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One more o.d.e. question

  1. Aug 27, 2005 #1
    for this O.D.E.
    y`= (1-2y-4x)/(1+y+2x)
    how do you treat it?
     
  2. jcsd
  3. Aug 27, 2005 #2

    TD

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    [tex]y' = \frac{{1 - 2y - 4x}}{{1 + y + 2x}} = \frac{{ - 2\left( { - {\raise0.5ex\hbox{$\scriptstyle 1$}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle 2$}} + y + 2x} \right)}}{{1 + y + 2x}}[/tex]

    Now try the substitution [tex]u = y + 2x[/tex], it should become separable.
     
  4. Aug 29, 2005 #3
    how'd you think of making that substitution?
     
  5. Aug 29, 2005 #4

    TD

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    It's a typical strategy for ODE's of the form:

    [tex]y' = \frac{{ax + by + c}}{{a'x + b'y + c'}}[/tex]

    Consider the nominator and denominator as 2 lines.

    Case 1: they intersect -> substitute so you move the intersection to the origine to lose the constant term, making it homogenous

    Case 2: they're parallel -> there is a k that you can factor out (like in this case) to get the same coefficients in x and y, substitue for that expression.
     
  6. Aug 29, 2005 #5
    wow!!! thank you very much!!! :)
     
  7. Aug 29, 2005 #6

    TD

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    No problem :smile:

    In a bit more detail, in case 1 (intersecting lines), search the point of intersection, I'll call it [itex]\left( {x_0 ,y_0 } \right)[/itex]

    Then do the substitution:

    [tex]\left\{ \begin{array}{l}
    x = u + x_0 \\
    y = v + y_0 \\
    \end{array} \right[/tex]

    The DE will become homogenous. The other case (parallel) is the one that applies to your problem and it will make the DE separable.
     
  8. Aug 30, 2005 #7
    ok~ great!!! :)
     
  9. Aug 31, 2005 #8

    HallsofIvy

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    The way I would have done y`= (1-2y-4x)/(1+y+2x) is to rewrite it as

    (1+ y+ 2x)dy+ (4x+ 2y-1)dx= 0. This is an "exact" equation.
     
  10. Aug 31, 2005 #9
    then (1+y+2x)dy=-(4x+ 2y-1)dx
    but then when you integrate both sides,
    say on the left side, what do you do with "2x"? is it still possible to integrate that?
     
  11. Aug 31, 2005 #10

    HallsofIvy

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    No, I did not intend that you try to integrate the two parts separately- not only is there a "2x" in the "dy" term but there is a "2y" in the "dx" term.

    I said this was an exact equation. That means that the "differential form"
    (1+ y+ 2x)dy+ (4x+ 2y-1)dx (and I want them on the same side of the equation!) IS an exact differential- that is, there exist some function F(x,y) so that
    dF= (1+ y+ 2x)dy+ (4x+ 2y- 1)dx. That, in turn, means that
    [tex]\frac{\partial F}{\partial x}= 4x+ 2y- 1[/tex] and
    [tex]\frac{\partial F}{\partial y}= 1+ y+ 2x[/tex].
    Since the "partial derivative of F with respect to x" treats y as if it were a constant, an "anti-partial-derivative" of that first would give F(x,y)=2x2+ 2xy- x+ g(y). Notice that, since we are treating y as a constant here, the "constant of integration could be any function of y- that's the "g(y)".

    Now differentiate that with respect to y:
    [tex]2x+ g'(y)= \frac{\partial F}{\partial y}= 1+ y+ 2x[/tex].
    Fortunately for us, the "2x" terms on each side cancel (that's because this was an "exact differential") leaving g'(y)= 1+ y so that g(y)= y+ (1/2)y2+ C.
    Putting that into the previous function, F(x,y)= 2x2+ 2xy- x+ y+ (1/2)y2+ C.

    The original equation was "dF= 0" so F(x,y)= constant. Including the "C" above in that constant, 2x2+ 2xy- x+ y+ (1/2)y2= C is the general solution to the differential equation.
     
  12. Aug 31, 2005 #11

    TD

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    Of course, I forgot to check whether it was exact :blushing:

    If it's not though, the method I described above works for any of these types in general.
     
  13. Aug 31, 2005 #12
    thank you very much for explaining!!! :)
     
  14. Sep 3, 2005 #13
    hmm... here's my work:
    suppose v= y+2x
    => v`= y`+2
    so y`= (1-2v)/(1+v)
    => v`= 3/(1+v)
    => (1+v)dv = 3dx
    => v+(v^2)/2 = 2x+c
    => y+2x +(y^2 +4xy + 4x^2)/2 = 3x +c
    however, the correct answer should be 2y - 2x + (y+2x)^2=c
    does anybody know where my calculations went wrong?
     
    Last edited: Sep 3, 2005
  15. Sep 3, 2005 #14

    TD

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    If

    [tex]y' = \frac{{1 - 2y - 4x}}{{1 + y + 2x}} = \frac{{ - 2\left( { - {\raise0.5ex\hbox{$\scriptstyle 1$}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle 2$}} + y + 2x} \right)}}{{1 + y + 2x}}[/tex]

    Then using [itex]u = y + 2x \Leftrightarrow y = u - 2x \Leftrightarrow y' = u' - 2[/itex] gives

    [tex]u' - 2 = \frac{{ - 2\left( { - {\raise0.5ex\hbox{$\scriptstyle 1$}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle 2$}} + u} \right)}}{{1 + u}} \Leftrightarrow u' = \frac{{1 - 2u}}{{1 + u}} + 2 = \frac{3}{{u + 1}}[/tex]

    I believe you forgot that -2 after substituting y'.

    But as HallsofIvy said, you can solve this as an exact DE too.
     
  16. Sep 3, 2005 #15
    sorry, i mistyped~
    @@a
    the result is still the same as yours, though...
    so there's still some place that went wrong?
     
  17. Sep 3, 2005 #16

    TD

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    Your answer seems correct, it's just not in the same form I think.
    Solved for y, both solutions give (apart from a factor 2 for the constant, but that's still a constant):

    [tex]y = \pm \sqrt {6x + c + 1} - 2x - 1[/tex]
     
  18. Sep 4, 2005 #17
    ok, thank you very much!!! :)
     
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