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Bases of solutions

  1. Oct 24, 2007 #1
    If for example, I have.
    Y1=Cos(2Ln(x)) and Y2=Sin(2Ln(x)) and I have to reach the general solution.

    I know how to get to the general solution is the cauchy equation:
    y'' X^n + y' x + 4 y = 0
    According to the answer, n=2 => y'' X^2 + y' x + 4 y = 0
    How am I to know that that it is X^2 ?
    Or is it always X^2 in a Euler Cauchy equation?
    Thanks in advance.
    Last edited: Oct 24, 2007
  2. jcsd
  3. Oct 25, 2007 #2


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    An "Euler-Cauchy" equation (often called an "equipotential" equation) is the simplest kind of linear equation with variable coefficients. Part of the definition says that the variable derivative of each derivative is simply a power of x with the exponent being the same as the order of the derivative. So, yes. The function multiplying the second derivative MUST be some constant times x2. (Surely the book does not give it as "X^2"- text book authors do no usually interchange small letters and capital letters!) Since you are given two independent solutions, the differential equation they solve will be second order. The simplest way to solve that problem would be to write the very general x^2y" + Bxy' + Cy= 0, plug in the two given solutions and you will have two linear equations for B and C.
    ax^2 y'+ b

    I responded, two days ago, thinking the problem was to find the differential equation. Rereading it I see that you wanted to know how to find the "general solution".

    That's much easier. The set of all solutions to any "homogenous linear nth order differential equation" form a vector space of dimension n. In this case, the equation if of order 2 so the "solution space" has dimension 2. That means that any set of 2 "independent vectors" (independent solutions) forms a basis and so any vector (solution) can be written as a linear combination of the two solutions. If cos(2ln(x)) and sin(2ln(x)) are independent (they are, but you should show that, perhaps by looking at the Wronskian) then any solution can be written in the form C1cos(2ln(x))+ C2sin(2ln(x)) (the "general solution").
    Last edited by a moderator: Oct 28, 2007
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