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Solving general linear ODE's

  1. Jan 16, 2008 #1
    1. The problem statement, all variables and given/known data

    Is there a standard method for finding the general solution to any given linear homogeneous ODE?

    I tried working it out myself. Here's what I tried. I'm almost positive it's wrong because it doesn't take into account most of the coefficients. But maybe I'm on the right track?

    3. The attempt at a solution
    [tex]a_0(x)y + a_1(x)y' + \cdots + a_{n-1}(x)y^{(n-1)} + y^{(n)} = b(x)\qquad\qquad\qquad(1)[/tex]​

    Now, recall that the product rule for derivatives of order [tex]n[/tex] is

    [tex]\frac{d^n}{dx^n}[f(x)g(x)] = {\sum^n_{k=0} \binom{n}{k} f^{(k)}(x)g^{(n-k)}}(x).[/tex]​

    We want to find some function [tex]u(x)[/tex] that, when multiplied to both sides of [tex](1)[/tex], will put the left side in the form of the [tex]n[/tex]th degree product rule. In other words,

    [tex]u(x)a_0(x)y + u(x)a_1(x)y' + \cdots + u(x)y^{(n)} = \frac{d^n}{dx^n}[u(x)y]\qquad\qquad\qquad(2)[/tex]​

    After expanding [tex](2)[/tex], we can set each term of the left side equal to the corresponding term on the right (e.g. [tex]u(x)a_2(x) y' = u^{(n-1)}(x)y'[/tex] ) and, after canceling out the all the [tex]y[/tex]'s, form a system. Also note that since [tex]u(x)y^{(n)}[/tex] appears on both sides of [tex](2)[/tex] it cancels out. From here until I say otherwise, all functions will be referenced, for convenience, without the "[tex](x)[/tex]" appended to it.

    [tex]
    \left \{
    \begin{array}{lcl}
    u^{(n)} &=& u a_0\\
    u^{(n-1)} &=& \dbinom{n}{1} u a_1\\
    & \vdots\\
    u'' &=& \dbinom{n}{n-2} u a_{n-2}\\
    u' &=& \dbinom{n}{n-1} u a_{n-1}\\
    \end{array}
    \right .
    \qquad\qquad\qquad(3)[/tex]​


    Rearranging [tex](3)[/tex], we have

    [tex]
    \left \{
    \begin{array}{lcl}
    \dfrac{u^{n}}{u^{(n-1)}} &=& \dfrac{a_1}{a_0} \\
    \\
    \dfrac{u^{(n-1)}}{u^{(n-2)}} &=& \dbinom{n}{1} \dfrac{a_2}{a_1}\\
    & \vdots\\
    \dfrac{u'''}{u''} &=& \dbinom{n}{n-2} \dfrac{a_{n-2}}{a_{n-3}}\\
    \\
    \dfrac{u''}{u'} &=& \dbinom{n}{n-1} \dfrac{a_{n-1}}{a_{n-2}}\\
    \end{array}
    \right .
    \qquad\qquad\qquad(4)[/tex]​


    Solving each equation in [tex](4)[/tex], we get

    [tex]u^{(k)} = e^{\binom{n}{n-k} {\int \frac{a_{n-k}}{a_{n-(k+1)}}dx }}[/tex].​


    Setting [tex]k=0[/tex],

    [tex]u(x) = e^{\int \frac{a_n(x)}{a_{n-1}(x)}dx}\qquad\qquad\qquad(5)[/tex]​


    Now, going back to [tex](1)[/tex] and [tex](2)[/tex], since, by definition,

    [tex]\frac{d^n}{dx^n}[u(x)y] = u(x)b(x),[/tex]​

    we can plug in [tex](5)[/tex],
    [tex]\frac{d^n}{dx^n}\left[ e^{\int \frac{a_n(x)}{a_{n-1}(x)}dx} y \right] = e^{\int \frac{a_n(x)}{a_{n-1}(x)}dx}b(x)\qquad\qquad\qquad(6)[/tex]​

    and then antidifferentiate, which, at the moment, I am feeling too lazy to actually do.
     
    Last edited: Jan 16, 2008
  2. jcsd
  3. Jan 16, 2008 #2

    rock.freak667

    User Avatar
    Homework Helper

    Are there not various methods to solve first order ODE's with constant coefficients such as variables are separable,integrating factor and such?
     
  4. Jan 16, 2008 #3
    Yes, but I was wondering about methods to solve general degree ODE's with variable coefficients.
     
  5. Jan 17, 2008 #4
    You have introduced one function [tex]u(x)[/tex] and you also have n equations in equation (3).

    In general your solution for [tex]k=0[/tex], i.e. equation (5), does not satisfy all members of equation (3).
    Thus there is no [tex]u(x)[/tex] in general.
     
  6. Jan 17, 2008 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Series solutions (including Frobenius' method) are the most general methods of solving linear equations.
     
  7. Jan 17, 2008 #6
    Yeah. I had a feeling it was something like that.
     
  8. Jan 17, 2008 #7
    Hm. Darn. I don't like using series solutions. They're so cumbersome. Ah, well.

    Also, I'm assuming the laplace and fourier transforms, in general, can't be used to solve linear ODE's with variable coefficients? I know that there are a few specific cases where they work, but, yeah.
     
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