Reduction of Order Method

When one solution of a homogeneous linear equation is known, the reduction of order method finds a second linearly independent solution by reducing the order of the differential equation.

DefinitionReduction of Order

Given the second-order equation $y'' + p(x)y' + q(x)y = 0$ with one known solution $y_1(x) \neq 0$ , a second solution can be found in the form:

$y_2(x) = v(x)y_1(x)$

where $v(x)$ satisfies a first-order equation.

Derivation

Substituting $y = vy_1$ into the differential equation:

$y' = v'y_1 + vy_1'$ $y'' = v''y_1 + 2v'y_1' + vy_1''$

The equation becomes:

$(v''y_1 + 2v'y_1' + vy_1'') + p(v'y_1 + vy_1') + q(vy_1) = 0$

Since $y_1'' + py_1' + qy_1 = 0$ , the $v$ terms cancel:

$v''y_1 + 2v'y_1' + pv'y_1 = 0$

$v''y_1 + v'(2y_1' + py_1) = 0$

Let $w = v'$ :

$w'y_1 + w(2y_1' + py_1) = 0$

This is separable:

$\frac{dw}{w} = -\frac{2y_1' + py_1}{y_1}dx$

ExampleFinding Second Solution

Given $y'' - 2y' + y = 0$ with known solution $y_1 = e^x$ , find $y_2$ .

Here $p(x) = -2$ . Using the formula:

$w' = -w\left(\frac{2e^x - 2e^x}{e^x}\right) = 0$

So $w = c_1$ , thus $v' = c_1$ , giving $v = c_1x + c_2$ .

Taking $c_1 = 1, c_2 = 0$ :

$y_2 = xe^x$

General solution: $y = c_1e^x + c_2xe^x = (c_1 + c_2x)e^x$

Standard Formula

For $y'' + p(x)y' + q(x)y = 0$ with known solution $y_1$ :

$y_2 = y_1\int \frac{e^{-\int p(x)dx}}{y_1^2}dx$

ExampleVariable Coefficients

Solve $x^2y'' - 2xy' + 2y = 0$ given $y_1 = x$ .

First convert to standard form: $y'' - \frac{2}{x}y' + \frac{2}{x^2}y = 0$

So $p(x) = -2/x$ . Using the formula:

$y_2 = x\int \frac{e^{-\int (-2/x)dx}}{x^2}dx = x\int \frac{e^{2\ln x}}{x^2}dx = x\int \frac{x^2}{x^2}dx = x\int dx = x^2$

General solution: $y = c_1x + c_2x^2$

Remark

The reduction of order method works for equations of any order, not just second-order. If an $(n-1)$ -dimensional solution space is known for an n-th order equation, the method can find the remaining solution.

ExampleEuler Equation

For Euler equations $x^2y'' + axy' + by = 0$ , if we know one solution (often found by trying $y = x^r$ ), reduction of order finds all others systematically.

Given $x^2y'' - xy' + y = 0$ with $y_1 = x$ :

$y_2 = x\int \frac{e^{\int (1/x)dx}}{x^2}dx = x\int \frac{x}{x^2}dx = x\ln x$

The reduction of order method is particularly valuable for variable-coefficient equations where other methods fail, transforming an n-th order problem into an $(n-1)$ -th order one.