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Nonlinear DE

  1. Dec 14, 2011 #1
    How do I solve the DE

    [itex]\dot{y}[/itex]=[itex]\frac{t+3y-5}{t-y-1}[/itex]

    I tried using a substitution of the form u=ax+by+c, however was
    unsucessful. Any suggestions?
     
  2. jcsd
  3. Dec 14, 2011 #2

    Simon Bridge

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    I multiplied both sides by the denominator and grouped like terms.
    This suggested a substitution [itex]z=\dot{y} + 1[/itex] - which could be solved for z in terms of t.

    There's probably a better way.
     
  4. Dec 14, 2011 #3
    Simon,

    Thank you for the suggestion. I tried your suggestion, however do
    not seem to be getting anywhere. When I group like terms, I get

    [itex](\dot{y}-1)x-(\dot{y}+3)-\dot{y}+5=0[/itex]

    If I use

    [itex] z=\dot{y}-1 [/itex],

    I get no where, plus I lose my derivatives.

    If I use

    [itex] \dot{z}=\dot{y}-1 [/itex],

    I get

    [itex] x \dot{z} - (\dot{z}+4) y - \dot{z} + 6=0 [/itex]

    which is worse than before. Im still lost.
     
  5. Dec 14, 2011 #4

    Simon Bridge

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    Hmmm ... did I mess up? Did I misplace 2y ... I just scribbled it on my window, lets see:

    [tex](t-y-1)\dot y = t + 3y -5[/tex]... expand
    [tex]t \dot y - y\dot y -\dot y -t -3y +5 = 0[/tex]... group terms
    [tex]t(\dot y -1) - y(\dot y +3) - \dot y +5 = 0[/tex]... yep, looks like I did!

    How about [itex]z=\dot y + 3[/itex] ...

    note: [itex]t(\dot y -1) = t(\dot y +3) - 4t[/itex]

    dammit! I still get a stray y.
     
  6. Dec 14, 2011 #5

    AlephZero

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    First get rid of the constants on the right hand side. If you solve
    t + 3y - 5 = 0
    t - y - 1 = 0
    you get t = 2, y = 1

    So let T = t + 2, Y = t + 1
    DY/DT = (T + 3Y)/(T - Y)
    Then you can separate the variables by letting Y(T) = Tu(T)
     
  7. Dec 14, 2011 #6
    Thank you!
     
  8. Dec 14, 2011 #7

    Simon Bridge

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    Yes - thank you and my apologies to namu.

    aside:
    did you know that "namu" is Maori for sand-fly?
     
  9. Dec 15, 2011 #8
    Hi!

    In fact this ODE is of the homogeneous kind (attached page) :
     

    Attached Files:

  10. Dec 18, 2011 #9
    No, i didn't know that. That is interesting. Namu is my name. Its often
    included in buddhist chants. Also in korean, namu means tree. It is interesting
    how one word has many different meanings in different cultures.
     
  11. Dec 18, 2011 #10
    Yes, I saw that once I realized the clever substitution that was pointed out. Thank you for the work.
     
  12. Dec 18, 2011 #11

    Simon Bridge

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    That's cool (the name)!

    I can see why a tree would come up in chants quite a lot. Trees are a feature in the life of Siddartha don't they?

    "Namu" can be a nickname, and they have spiritual connotations to the Maori too.
    This has been a neat window between cultures, thanks.
     
  13. Dec 18, 2011 #12
    First, you need to make an "additive" substitution:
    [tex]
    t = \tilde{t} + \alpha, y = \tilde{y} + \beta
    [/tex]
    that will cancel the free terms in the numerator and denominator of the fraction. Effectively, you need to solve the following linear system:
    [tex]
    \alpha + 3 \beta - 5 = 0, \ \alpha - \beta - 1 = 0
    [/tex]
    After that, considering [itex]\tilde{y} = f(\tilde{t})[/itex], you get a homogeneous ODE:
    [tex]
    \dot{\tilde{y}} = \frac{\tilde{t} + 3 \tilde{y}}{\tilde{t} - \tilde{y}} = \frac{1 + 3 \frac{\tilde{y}}{\tilde{t}}}{1 - \frac{\tilde{y}}{\tilde{t}}} = f(\frac{\tilde{y}}{\tilde{t}}), f(z) = \frac{1 + 3 z}{1 - z}
    [/tex]
    You can separate the variables by making one more subtitution:
    [tex]
    z(\tilde{t}) = \frac{\tilde{y}}{\tilde{t}} \Rightarrow \tilde{y} = \tilde{t} \, z, \ \dot{\tilde{y}} = \tilde{t} \, \dot{z} + z
    [/tex]
     
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