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Math Help - 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.

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

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

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

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

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

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

     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.

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

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

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

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

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

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

     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 y' then

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

    so

     <br />
\frac{y'^2}{2} = - \frac{a y^2}{2} - \frac{b}{y} + c_1<br />

    You can solve for 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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    y' = \sqrt {- a y^2} - \sqrt {\frac{2b}{y}} + c_1

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

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

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

     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

    <br />
\frac{y'^2}{2} = - \frac{a y^2}{2} - \frac{b}{y} + c_1

    to

    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

    <br />
\frac{y'^2}{2} = - \frac{a y^2}{2} - \frac{b}{y} + c_1

    to

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

    <br />
\frac{y'^2}{2} = - \frac{a y^2}{2} - \frac{b}{y} + c_1 \rightarrow

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

    \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

    \chi \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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