Method of Reduction of Order

We are going to show that in order to find a fundamental system for Equation (8.2.1), it is sufficient to have the knowledge of a solution of Equation (8.2.1). In other words, if we know one (non-zero) solution $y_1$ of Equation (8.2.1), then we can determine a solution $y_2$ of Equation (8.2.1), so that $\{y_1, y_2\}$ forms a fundamental system for Equation (8.2.1). The method is described below and is usually called the method of reduction of order.

Let $y_1$ be an every where non-zero solution of Equation (8.2.1). Assume that $y_2 = u(x) y_1$ is a solution of Equation (8.2.1), where $u$ is to be determined. Substituting $y_2$ in Equation (8.2.1), we have (after a bit of simplification)

$$u^{\prime\prime} y_1 + u^\prime (2 y_1^\prime + p y_1) + u (y_1^{\prime\prime} + p y_1^\prime + q y_1) = 0.$$

By letting $u^\prime = v,$ and observing that $y_1$ is a solution of Equation (8.2.1), we have

$$v^\prime y_1 + v ( 2 y_1^\prime + p y_1) = 0$$

which is same as

$$\frac{d}{dx}(v y_1^2) = - p ( v y_1^2).$$

This is a linear equation of order one (hence the name, reduction of order) in $v$ whose solution is

$$v y_1^2 = exp( - \int_{x_0}^x p(s) ds), \;\; x_0 \in I.$$

Substituting $v = u^\prime$ and integrating we get

$$u = \int_{x_0}^x \frac{1}{y_1^2(s)} exp( - \int_{x_0}^s p(t) dt) ds, \; x_0 \in I$$

and hence a second solution of Equation (8.2.1) is

$$y_2 = y_1 \int_{x_0}^x \frac{1}{y_1^2(s)} exp( - \int_{x_0}^s p(t) dt) ds.$$

It is left as an exercise to show that $y_1, y_2$ are linearly independent. That is, $\{y_1, y_2\}$ form a fundamental system for Equation (8.2.1).

We illustrate the method by an example.

EXAMPLE 8.2.10   Given that e $y_1 = \displaystyle\frac{1}{x}, \; x \geq 1$ is a solution of

$$x^2 y^{\prime\prime} + 4 x y^\prime + 2 y = 0,$$ (8.2.11)

determine another solution $y_2$ of (8.2.11), such that the solutions $y_1, y_2, \;$ for $x \geq 1$ are linearly independent.
Solution: With the notations used above, note that $x_0 = 1, \; p(x) = \displaystyle \frac{4}{x}, \;$ and $y_2(x) = u(x) y_1(x),$ where $u$ is given by

$$u = \int_1^x \frac{1}{y_1^2(s)} \exp\left( - \int_1^s p(t) dt \right) ds$$

$$= \int_1^x \frac{1}{y_1^2(s)} \exp\bigl( \ln (s^4) \bigr) ds$$

$$= \int_1^x \frac{s^2}{s^4}ds = 1 - \frac{1}{x};$$

where $A$ and $B$ are constants. So,

$$y_2(x) = \frac{1}{x} - \frac{1}{x^2}.$$

Since the term $$\frac{1}{x}$$ already appears in $y_1,$ we can take $y_2 = \displaystyle\frac{1}{x^2}.$ So, $$\frac{1}{x}$$ and $$\frac{1}{x^2}$$ are the required two linearly independent solutions of (8.2.11).

EXERCISE 8.2.11   In the following, use the given solution $y_1,$ to find another solution $y_2$ so that the two solutions $y_1$ and $y_2$ are linearly independent.

1. $y^{\prime\prime} = 0, \; y_1 = 1, \; x \geq 0.$
2. $y^{\prime\prime} + 2 y^\prime + y = 0, \; y_1 = e^x, \; x \geq 0.$
3. $x^2 y^{\prime\prime} - x y^\prime + y = 0, \; y_1 = x, \; x \geq 1.$
4. $x y^{\prime\prime} + y^\prime = 0, \; y_1 = 1, \; x \geq 1.$
5. $y^{\prime\prime} + x y^\prime - y = 0, \; y_1 = x, \; x \geq 1.$