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Math Help - First-order nonlinear ODE (approach is not clear to me)

  1. #1
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    First-order nonlinear ODE (approach is not clear to me)

    I've been given the following ODE to be solved:

    (yx^2)y' + (2y^2x + 3x^3) = 0

    But I don't really know how to start. A direct integration is (obviously enough) not possible and it's not in the form to be solved by an integrating factor (or so I understand).

    Then it could be solved by separation of variables (though I can't see what the separation itself would be), or by using the exact equation technique (though I'm not even sure how to check if it's in the appropriate form).

    How do I get started?
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  2. #2
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    At the moment this is unsolvable exactly.

    Are you sure you haven't got a "square" in the wrong spot. If it was

    (yx^2)\frac{dy}{dx} + (2yx^2 + 3x^3) = 0 this would be separable...
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  3. #3
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    Quote Originally Posted by Rafael Almeida View Post
    I've been given the following ODE to be solved:

    (yx^2)y' + (2y^2x + 3x^3) = 0
    \displaystyle\frac{dy}{dx}=-\frac{2{{y}^{2}}x+3{{x}^{3}}}{{{x}^{2}}y}=-2\cdot \frac{y}{x}-3\cdot \frac{x}{y}.

    homogeneous ODE.
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  4. #4
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    I just double, no, triple-checked it here, and the square is indeed at the "y", as I posted it. What might have happened is that the professor mistyped it when writing the exam (I'm doing a post-mortem), but that would be at least unexpected since I don't remember anybody complaining about it during the exam nor did he make any corrections.

    So, the equation as I originally posted is not solvable at all? What does "unsolvable exactly" mean? Because, according to Wolfram Alpha (solve - Wolfram|Alpha[(y+x^2)dy/dx+%2B+2y^2+x+%2B+3x^3+%3D%3D+0]), it does give me a solution.

    PS: I forgot to mention, the original problem says to solve the IVP given by the equation I posted and the condition y(2) = 2
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  5. #5
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    The LHS is almost a product rule. Try this:

    yy'x^{2}+2y^{2}x+3x^{3}=0,\;\quad y(2)=2 is the original DE.

    Now (x^{2}y^{2})'=2xy^{2}+2x^{2}yy'. Hence,

    2yy'x^{2}+2y^{2}x+3x^{3}=yy'x^{2},\;\quad y(2)=2.

    Let u=x^{2}y^{2}. Then

    u'=2xy^{2}+2x^{2}yy'=2\dfrac{u}{x}+2x^{2}yy', from which we may obtain

    \dfrac{u'}{2}-\dfrac{u}{x}=x^{2}yy'. Substitute this into the DE as we re-wrote it, and you obtain the following:

    u'+3x^{3}=\dfrac{u'}{2}-\dfrac{u}{x},\;\quad u(2)=16.

    This equation is first-order linear.
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  6. #6
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    Another approach. For (yx^2)y' + (2y^2x + 3x^3) = 0 try letting u = y^2. It should make it linear in u.
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