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ODE: Can't find mistake

  1. Dec 25, 2005 #1
    Hi all,

    I've been just solving this one:

    [tex]
    y' + \frac{y}{x} = 3\sqrt[3]{\left(xy\right)^2}\arctan x
    [/tex]

    The problem is, one of the solutions I got doesn't pass the original equation and I can't find the mistake. Here it is:

    After substituting

    [tex]
    z = \sqrt[3]{y}
    [/tex]

    and thus getting

    [tex]
    3z^2z' + \frac{z^3}{x} = 3\sqrt[3]{x^2}z^2\arctan x
    [/tex]

    dividing with [itex]z^2[/itex] (and so getting the condition for y not to be particular solution [itex]y \equiv 0[/itex])

    [tex]
    3z' + \frac{z}{x} = 3x^{\frac{2}{3}}\arctan x
    [/tex]

    To solve this, I first solved the homogenous equation

    [tex]
    3z' + \frac{z}{x} = 0
    [/tex]

    and few steps I won't write here I got

    [tex]
    \log |z|^3 = \log |x| + C
    [/tex]

    [tex]
    |z|^3 = e^{C}|x|
    [/tex]

    [tex]
    z_1 = C\sqrt[3]{x}
    [/tex]

    [tex]
    z_2 = -C\sqrt[3]{x}
    [/tex]

    Well, I think this is the problematic step although I think it's ok.

    To get the proper [itex]C = C(x)[/itex] for the non-homogenous equation I expressed [itex]z[/itex] in terms of

    [itex]C(x)[/itex] and put it to the equation. It involved another ODE at the end of which I got

    [tex]
    \log |C| = \log e^{C}\frac{1}{\sqrt[3]{x^2}}
    [/tex]

    [tex]
    C = Q\frac{1}{\sqrt[3]{x^2}}
    [/tex]

    I know that I should actually also write that

    [tex]
    C_2 = -Q\frac{1}{\sqrt[3]{x^2}}
    [/tex]

    but this won't give me anything new since changing the sign in front of [itex]C[/itex] in expression

    [tex]
    z = \pm C\sqrt[3]{x}
    [/tex]

    will give all possibilites.

    Concerning [itex]Q[/itex], I got

    [tex]
    Q = \left(\frac{x^2 + 1}{2}\arctan x - \frac{x}{2} + R\right)
    [/tex]

    and so

    [tex]
    C = \frac{1}{\sqrt[3]{x^2}}\left(\frac{x^2 + 1}{2}\arctan x - \frac{x}{2} + R\right)
    [/tex]

    [tex]
    z = \pm\frac{1}{\sqrt[3]{x}}\left(\frac{x^2 + 1}{2}\arctan x - \frac{x}{2} + R\right)
    [/tex]

    and finally

    [tex]
    y = z^3 = \pm\frac{1}{x}\left(\frac{x^2 + 1}{2}\arctan x - \frac{x}{2} + R\right)
    [/tex]



    Well, with the plus-signed solution, it satisfies the original equation while with the minus sign it doesn't. Which is
    something that I think can be seen already from the original ODE since the left side will get negative sign while the right
    side doesn't depend on the sign of [itex]y[/itex]


    Anyway, can you see where I did a mistake?


    Thank you very much!
     
    Last edited: Dec 25, 2005
  2. jcsd
  3. Dec 25, 2005 #2

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    Hey Twoflower. What up? Me, when I got to:

    [tex]3z^{'}+\frac{z}{x}=3x^{2/3}\text{ArcTan[x]}[/tex]

    I'd treat it like a regular first order ODE and solve for the integrating factor. You know:

    [tex]\sigma=x^{1/3}[/tex]

    so that:

    [tex]d\left[x^{1/3}z\right]=x\text{Arctan[x]}[/tex]

    leaving:

    [tex]x^{1/3}z=-\frac{x}{2}+\frac{1}{2}\text{Arctan[x]}+
    \frac{x^2}{2}\text{Arctan[x]}+c[/tex]
     
  4. Dec 25, 2005 #3

    Thank you Saltydog. Your solution is basically the same I got excepting I also have the same solution also with the minus sign.

    We had been told integrating factor as an alternative to method of variation of parameters I used here since I like it more.

    You know, there has to be some flaw in my approach since the minus-sign solution doesn't satisfy the ODE while the same solution, only with positive sign, does. And I can't see, why should I exclude the minus-sign solution...
     
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