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Non-Linear ODE’s

  1. Jan 25, 2004 #1
    I know that there is a general analytic method to solve the following non-linear differential equation

    [tex]
    \frac {dy} {dx} = ay^2 + by + c
    [/tex]

    … where a, b and c are constants. It is just a Riccati equation generalized to constant coefficients. I am wondering if there is a analytic method to solve the following non-linear differential equation where a, b, c and d are constants…

    [tex]
    \frac {dy} {dx} = ay^3 + by^2 + cy + d
    [/tex]

    Thanks in advanced.
     
    Last edited by a moderator: Jan 25, 2004
  2. jcsd
  3. Jan 25, 2004 #2
    Found something extra...

    Actually I have found the answer and something extra that is quite interesting. The extra thing I have discovered tells us about the solvability of such differential equations…

    Consider the general differential equation below, which the differential equations in my first post are just specific cases off…

    [tex]
    \frac {dy} {dx} = \sum_{n=0}^k a_{n} y^{n}
    [/tex]

    The right hand side of the equation is a polynomial of kth degree in y with constant coefficients a_n. Let’s say we know the zeroes of the polynomial so we can rewrite it as…

    [tex]
    \frac {dy} {dx} = \prod_{n=1}^k ( y - r_{n} )
    [/tex]

    where r_n are the zeroes of the polynomial. Rearranging and using the method of partial fractions we arrive at the following…

    [tex]
    \frac {dy} {dx} \prod_{n=1}^k \frac {1} {y - r_{n}} = 1
    [/tex]

    [tex]
    \frac {dy} {dx} \sum_{n=1}^k \frac {p_{n}} {y - r_{n}} = 1
    [/tex]

    where the numbers p_n only depend on the zeroes of the polynomial and therefore are constant. We can integrate the equation and after some manipulation arrive at the following…

    [tex]
    dy \sum_{n=1}^k \frac {p_{n}} {y - r_{n}} = dx
    [/tex]

    [tex]
    \int dy \sum_{n=1}^k \frac {p_{n}} {y - r_{n}} = \int dx
    [/tex]

    [tex]
    \sum_{n=1}^k p_{n} ln(y - r_{n}) = x + C
    [/tex]

    [tex]
    ln( \prod_{n=1}^k (y - r_{n})^{p_{n}} ) = x + C[/tex]

    [tex]
    \prod_{n=1}^k (y - r_{n})^{p_{n}} = e^{x+C}
    [/tex]

    where C is the constant of integration. We can now see that the solution, y, depends on the roots r_n (even p_n depends on the roots). So we can say that if we know the roots of the polynomial in y we can solve the differential equation. There are general methods of finding the roots of polynomial equations upto order four, quartic equations, so it can be stated that if the polynomial in y is of order four or below the differential equation can be solved. What about higher orders? Well we can’t hope to find an analytic solution if the polynomial in y is of order five or higher. If we could find an analytic solution that means we would know the roots of the polynomial and would have done it in a finite number of operations. But this is a violation of Abel’s impossibility theorem, which states that the roots of a polynomial of order five or higher cannot be found by a finite number of operations. So in conclusion if the order of the polynomial in y is less than five then an analytic solution can be found, otherwise an analytic solution does not exist.
     
    Last edited by a moderator: Jan 25, 2004
  4. Jan 25, 2004 #3

    selfAdjoint

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    Neat! But an analytic solution MAY exist for degree 5 and higher, it doesn't just GENERALLY exist. God is in the details!
     
  5. Jan 25, 2004 #4
    Yes you are quite right. An analytic solution exists if the polynomial in y can be fully factored, in general quintic and higher order polynomial equations cannot be fully factored but there are some specific cases where they can be.
     
  6. Mar 30, 2004 #5
    The general solution

    Well, guys;
    Look what Maple 9 says (see the attachment AbelConst.gif).
    So, if you can determine the integral (use Maple 9 for it in your particular case, it can't be determined for any set {a,b,c,d}), you'll be able to solve the equation.
    For example, see attachment AbelEx1.gif.
    Good luck,
    Max.
     

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    Last edited: Apr 2, 2004
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