How Does Reduction of Order Help Solve Non-Homogeneous Differential Equations?

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SUMMARY

The method of reduction of order is utilized to solve non-homogeneous differential equations with non-constant coefficients, specifically in the form y'' + py' + qy = g. Given a known solution Y1 to the corresponding homogeneous equation, the substitution y = vY1 and w = v' leads to the first-order equation Y1w' + (2Y1' + pY1)w = g. By deriving the first and second derivatives of y and substituting them back into the original equation, one can demonstrate that y is indeed a solution to the non-homogeneous equation.

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Homework Statement


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

The Attempt at a Solution


[/B]
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?
 
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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.
 
Panphobia said:

Homework Statement


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

The Attempt at a Solution


[/B]
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?
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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