# Proof of reduction of order

1. Mar 8, 2015

### Panphobia

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. Mar 9, 2015

### Zondrina

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.

3. Mar 9, 2015

### HallsofIvy

Staff Emeritus
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.