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Non-Linear ODE

  1. Jun 17, 2015 #1

    joshmccraney

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    Hi PF!

    Can any of you help me reduce this ODE to find a solution?

    $$y y''+2y'^2+xy'+\frac{1}{2}y = 0 \implies \\ (y y')'+y'^2+xy'+\frac{1}{2}y = 0 \implies\\ (yy')'+(xy)'+y'^2-\frac{1}{2}y=0$$

    but here I am stopped. Am I even going the correct route? I know I can re-write this equation as terms primed, because a quadratic solves this ODE. Any help is awesome!
     
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  3. Jun 17, 2015 #2

    RUber

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    Are x and y both functions of t, so (xy)' = xy' + x'y? Or is y a function of x, so (xy)' = xy' +y ?
     
  4. Jun 17, 2015 #3

    pasmith

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    I can see that if [itex]y'' = 2a[/itex] then every term of the ODE is a multiple of [itex]x^2[/itex], so certainly [itex]y = ax^2[/itex] is a solution (for two particular values of [itex]a[/itex]).

    You can of course just substitute [itex]y = ax^2 + bx + c[/itex] and see what constraints there are on [itex](a,b,c)[/itex].
     
  5. Jun 17, 2015 #4

    joshmccraney

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    Sorry, ##y## is the dependent variable and ##x## is the independent variable.
     
  6. Jun 17, 2015 #5

    joshmccraney

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    And yes, I can verify that a quadratic solves the ODE, but I'm trying to see how to reduce this into an equation of primes. I'm curious and know it can be done, I just can't see how. Any ideas?
     
  7. Jun 17, 2015 #6

    pasmith

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    Substituting [itex]v = 2y' + x[/itex] yields [itex]\frac12 yv' + y'v = 0[/itex]. Thus either [itex]y = 0[/itex] or [itex]v = 0[/itex] or [tex]\frac12 \frac{d(\ln v)}{dx} + \frac{d(\ln y)}{dx} = 0.[/tex]

    (EDIT: Actually just multiplying [itex]\frac12 yv' + y'v = 0[/itex] by [itex]2y[/itex] suffices.)
     
    Last edited: Jun 18, 2015
  8. Nov 24, 2015 #7

    joshmccraney

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    This was awesome, pasmith! How would you go about solving this one ##y y''+2y'^2+xy'+Ay = 0## if ##A## could be any non-zero real number you wanted it to be?
     
  9. Nov 25, 2015 #8

    pasmith

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    You may still be able to obtain quadratic solutions (if [itex]y[/itex] is a quadratic then so is [itex]yy'' + 2y'^2 + xy' + Ay[/itex]) but setting [itex]v = 2y' + x[/itex] yields [tex]
    \frac{d}{dx}(y^2 v) = (1 - 2A)y^2[/tex] which only helps if [itex]A = \frac12[/itex].
     
    Last edited: Nov 25, 2015
  10. Nov 29, 2015 #9

    joshmccraney

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    Yea, it seems ##A=1/2## is the only way this works. It's very clever, though!
     
  11. Nov 29, 2015 #10
    Just out of curiosity is it all possible to solve this via power series? Or does the y*y" make it a no go?
     
  12. Dec 1, 2015 #11

    pasmith

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    Try it. If [itex]y = \sum_{n=0}^\infty a_nx^n[/itex] then [itex]yy''[/itex] term gives you [tex]a_0a_{n+2}(n+2)(n+1) + a_1a_{n+1}(n+1)n + \dots + a_{k}a_{n+2-k}(n+ 2 - k)(n + 1 - k) + \dots + 2a_na_2[/tex] as the coefficient of [itex]x^n[/itex] and [itex]y'^2[/itex] gives you [tex]
    (n+1)a_1a_{n+1} + \dots + (k+1)(n - k + 1)a_{k+1}a_{n - k + 1} + \dots + (n+1)a_1a_{n+1}[/tex] so given [itex]a_0[/itex] and [itex]a_1[/itex] you can obtain [itex]a_2[/itex], then [itex]a_3[/itex], and so on.
     
  13. Dec 1, 2015 #12
    Thanks! Will try it out on hand and later run it trough mathematica.
     
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