"=y'^{2}[/tex]
    It starts by making z=y'.
    Then since you need [tex]\frac{dz}{dy}[/tex], you can do this:
    [tex]\frac{dz}{dx}=\frac{dx}{dy}\frac{dy}{dx}[/tex]
    then since [tex]y"=\frac{dz}{dx}[/tex]
    you get this: [tex]y\frac{dz}{dy}=z[/tex]

    Does anyone recognize this trick? If so can I apply it here?
     
    Last edited: Oct 17, 2011
  2. jcsd
  3. Oct 17, 2011 #2

    Dick

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    (y^2)'=2yy', right? So your ODE is y''=(y^2)'. Integrate both sides.
     
  4. Oct 18, 2011 #3
    I managed to solve it. Thanks for the tip.
     
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