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Reduce order of ODE y =c-2y'2/y

  1. Sep 6, 2008 #1
    reduce order of ODE y"=c-2y'2/y

    I'm stumped trying to come up with the appropriate substitution for:


    c is a constant and the dependent variable does not appear in the equation. According to the paper I am working from the equation can be reduced to a Bernoulli-type problem and solved, but I can't seem to replicate their work. Any hints?

    y'=0 at the far boundary and y(0)=y0
    Last edited: Sep 6, 2008
  2. jcsd
  3. Sep 6, 2008 #2
    Re: reduce order of ODE y"=c-2y'2/y

    Any second-order equation of the form

    [tex] y^{\prime \prime} + \alpha(y) y^{\prime 2} + \beta(y) = 0[/tex]

    (where the derivative is with respect to 'x') may be converted into a first order equation of the form

    [tex] \frac{du}{dy} + 2 \alpha(y) u + 2 \beta(y) = 0 [/tex]

    with the simple substitution

    [tex] u = y^{\prime 2}[/tex].

    Just plug that into the second equation and you'll see it works, bearing in mind that

    [tex]\frac{d y^{\prime 2}}{dy} = 2 y^{\prime \prime}[/tex]

    After some tedious calculation, it turns out that your particular equation may be expressed (in first-order terms) as

    [tex] y^{\prime 2} = \frac{2c}{5} y + \frac{\lambda}{y^4} [/tex]

    (where lambda is some constant). To demonstrate this, differentiate the above with respect to 'y', and you get

    [tex] 2 y^{\prime \prime} = \frac{2c}{5} - \frac{4 \lambda}{y^5}[/tex]

    Now make the substitution

    [tex] \frac{\lambda}{y^4} = y^{\prime 2} - \frac{2c}{5} y [/tex]

    to get

    [tex] 2 y^{\prime \prime} = \frac{2c}{5} - \frac{4}{y}(y^{\prime 2} - \frac{2c}{5} y) [/tex]

    which expands to

    [tex] 2 y^{\prime \prime} = \frac{2c}{5} - \frac{4}{y} y^{\prime 2} + \frac{8c}{5} [/tex]

    Now add the constant terms and divide by 2 on both sides to get your original equation.
    Last edited: Sep 6, 2008
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