# Reduction of order

### Reduction of order

#### Lessons

A linear homogeneous second order differential equation is of the form:

$a(x) y''+b(x) y'+c(x)y=0$

There will be two solution to the above differential equation:

$y_1 (x)=f(x)$
$y_2 (x)=g(x)$

And all general solutions will be of the form:

$y(x)=c_1 f(x)+c_2 g(x)$

The method of Reduction of order assumes that one solution is already know (i.e. we know that $y_1 (x)=f(x)$ is a solution the above differential equation) and uses a specific method to find another solution.

Note:

We will use interchangeably $v(x)=v$, and other shorthand notation of that form.

Steps to solve Reduction of Order problems:
1) Assume that your second solution is of the form $y_2 (x)=v(x)f(x)$, where $y_1 (x)=f(x)$ is the solution we know.
2) Find $y_2'$, and $y_2'$'
3) Insert $y_2, y_2'$, and $y_2''$ into $a(x) y''+b(x) y'+c(x)y=0$
4) Get rid of the solution with $y_1 (x)=f(x)$, and rearrange your equation to isolate $v'$ and $v''$
5) Substitute $w=v'$, and $w'=v''$
6) Solve for $w$ in the new first order differential equation
7) Substitute $w=v'$ back in and then solve for $v$
8) The second solution to the differential equation is $y_2 (x)=v(x)f(x)$, with the $v$ found from the previous steps.
• Introduction
a)
What is the Reduction of Order Method?

b)
A quick recap of the steps in the Reduction of Order Method

• 1.
Using Reduction of Order
Find the general solution to the following homogeneous linear second order differential equation:
$xy''+y'=0$

Where it is known that $y_1 (x)=\tilde{c}_1$ is a solution
Where $\tilde{c}_1$ is a constant.

• 2.
Find the particular solution to the following homogeneous linear second order differential equation:
$x^2 y''-xy'+y=0$

Where it is known that $y_1 (x)=x$ is a solution. The initial values are $y(1)=2$, and $y' (1)=1$
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