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Math Help - Solve the following ODE .. #6

  1. #1
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    Solve the following ODE .. #6

    Hello .

    Question:

    Solve the following ODE :
    y^2dx + (x^2+3y+4y^2)dy = 0


    Solution:

    It is not Linear or Bernoulli in x or y.
    It is not Exact.
    It is not Separable.
    It can not be converted it to an exact equation by the integrating factor.

    I tried to substitute t=y^2 \implies dy=\dfrac{dt}{2\sqrt{t}} to get:

    2t^{3/2}dx+(x^2+3\sqrt{t}+4t)dt=0

    Which is ,again, I do not know what is its type!

    Maybe it can be solved by inspection, but I can not figure it out.
    Any help?
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  2. #2
    MHF Contributor chisigma's Avatar
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    After some algebraic manipulations the DE becomes...

    \displaystyle \frac{dx}{dy} = -\frac{x^{2}}{y^{2}} - \frac{3}{y} - 4 (1)

    ... where y is the independent variable. The (1) is a Riccati's equation, the solution of which is in most cases a little hard...

    Kind regards

    \chi \sigma
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  3. #3
    A Plied Mathematician
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    Mathematica obtains the following solution of the Ricatti equation that chisigma mentioned:

    Let

    f_{1}(y)=\sqrt{y}J_{i\sqrt{15}}\left(\dfrac{2\sqrt  {3}}{\sqrt{y}}\right)+\sqrt{3}J_{-1+i\sqrt{15}}\left(\dfrac{2\sqrt{3}}{\sqrt{y}}\rig  ht),

    f_{2}(y)=-\sqrt{3}J_{1+i\sqrt{15}}\left(\dfrac{2\sqrt{3}}{\s  qrt{y}}\right)+\sqrt{y}J_{-i\sqrt{15}}\left(\dfrac{2\sqrt{3}}{\sqrt{y}}\right  )C_{1},

    f_{3}(y)=\sqrt{3}J_{-1-i\sqrt{15}}\left(\dfrac{2\sqrt{3}}{\sqrt{y}}\right  )C_{1}-\sqrt{3}J_{1-i\sqrt{15}}\left(\dfrac{2\sqrt{3}}{\sqrt{y}}\right  )C_{1}.

    Then

    x(y)=-\dfrac{\sqrt{y}\left(f_{1}(y)+f_{2}(y)+f_{3}(y)\ri  ght)}{2\left(J_{i\sqrt{15}}\left(\dfrac{2\sqrt{3}}  {\sqrt{y}}\right)+J_{-i\sqrt{15}}\left(\dfrac{2\sqrt{3}}{\sqrt{y}}\right  )C_{1}\right)}.

    WolframAlpha gives this.
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  4. #4
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    Just to add a bit to this thread. If you let

    x = f(y) \dfrac{u'}{u}

    in Chisigma equation (1) and pick f(y) accordingly, you can linearize your ODE.
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  5. #5
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    Just to add (part 2). Under the change of variables

    x = \dfrac{12}{X^2},\;\;\; y = \dfrac{XY}{\sqrt{6}}

    this linear ODE becomes

    X^2 Y'' + X Y' +\left(X^2 + 15\right)Y = 0

    which explains the odd Bessel function solutions that Adrian picked up.
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  6. #6
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    This is too advanced for me.
    Thanks.
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