Reduction of Order: Finding a Second Solution from a Known One

A second-order equation needs two independent solutions, but sometimes only one can be found by inspection. This section shows how to manufacture the second from the first. The substitution $y_{2} = v (x) y_{1} (x)$ reduces the equation to a first-order equation for v', and the result is a formula. Best of all, the method does not require constant coefficients.

Quick Reference

Item	Statement
Equation	y'' + P(x)y' + Q(x)y = 0, in standard form
Given	One solution $y_{1} (x)$ , not identically zero
Substitution	$y_{2} (x) = v (x) y_{1} (x)$
The formula	$v (x) = \int \frac{1}{y_{1}^{2}} e^{- \int P (x) d x} d x$
Wronskian produced	$W (y_{1}, y_{2}) = e^{- \int P (x) d x}$ , never zero
Result	${y_{1}, y_{2}}$ is a fundamental set
Restriction	None on the coefficients; $P$ and $Q$ may be any continuous functions
Practical note	Multiply $y_{2}$ by any convenient nonzero constant to tidy it

Motivation

We know that the general solution of the homogeneous equation

y'' + P(x)y' + Q(x)y = 0 \tag{i}

c_{1} y_{1} (x) + c_{2} y_{2} (x),

where $y_{1} (x)$ and $y_{2} (x)$ are two linearly independent solutions of (i), and $c_{1}$ and $c_{2}$ are constants.

Sometimes we can find one solution of (i) by inspection. Trying $y = x$ , $y = x^{2}$ , $y = e^{x}$ , or $y = 1$ in an equation with polynomial coefficients frequently works, and applied problems often supply one solution on physical grounds. What can we do in that case, and how can we find the other independent solution?

The Theorem

If $y_{1} (x)$ is a solution of

y'' + P(x)y' + Q(x)y = 0 ,

and if we set $y_{2} (x) = v (x) y_{1} (x)$ where

v (x) = \int \frac{1}{y_{1}^{2}} e^{- \int P (x) d x} d x,

then $y_{2} (x)$ is a solution of this equation and ${y_{1} (x), y_{2} (x)}$ is a fundamental set of solutions.

Notice that this method is not restricted to equations with constant coefficients. It applies to any second-order linear homogeneous equation whose coefficients are continuous, which is precisely why it remains useful long after the characteristic equation method runs out.

Proof

We assume that $y_{2} (x) = v (x) y_{1} (x)$ is a solution of y'' + P(x)y' + Q(x)y = 0, and try to determine the unknown function $v (x)$ .

Differentiating twice with the product rule,

\begin{aligned} y_{2}'(x) &= v'y_{1} + vy_{1}' \\ y_{2}''(x) &= v''y_{1} + 2v'y_{1}' + vy_{1}'' \end{aligned}

Substituting into (i),

v''y_{1} + 2v'y_{1}' + vy_{1}'' + P(x)\left[v'y_{1} + vy_{1}'\right] + Q(x)vy_{1} = 0

or, after rearranging by powers of derivatives of $v$ ,

v''y_{1} + v'\left(2y_{1}' + P(x)y_{1}\right) + v\left[\underbrace{y_{1}'' + P(x)y_{1}' + Q(x)y_{1}}_{=\,0}\right] = 0 . \tag{ii}

Since $y_{1}$ is a solution of (i), the expression in square brackets vanishes. This is the pivotal step, and it is why the method needs a known solution: without it, nothing cancels.

Equation (ii) reduces to

v''y_{1} + v'\left(2y_{1}' + P(x)y_{1}\right) = 0 .

Note that $v$ itself no longer appears, only v' and v''. So this is really a first-order equation in the unknown v', which is where the name "reduction of order" comes from.

Dividing by v'y_1,

\frac{v''}{v'} = -2\frac{y_{1}'}{y_{1}} - P(x) .

Integrating both sides, and recognizing \displaystyle\int\frac{v''}{v'}dx = \ln v' and \displaystyle\int\frac{y_1'}{y_1}dx = \ln y_1,

\ln v' = -2\ln y_{1} - \int P(x)\,dx

\begin{aligned} v' &= \exp\left(\ln\frac{1}{y_{1}^{2}} - \int P(x)\,dx\right) \\ &= \exp\left(\ln\frac{1}{y_{1}^{2}}\right)\exp\left(-\int P(x)\,dx\right) \\ &= \frac{1}{y_{1}^{2}}\,e^{-\int P(x)\,dx} \end{aligned}

and thus

v = \int \frac{1}{y_{1}^{2}} e^{- \int P (x) d x} d x .

It remains to show that $y_{1}$ and $y_{2}$ are independent, that is, that $W (y_{1}, y_{2}) \neq 0$ :

\begin{aligned} W\left(y_1, y_2\right) = W\left(y_1, vy_1\right) &= \begin{vmatrix} y_1 & vy_1 \\ y_1' & v'y_1 + v y_1' \end{vmatrix} \\ &= v'y_1^{2} + vy_1'y_1 - vy_1y_1' \\ &= v'y_1^{2} \\ &= \frac{1}{y_1^{2}}e^{-\int P(x)\,dx}\,y_1^{2} \\ &= e^{-\int P(x)\,dx} \end{aligned}

Because the exponential function is never zero, $W (y_{1}, y_{2})$ is never zero, so the two solutions are independent.

Note that this is exactly Abel's formula from Section 3.3, with the constant equal to $1$ . The two results are two faces of the same computation.

Do not memorize the formula; understand the substitution. Many students find it easier to substitute $y = v (x) y_{1} (x)$ directly into the given equation, watch the $v$ terms cancel, and solve the resulting first-order equation for v'. That route is self-correcting, whereas a misremembered formula is not.

Worked Examples

By simple inspection, we can discover that $y = x$ is one solution of x^{2}y'' + xy' - y = 0 for $x > 0$ . Find the general solution of this equation.

Solution

First check the given solution: with $y = x$ we have y' = 1 and y'' = 0, so

x^{2} \cdot 0 + x \cdot 1 - x = 0 . ✓

We write a second linearly independent solution as $y_{2} = v (x) y_{1} (x) = v (x) x$ . To use the formula we need the equation in standard form, so divide by $x^{2}$ :

y'' + \frac{1}{x}y' - \frac{1}{x^{2}}y = 0

giving $P (x) = \frac{1}{x}$ . Now

\begin{aligned} v &= \int\frac{1}{y_1^{2}}e^{-\int P(x)\,dx}\,dx \\ &= \int\frac{1}{x^{2}}e^{-\int\frac{1}{x}dx}\,dx \qquad \left(y_1 = x\right) \\ &= \int\frac{1}{x^{2}}e^{-\ln x}\,dx \\ &= \int\frac{1}{x^{2}}e^{\ln\frac{1}{x}}\,dx \qquad \left(B\ln A = \ln A^{B}\right) \\ &= \int\frac{1}{x^{3}}\,dx \qquad \left(e^{\ln A} = A\right) \\ &= -\frac{1}{2}x^{-2} = -\frac{1}{2x^{2}} \end{aligned}

So the second solution is

y_{2} = v (x) \cdot x = - \frac{1}{2 x^{2}} \cdot x = - \frac{1}{2 x} .

Notice that if $y$ is a solution of a homogeneous equation, then $k y$ is also a solution for any constant $k$ . Multiplying by $- 2$ , we may just as well write $y_{2} = \frac{1}{x}$ .

Therefore the general solution of the given differential equation is

y = c_{1} x + \frac{c_{2}}{x} .

Given that $y_{1} = e^{x}$ solves y'' - 2y' + y = 0, find a second independent solution.

Solution

The equation is already in standard form with $P (x) = - 2$ . Then

v = \int \frac{1}{e^{2 x}} e^{- \int (- 2) d x} d x = \int e^{- 2 x} e^{2 x} d x = \int d x = x .

So $y_{2} = x e^{x}$ , and the general solution is $y = c_{1} e^{x} + c_{2} x e^{x}$ .

This confirms from a completely different direction the rule from Section 4.2 that a repeated root $r_{1}$ contributes both $e^{r_{1} x}$ and $x e^{r_{1} x}$ . The factor of $x$ is not a convention; it falls out of the integral.

Given that $y_{1} = x^{2}$ solves x^{2}y'' - 3xy' + 4y = 0 for $x > 0$ , find the general solution.

Solution

Verify first: y_1' = 2x, y_1'' = 2, and $x^{2} (2) - 3 x (2 x) + 4 x^{2} = 2 x^{2} - 6 x^{2} + 4 x^{2} = 0$ . ✓

In standard form, $P (x) = - \frac{3}{x}$ , so

e^{- \int P d x} = e^{\int (3 / x) d x} = e^{3 \ln x} = x^{3} .

Therefore

v = \int \frac{x^{3}}{{(x^{2})}^{2}} d x = \int \frac{x^{3}}{x^{4}} d x = \int \frac{d x}{x} = \ln x .

The second solution is $y_{2} = x^{2} \ln x$ , and the general solution is

y = c_{1} x^{2} + c_{2} x^{2} \ln x .

The appearance of a logarithm here is characteristic. It signals a repeated root in the associated Cauchy-Euler problem of Section 4.5, and it plays the same role that the factor of $x$ plays for constant coefficients.

Given that $y_{1} = \sin x$ solves y'' + y = 0, recover the second solution.

Solution

Here $P (x) = 0$ , so $e^{- \int P d x} = 1$ and

v = \int \frac{d x}{\sin^{2} x} = \int \csc^{2} x d x = - \cot x .

Therefore

y_{2} = - \cot x \cdot \sin x = - \frac{\cos x}{\sin x} \cdot \sin x = - \cos x .

Discarding the sign gives $y_{2} = \cos x$ , as expected.

Exercises

Given that $y_{1} = x$ solves \left(1-x^{2}\right)y'' - 2xy' + 2y = 0 on $- 1 < x < 1$ , find a second solution.

Solution

In standard form, $P (x) = \frac{- 2 x}{1 - x^{2}}$ , so

- \int P d x = \int \frac{2 x}{1 - x^{2}} d x = - \ln (1 - x^{2}),

giving $e^{- \int P d x} = \frac{1}{1 - x^{2}}$ .

Therefore

v = \int \frac{1}{x^{2} (1 - x^{2})} d x = \int (\frac{1}{x^{2}} + \frac{1}{1 - x^{2}}) d x = - \frac{1}{x} + \frac{1}{2} \ln \frac{1 + x}{1 - x},

using the partial fractions $\frac{1}{x^{2} (1 - x^{2})} = \frac{1}{x^{2}} + \frac{1}{1 - x^{2}}$ .

The second solution is

y_{2} = x v = - 1 + \frac{x}{2} \ln \frac{1 + x}{1 - x} .

This is Legendre's equation with $n = 1$ , whose second solution is the classical function $Q_{1} (x)$ .

Given that $y_{1} = e^{2 x}$ solves y'' - 4y' + 4y = 0, find the general solution.

Solution

With $P (x) = - 4$ ,

v = \int \frac{e^{4 x}}{e^{4 x}} d x = \int d x = x,

so $y_{2} = x e^{2 x}$ and $y = c_{1} e^{2 x} + c_{2} x e^{2 x}$ , matching the repeated-root rule.

Given that $y_{1} = \frac{\sin x}{\sqrt{x}}$ solves x^{2}y'' + xy' + \left(x^{2} - \tfrac{1}{4}\right)y = 0 on $x > 0$ , find a second solution.

Solution

In standard form, $P (x) = \frac{1}{x}$ , so $e^{- \int P d x} = \frac{1}{x}$ . Since $y_{1}^{2} = \frac{\sin^{2} x}{x}$ ,

v = \int \frac{1 / x}{\sin^{2} x / x} d x = \int \frac{d x}{\sin^{2} x} = - \cot x .

Therefore

y_{2} = - \cot x \cdot \frac{\sin x}{\sqrt{x}} = - \frac{\cos x}{\sqrt{x}},

or simply $\frac{\cos x}{\sqrt{x}}$ after dropping the sign. This is Bessel's equation of order $\frac{1}{2}$ , one of the rare Bessel equations with elementary solutions.

Frequently Asked Questions

Why does the substitution

y_{2} = v y_{1}

lower the order?

Because the coefficient of $v$ in the resulting equation is exactly y_1'' + Py_1' + Qy_1, which is zero since $y_{1}$ solves the equation. With $v$ itself eliminated, only v' and v'' remain, so the equation is first order in the unknown v'.

Must the equation be in standard form?

Yes. The formula reads $P (x)$ off an equation whose leading coefficient is $1$ . Applying it to x^{2}y'' + xy' - y = 0 with $P = x$ instead of $P = 1 / x$ gives a completely wrong answer. Divide by the leading coefficient before you start.

What happened to the constants of integration?

They were dropped, and legitimately so. A constant of integration in the outer integral adds a multiple of $y_{1}$ to $y_{2}$ ; a multiplicative constant scales $y_{2}$ . Neither changes the span of ${y_{1}, y_{2}}$ , and both are absorbed into $c_{1}$ and $c_{2}$ in the general solution. Take the simplest antiderivative available.

What if

y_{1}

is zero at some point of the interval?

The formula divides by $y_{1}^{2}$ , so it breaks down there. Work on a subinterval where $y_{1}$ is nonzero, then extend the resulting $y_{2}$ by continuity. In the $\sin x$ example, $v = - \cot x$ is undefined at multiples of $π$ , yet the final answer $y_{2} = - \cos x$ is perfectly well behaved everywhere.

Does this work for equations of order three or higher?

The same idea applies but is much messier. Substituting $y = v y_{1}$ into an $n$ th order equation reduces it to order $n - 1$ in v', which for $n = 3$ still leaves a second-order equation to solve. There is no clean formula, and in practice other methods are preferred at higher order.

How is this related to Abel's formula?

They are the same result. The Wronskian computed in the proof, $W = e^{- \int P d x}$ , is precisely Abel's formula with $c = 1$ . Conversely, Section 3.3 solved the equation W = y_1y_2' - y_2y_1' for $y_{2}$ , which is another route to the same second solution. Either derivation may be used.

The Use of One Known Solution to Find Another

Quick Reference

Motivation

The Theorem

Worked Examples

Exercises

Frequently Asked Questions

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