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1st order nonlinear ODE

  1. Sep 30, 2010 #1
    Can anyone help with the following:

    dy/dx = ay / (bx2 +xy )

    a,b constants

    thanks,
     
  2. jcsd
  3. Sep 30, 2010 #2
    The key is to consider the unknown function x(y) instead of y(x)
     

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  4. Sep 30, 2010 #3
    Hi. Thanks for that. However, I'm not convinced your use of the Ei is correct if you define:

    [tex]Ei(x)=\int_{-\infty}^x \frac{e^t}{t}dt[/tex]

    If I start with the expression:

    [tex]d(e^y t)=-\frac{e^y}{y}[/tex]

    then integrating from y_0, t_0 to y,t:

    [tex]\int_{\substack{t=t_0 \\ y=y_0}}}^{y,t}d(e^y t)=-\int_{y_0}^y \frac{e^y}{y}dy[/tex]

    and get:

    [tex]x(y)=\frac{e^y}{\frac{e^{y_0}}{x_0}-\int_{y_0}^y \frac{e^y}{y}dy}[/tex]

    Now, that agrees with numerical calculations but I'm not sure how I could express that in terms of Ei(x) though.
     
  5. Oct 1, 2010 #4
    Of course, you cannot express that in terms of Ei(x), but in terms of Ei(y).
    May be, writing "in terms of" isn't the good wording. What I mean is that Ei(y) is the special function involved in the formula for x(y), as it was shown.
    But I never said that Ei(x) is involved in an hypothetical formula for y(x). On the contrary, I said that the analytical inversion of x(y) in order to obtain y(x) is probably utopian with a finite number of elementary functions and even with classical special functions.
     
  6. Oct 1, 2010 #5
    thanks guys! - I appreciate the help
     
  7. Oct 1, 2010 #6
    Ok, I messed that up. I meant Ei(y) but I see as long as you define Ei(y) appropriately, I see what you mean.

    Here's the numerical check of:

    [tex]y'=\frac{y}{x^2+xy},\quad y(1)=1[/tex]

    Code (Text):

    mysol = NDSolve[{Derivative[1][y][x] ==
          y[x]/(x^2 + x*y[x]), y[1] == 1}, y,
        {x, 1, 5}];

    p1 = Plot[y[x] /. mysol, {x, 1, 5}];

    myx[y_] := Exp[y]/(Exp[1] -
         NIntegrate[Exp[u]/u, {u, 1, y}]);

    mytable = Table[{myx[y], y},
        {y, 1, 1.6, 0.01}];

    p2 = ListPlot[mytable, Joined -> True];

    Show[{p1, p2}]
     
    Notice how I tabulate the numbers in terms of {x(y),y} to retrieve the inverse numerically. The two plots superimpose nicely.
     
  8. Oct 1, 2010 #7
    For me where was no doubt.
    Nevertheless its a good idea to check it numerically. Well donne !
     
  9. Oct 13, 2010 #8
    Is this a research question, or just homework? If it's serious, I may be able to get somewhere with an analytic (explicit) solution... possibly.
     
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