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2nd order nonlinear DE need help bad

  1. Nov 9, 2004 #1
    this is the problem: xy'' -x(y')^2 = y'

    my book says that i need to substitute u=y' and du/dx=y''...
    so i get:

    x(du/dx)-xu^2 = u

    so next the book says i need to separate the x's/dx' to one side and u's/du's to the other. however, i cannot do it

    am i using the correct technique?


    thanks-
    Justin
     
  2. jcsd
  3. Nov 9, 2004 #2

    Galileo

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    edit: Nevermind
     
    Last edited: Nov 9, 2004
  4. Nov 9, 2004 #3
    this problems driving me crazy :surprised
     
  5. Nov 9, 2004 #4

    ehild

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    Substitute xy'=u instead.

    ehild
     
  6. Nov 9, 2004 #5
    i dont see how that would work....if u=xy' then du/dx=y' +xy''
    how would thos substitutions work?
     
  7. Nov 9, 2004 #6
    if you want you can work it like this:

    x(du/dx)-xu^2=u => (u^-2)(du/dx)-(u^-1)/x=1 then use this

    substitution, I think this is "Bernoulli equation" z=-(u^-1) and dz/dx=(u^-2)(du/dx)

    => dz/dx +z/x=1. I hope this help you.
     
  8. Nov 9, 2004 #7

    James R

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    Your equation is:

    [tex]xy'' -x(y')^2 = y'[/tex]

    Since y does not appear in the equation, start by putting:

    [tex]y' = u[/tex]

    which means the equation is:

    [tex]xu' - xu^2 = u[/tex]
    [tex]u' - u^2 = \frac{u}{x}[/tex]

    Now put

    [tex]u(x) = xv(x)[/tex]

    So that

    [tex]u' = v'x + v[/tex]

    The differential equation then becomes separable, as follows:

    [tex]v'x + v - v^2 x^2 = v[/tex]
    [tex]xv' = v^2 x^2[/tex]
    [tex]dv/v^2 = x dx[/tex]

    Integrating gives:

    [tex]-\frac{1}{v} = \frac{x^2}{2} + c[/tex]

    where c is an undetermined constant. Rewriting v in terms of u gives:

    [tex]-\frac{x}{u} = \frac{x^2}{2} + c[/tex]

    Rearranging, we get:

    [tex]-\frac{u}{x} = \frac{1}{x^2/2 + c}[/tex]

    or

    [tex]y' = -\frac{x}{x^2/2 + c}[/tex]

    This can now be easily integrated to give the solution:

    [tex]y = -\ln(x^2/2 + c) + d[/tex]

    where d is a second constant of integration.
     
  9. Nov 9, 2004 #8

    ehild

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    [tex]y' = \frac{u}{x}\mbox { , } y''=\frac{u'x-u}{x^2}[/tex]

    Plugging in y' and y'' the differential equation becomes

    [tex]\frac{xu'-u}{x}-\frac{u^2}{x}=\frac{u}{x}\rightarrow u'x=u^2+2u[/tex]

    This is easily separable:

    [tex]\frac{du}{u^2+2u}=\frac{dx}{x}[/tex]

    and results in

    [tex]u=\frac{2x^2}{C-x^2} \mbox { that is } y'=\frac{2x}{C-x^2}[/tex]

    This is the same solution James got but his method is simpler, I have to admit :smile: You see there are many ways to go....

    ehild
     
  10. Nov 10, 2004 #9
    Thank You

    thanks guys! its very clear to me now :) my book had one very limited example which made it sound like i only need to make one substitution. however, i clearly see it now!

    thanks,
    Justin
     
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