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  1. Default Differential equation

    y*(y'') - y' * (y^2) - (y'')^2 = 0

    ... it doesn't look hard but I just can't get it :| kinda blocked, can I get some help please

  2. Default Re: Differential equation

    it is non-linear...does that help? lol do you'll end up with one of those weird graphs with saddle points etc...

  3. Default Re: Differential equation

    (y'')^2 -y(y'') +(y^2)y' = 0

    Use quadratic formula:
    y'' = [y √(y^2 -4(y^2)y'] /2

    (2y''-y)^2 = y^2 (1 -4y')
    Note: At this stage, it is clear that if y=0, then 2y''-y =0, so y''=0. This corresponds to the trivial
    solution of y=0.

    Simplify further:
    (2y''/y^2 -1)^2 = 1 -4y'

    At this stage, note that the equation holds when both sides equal the same constant.
    (2y''/y^2 -1)^2 = 1 -4y' = C

    You end up with two easier equations:
    A) 1 -4y' = C
    B) 2y''- (1√C)*y^2 = 0

  4. Default Re: Differential equation

    It kinda reads like you completed the square there Rick. You can skip the quadratic formula if you want.
    (y''-1/2y)^2 = y''-y y''+1/4y^2
    So the original is (y''-1/2y)^2 + (y'-1/4)y^2 = 0



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