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Nonlinear ODE

  1. Feb 23, 2012 #1
    Given (y')^2 *y'' -8y=0 and ICs y(0)=1 y'(0)= -2

    I used substitution dy/dx = v and d^2y/dx^2 = vdv/dy and got to:

    v^2*v*v' = 8y
    V^3 dv/dy = 8y
    v^4/4 =4y + c
    V^4 =4y + c

    And IC y'(0) = v(0) =-2
    So c = 16

    v = 2(y^2+1)^(1/4)
    dy/dx = 2(y^2+1)^(1/4)

    (y^2+1)^(-1/4)dy = 2dx

    But here I got stuck. :( did I do the steps wrong? I then tried trig substitution and got:

    Sec(u)^3/2 du = 2dx

    But couldn't integrate either. This is a DE class so I didn't think it should be like that and maybe I made a mistake. Any help would be greatly appreciated. :]
     
  2. jcsd
  3. Feb 23, 2012 #2

    HallsofIvy

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    The integral of 8y is NOT 4y+ C.

    How did you get this? If you multiply both sides of v^4/4= 4y+ C (which was wrong) by 4 you would get y^4= 16y+ 4c (which you could write as y^4= 16y+ C' since 4 times an unknown constant is still an unknown constant).

     
  4. Feb 23, 2012 #3
    Oops yes, I had before putting in first IC

    V^4 = 16y^2 + C.

    I jumped a bit a head of myself while typing this in. XD I just had this on an exam a few hours ago.
     
    Last edited: Feb 23, 2012
  5. Feb 23, 2012 #4
    From:
    V^3 dv/dy = 8y, I used method of separable variables.

    V^3 dv = 8ydy then integrate both sides:

    V^4/4 = 8y^2/2 + c
     
  6. Feb 23, 2012 #5
    I accidentally left the squared out on that part. XD it's not easy typing on phone.
     
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