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Math Help - Example of Fundamental Solutions of Linear Homogeneous Equations

  1. #1
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    Example of Fundamental Solutions of Linear Homogeneous Equations

    An example task from Boyce & DiPrima's "Elementary Differential Equations and Boundary Value Problems" sounds as follow (p. 145, Example 1):

    "Find the longest interval in which the solution of the initial value problem

    (t^2 - 3t)y'' + ty' - (t + 3)y = 0, y(1) = 2, y'(1) = 1

    is certain to exist.

    If the given differential equation is written in the form of Eq. (4), then p(t) = 1/(t-3), q(t) = -(t+3)/t(t-3), and g(t) = 0.

    ..."

    The example does not stop at this point and I think I understand the rest of it, however I don't understand how the values of p(t) and q(t) are determined, how is this accomplished?

    The Eq. (4) which is refered to is part of Theorem 3.2.1. which sounds as follows:

    "Consider the initial value problem

    y'' + p(t)y' + q(t)y = g(t), y(t0) = y0, y'(t0) = y'0, (4)

    where q, p, and g are continuous on an open interval I that contains the point t0. Then there is
    exactly one solutions y = (t) of this problem, and the solution exists throughout the interval I."

    So again my question is, how are p(t) and q(t) determined?


    Thanks in advance!
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  2. #2
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    Quote Originally Posted by posix_memalign View Post
    An example task from Boyce & DiPrima's "Elementary Differential Equations and Boundary Value Problems" sounds as follow (p. 145, Example 1):

    "Find the longest interval in which the solution of the initial value problem

    (t^2 - 3t)y'' + ty' - (t + 3)y = 0, y(1) = 2, y'(1) = 1
    We want to divide through by t^2-3t=t(t-3) in order to do that we need to know the function is non-zero. We also have that 1 needs to be contained in this interval by the initial value condition. Therefore, it is the interval (-\infty,3).
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