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Proof of reduction of order

  1. Mar 8, 2015 #1
    1. The problem statement, all variables and given/known data
    The method of reduction can be used to solve non-homogeneous equations with non-constant coefficients given by y''+py'+qy=g
    If Y1 is a known solution to the corresponding homogeneous differential equation, let

    y = vY1, w = v'

    Show that if w satisfies the linear first-order equation

    Y1w' + (2Y1' + pY1)w = g
    then y is a solution of the original differential equations

    3. The attempt at a solution

    I am not exactly sure what I am supposed to prove here, am I supposed to prove that w satisfies that equation and then take that y is a solution as a given, or am I supposed to take that w satisfies that equation, solve it then prove y is a solution?
     
  2. jcsd
  3. Mar 9, 2015 #2

    Zondrina

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    Homework Helper

    Imagine for a moment you have an ordinary, second order, linear, homogeneous differential equation in Cauchy Euler form:

    $$ax^2y'' + bxy' + cy = 0$$

    This will be used to explain reduction order. In standard form, the equation takes the form:

    $$y'' + \frac{b}{ax}y' + \frac{c}{ax^2}y = 0$$
    $$y'' + p(x)y' + q(x)y = 0$$

    The point ##x_0 = 0## turns out to be a regular singular point since ##p(x)## and ##q(x)## are not analytic at ##0##, but ##(x - x_0)p(x)## and ##(x - x_0)^2q(x)## are both analytic at ##0##. The solution may be undefined at ##0## as a result, so assume ##x \neq 0##. So for ##x > 0##, we seek solutions of the form ##y = x^r##.

    Taking derivatives ##y'## and ##y''##, and plugging them back into the original differential equation, the indicial equation is obtained:

    $$ar(r - 1) + br + c = 0$$

    Suppose the indicial roots are equivalent, that is ##r_1 = r_2##. Then one solution of the Cauchy Euler equation takes the form:

    $$y_1 = x^r$$

    Where ##r_1 = r_2 = r##. In order to determine a second solution, employ reduction of order. So set:

    $$y_2 = u(x)y_1$$

    We must now determine ##u(x)## in order for ##y_2## to be a solution. The way you do this is by taking derivatives once more:

    $$y_2 = u(x)y_1$$
    $$y_2' = uy_1' + u'y_1$$
    $$y_2'' = uy_1'' + 2u'y_1' + u''y_1$$

    Subbing these derivatives back into the Cauchy Euler equation and doing some math will allow you to find ##y_2##.

    The process of reduction order is the same for other equations. You start with one solution ##y_1## and determine a second solution ##y_2 = u(x)y_1## by finding a function ##u(x)## that makes ##y_2## a solution.
     
  4. Mar 9, 2015 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    What you are to do is take the first and second derivatives of y= vY1, insert them into the given second order equation, and show that this can be reduced to the correct first order equation for w.
     
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