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Unseparable O.D.E.

  1. Aug 27, 2005 #1
    another O.D.E. question:
    xyy`= 2y^2 +4x^2
    also unseparable...
     
  2. jcsd
  3. Aug 27, 2005 #2

    Galileo

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    I`m a bit rusty on D.E.'s, but you can write it as:

    [tex]y'=2(y/x)+4(x/y)[/tex]

    IIRC, whenever you have y'=F(y/x), the equation also called homogeneous (different from the homogeneous linear ODE definition), this one is invariant under 'zoom'.

    If you make the subsitution z=y/x, the equation becomes:
    [tex]z+xz'=2z+4/z[/tex]
    or
    [tex]xz'=z+4/z[/tex]
    which is separable.
     
    Last edited: Aug 27, 2005
  4. Aug 28, 2005 #3

    GCT

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    it's separable, use v=y/x
     
  5. Aug 29, 2005 #4
    why'd you think of v=y/x?
     
  6. Aug 29, 2005 #5

    TD

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    Galileo explained. It's because the DE is homogeneous (not with the 0 constant, but with the sum of all exponents of all terms is constant (here 2). You can then divide so the DE becomes a function of (y/x). Susbtitution will make it separable.
     
  7. Aug 29, 2005 #6
    so the first step when you see an O.D.E you have to check if it's separable
    => if not, then check if it's homogeneous, and use that substitution...
    thanks! :)
     
  8. Aug 29, 2005 #7

    TD

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    Checking if it's separable should indeed always be the first you do and when it's not, try to see if you can make it separable with a substitution.
    The substitution used here a very common one.
     
  9. Aug 30, 2005 #8
    what are the most common substitutions used?
     
  10. Aug 31, 2005 #9
    Well y = v(x)x is a common one and if I recall correctly, for second order isobaric equations you can make the substitution y = vx^m if there is a single value of m which makes the equation dimensionally consistent (you assign weights to y, dy, x and dx).
     
    Last edited: Sep 1, 2005
  11. Aug 31, 2005 #10
    ok~ thanks!!! :)
     
  12. Sep 3, 2005 #11
    hmmm... here's my work~
    (i forgot to include that the original condition is y(2)=4, sorry about that!)
    y`= 2y/x+4x/y
    suppose y=vx
    =>v=y/x
    so y`=v+v`x
    => v+xv`=2v +4/v
    => v`x=v+4/v
    => v`x=(v^2 +4)/v
    => v/(v^2 +4)dv=dx
    => 1/2[dv^2/(v^2 +4)] =dx
    => 1/2ln[absolute value (v^2 +4)] =x+ c`
    => ln[absolute value (y^2/(x^2) +4)]=2x+c`
    => y^2/(x^2) +4 =e^(2x +c`)
    => y^2 +4x^2 =c(x^2)*[e^(2x)]
    => y^2=c(x^2)*[e^(2x)]-4x^2
    now my problem is that when you take the square root on both sides of the equations, the right side should have a positive and negative root, but the correct answer should only have a positive square root~
    why?
    btw, what is meant by the different homogeneous definitions and "zoom"?

     
  13. Sep 3, 2005 #12

    Galileo

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    Whoopsie, you've lost an x.

    The actual answer also needs a square root since you'll get a relation between y^2 and x. Now if this equation is true both the positive and negative square roots must be valid solutions to the ODE, but a little bit of thought shows that no more than one can fit the same boundary condition (y(2)=4)
    A linear DE of, say, second order:
    [tex]y''+A(x)y'+B(x)y=r(x)[/tex]
    is called homogeneous if r(x)=0. Your ODE is a also called homogeneous, but it's in a different context, so it naturally hasn't the same meaning.

    Since the ODE can be written in terms of y/x. If we make the change of variables [itex]\hat y=ay[/itex] and [itex]\hat x=ax[/itex] for some positive a. So that we're scaling the y and x-axes by the same amount (this is what I meant by zoom), the ODE looks the same, since [itex]\hat y/\hat x=y/x[/itex].
     
  14. Sep 4, 2005 #13
    thank you very much!!! :)
    i'll try and be more careful on my calculations~
     
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