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Thread: Non linear 2nd order differential equation

  1. #1
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    Non linear 2nd order differential equation

    25 July 2009

    I am not able to find the general function that integrates the following 2nd order differential equation;

    y^2 y + a y^3 b = 0

    in which 0 ≤ a ≤ 1 is a constant;
    b > 0 is a constant.

    The equation relates to the behaviour (orbit/motion) of a massless point subjected to a particular central motion.
    I could easily find only the solutions associated with nil values for constants a and b.

    I would be deeply grateful to any friend who can help me resolve the problem.
    Many thanks for any help.
    Leo Klem
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  2. #2
    Newbie Deco's Avatar
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    Uhm, let me see if I know what you're asking.

    $\displaystyle y'' = -ay + \frac{b}{y^2} $

    $\displaystyle y' = -a \int {y} dy + b \int {y^{-2}} dy $

    $\displaystyle y' = -a(\frac {y^2}{2} + C_1) + b(-y^{-1} + C_2) $

    $\displaystyle y' = -a\frac {y^2}{2} - by^{-1} + C_t $

    $\displaystyle y = -a \int (\frac {y^2}{2})dy - b \int (y^{-1})dy + \int C_t dy$

    $\displaystyle y = -a (\frac {y^3}{6} + C_3) - b (\ln y + C_4) + C_ty + C_5 $

    $\displaystyle y = - \frac {ay^3}{6} - b \ln y + C_ty + C_T $

    It's either that or I'm totally off the mark.
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  3. #3
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    Quote Originally Posted by Deco View Post
    Uhm, let me see if I know what you're asking.

    $\displaystyle y'' = -ay + \frac{b}{y^2} $

    $\displaystyle y' = -a \int {y} dy + b \int {y^{-2}} dy $

    $\displaystyle y' = -a(\frac {y^2}{2} + C_1) + b(-y^{-1} + C_2) $

    $\displaystyle y' = -a\frac {y^2}{2} - by^{-1} + C_t $

    $\displaystyle y = -a \int (\frac {y^2}{2})dy - b \int (y^{-1})dy + \int C_t dy$

    $\displaystyle y = -a (\frac {y^3}{6} + C_3) - b (\ln y + C_4) + C_ty + C_5 $

    $\displaystyle y = - \frac {ay^3}{6} - b \ln y + C_ty + C_T $

    It's either that or I'm totally off the mark.
    You're off the mark on both integrations (but close on the first)!

    If you multiply your equation by $\displaystyle y'$ then

    $\displaystyle y'y'' = -ay y' + \frac{b y'}{y^2} $

    so

    $\displaystyle
    \frac{y'^2}{2} = - \frac{a y^2}{2} - \frac{b}{y} + c_1
    $

    You can solve for $\displaystyle y'$ and separate but I think the resulting integral will be difficult.
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  4. #4
    Newbie Deco's Avatar
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    $\displaystyle y' = \sqrt {- a y^2} - \sqrt {\frac{2b}{y}} + c_1$

    $\displaystyle y'= \sqrt {-a} \int y dy - \sqrt {2b} \int y^{-\frac {1}{2}} dy + \int C_1 dy $

    $\displaystyle y = \sqrt {-a} ( \frac {y^2}{2}) - \sqrt {2b} (2y^{\frac {1}{2}}) + C_1y + C_2$

    Something like that?

    But since a is positive, wouldn't this result in a complex answer?
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  5. #5
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    Quote Originally Posted by Deco View Post
    $\displaystyle y' = \sqrt {- a y^2} - \frac{2b}{y} + c_1$

    $\displaystyle y'= \sqrt {-a} \int y dy - \sqrt {2b} \int y^{-\frac {1}{2}} dy + \int C_1 dy $

    $\displaystyle y = \sqrt {-a} ( \frac {y^2}{2}) - \sqrt {2b} (2y^{\frac {1}{2}}) + C_1y + C_2$

    Something like that?
    How on earth did you go from

    $\displaystyle
    \frac{y'^2}{2} = - \frac{a y^2}{2} - \frac{b}{y} + c_1$

    to

    $\displaystyle y' = \sqrt {- a y^2} - \frac{2b}{y} + c_1$?
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  6. #6
    Newbie Deco's Avatar
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    Oops. Wow. Don't know what I was thinking.
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  7. #7
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    Non linear 2nd order equation.Thanks, with a clarification

    Many thanks to my two first interlocutors. I have to point out that the problem described by the differential equation regards a central motion. The integrating function, if it does exist, is actually useful in the explicit form given by
    y = f(x; a, b, C, D)

    in which x is the independent variable (it represent the "orbital" angle), a and b are the given equation constants, and C, D are the two integration constants.
    Solutions in the form of an expression of functions of y do actually create difficult problems of interpretation as to the kind of motion regarded.
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  8. #8
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Danny View Post
    How on earth did you go from

    $\displaystyle
    \frac{y'^2}{2} = - \frac{a y^2}{2} - \frac{b}{y} + c_1$

    to

    $\displaystyle y' = \sqrt {- a y^2} - \frac{2b}{y} + c_1$?
    It is evident that the correct procedure is...

    $\displaystyle
    \frac{y'^2}{2} = - \frac{a y^2}{2} - \frac{b}{y} + c_1 \rightarrow$

    $\displaystyle \rightarrow y^{'}= +/- \sqrt{\frac{- a y^{3} - 2 b + 2 y c_{1}}{y}} \rightarrow $

    $\displaystyle \rightarrow \int \sqrt{\frac{y}{-a y^{3} -2b + 2 y c_{1}}} dy = +/- x + c_{2}$ (1)

    May be that the integral in (1) is a little difficult to solve ... may be ...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  9. #9
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    What beyond the inverse function?

    Thanks to Chisigma. You got the correct procedure, but the solution is still away. It's the point where I had actually to stop. It's clearly a "solution" in the form of an inverse function, i.e., the argument expressed as a function of the dependent variable. The integral you indicate seems however difficult to solve, but in a few particular cases, such as - for instance - assuming C1 = 0.
    Could you please provide further suggestions?
    Nome a parte, sono italiano anch'io, ma non sono un matematico. Grato comunque.
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