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General solution of a linear ordinary differential equation

  1. Sep 30, 2008 #1
    Hi all.
    Can the general solution of a linear ordinary differential equation be expressed in terms of its initial conditions?
    It seems that I have seem this kind of representation.
    It makes "some sense" to me but I hope to know if there is some "proof" or explanation of why it can be?

    To be specific, for a n-th order ODE,
    the solution is something like
    y = y(xo)(something) + y'(xo)(something) + ... + y(n-1)(xo)(something)...

    why?
     
  2. jcsd
  3. Oct 1, 2008 #2
    Hallo. The equation you mention here looks only like a Taylor- Mac Laurin development of the y(x0 + epsilon) function arond y(x0). Your something should be a factor (1/N!). epsilon ^N with N = 0, 1, ... Why it is so ? Well a good question but I need help too. Certainly the explanation is related to the methods of the infinitesimal calculus. The fact is that such a development demonstrates the relationship between the function in or at x0 plus epsilon and the function in or at xo. So yes conditions somewhere else have an influence on what hapens here.
     
  4. Oct 1, 2008 #3

    HallsofIvy

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    Staff Emeritus
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    Given any linear, nth (homogeneous) ordinary differential equation, you can define its "fundamental" solutions (at x= x0) as the solutions satisfiying
    1) Y1(x0)= 1, Y1'(x0= 0, Y1"(x0)= 0, ...
    2)Y2(x0)= 0, Y2'(x0)= 1, Y2"(x0)= 0, ...
    .
    .
    .
    n)Yn(x0)= 0, Yn'(x0)= 0, Yn"(x0= 0, ..., Yn(n)(x0)= 1.

    Any solution to the differential equation can be written as a linear combination of those: Y(x)= C1Y1(x)+ C1Y2(x)+ ...+ CnYn(x).

    If you set x= x0 in that equation you get C1= Y(x0).
    If you differentiate both sides of the equation to get Y'(x)= C1Y1'(x)+ C2Y2'(x)+ ... and set = x0, you get C2=Y'(x0). Continuing in that way, you see that the coefficients are precisely the "initial values" for Y.
     
  5. Oct 1, 2008 #4
    i have problem understanding the following..
    "define its "fundamental" solutions (at x= x0) as the solutions satisfiying
    1) Y1(x0)= 1, Y1'(x0= 0, Y1"(x0)= 0, ...
    2)Y2(x0)= 0, Y2'(x0)= 1, Y2"(x0)= 0, ...
    .
    .
    .
    n)Yn(x0)= 0, Yn'(x0)= 0, Yn"(x0= 0, ..., Yn(n)(x0)= 1."

    why is this so...?
     
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