3.2 Reduction of Order

INTRODUCTION

In Section 3.1 we saw that the general solution of a homogeneous linear second-order differential equation

a2(x)y″ + a1(x)y′ + a0(x)y = 0 (1)

was a linear combination y = c1y1 + c2y2, where y1 and y2 are solutions that constitute a linearly independent set on some interval I. Beginning in the next section we examine a method for determining these solutions when the coefficients of the DE in (1) are constants. This method, which is a straightforward exercise in algebra, breaks down in a few cases and yields only a single solution y1 of the DE. It turns out that we can construct a second solution y2 of a homogeneous equation (1) (even when the coefficients in (1) are variable) provided that we know one nontrivial solution y1 of the DE. The basic idea described in this section is that the linear second-order equation (1) can be reduced to a linear first-order DE by means of a substitution involving the known solution y1. A second solution, y2 of (1), is apparent after this first-order DE is solved.

Reduction of Order

Suppose y(x) denotes a known solution of equation (1). We seek a second solution y2(x) of (1) so that y1 and y2 are linearly independent on some interval I. Recall that if y1 and y2 are linearly independent, then their ratio y2/y1 is nonconstant on I; that is, y2/y1 = u(x) or y2(x) = u(x)y1(x). The idea is to find u(x) by substituting y2(x) = u(x)y1(x) into the given differential equation. This method is called reduction of order since we must solve a first-order equation to find u.

The first example illustrates the basic technique.

EXAMPLE 1 Finding a Second Solution

Given that y1 = ex is a solution of y″ − y = 0 on the interval (−∞, ∞), use reduction of order to find a second solution y2.

SOLUTION

If y = u(x)y1(x) = u(x)ex, then the first two derivatives of y are obtained from the product rule:

y′ = uex + exu′, y″ = uex + 2exu′ + exu″.

By substituting y and y″ into the original DE, it simplifies to

y″ − y = ex(u″ + 2u′) = 0.

Since ex ≠ 0, the last equation requires u″ + 2u′ = 0. If we make the substitution w = u′, this linear second-order equation in u becomes w′ + 2w = 0, which is a linear first-order equation in w. Using the integrating factor e2x, we can write d/dx [e2xw] = 0. After integrating we get w = c1e−2x or u′ = c1e−2x. Integrating again then yields u = − c1e−2x + c2. Thus

y = u(x)ex = − e−x + c2ex. (2)

By choosing c2 = 0 and c1 = −2 we obtain the desired second solution, y2 = e−x. Because W(ex, e−x) ≠ 0 for every x, the solutions are linearly independent on (−∞, ∞). ≡

Since we have shown that y1 = ex and y2 = e−x are linearly independent solutions of a linear second-order equation, the expression in (2) is actually the general solution of y″ − y = 0 on the interval (−∞, ∞).

General Case

Suppose we divide by a2(x) in order to put equation (1) in the standard form

y″ + P(x)y′ + Q(x)y = 0, (3)

where P(x) and Q(x) are continuous on some interval I. Let us suppose further that y1(x) is a known solution of (3) on I and that y1(x) ≠ 0 for every x in the interval. If we define y = u(x)y1(x), it follows that

This implies that we must have

(4)

where we have let w = u′. Observe that the last equation in (4) is both linear and separable. Separating variables and integrating, we obtain

We solve the last equation for w, use w = u′, and integrate again:

By choosing c1 = 1 and c2 = 0, we find that y = u(x)y1(x) is a second solution of equation (3). We summarize the result.

THEOREM 3.2.1 Reduction of Order

Let y1(x) be a solution of the homogeneous differential equation (3) on an interval I and that y1(x) ≠ 0 for all x in I. Then

(5)

is a second solution of (3) on the interval I.

It makes a good review of differentiation to verify that the function y2(x) defined in (5) satisfies equation (3). Furthermore, if y2(x) is the solution in (5), then the functions y1(x) and y2(x) are linearly independent on any interval I on which y1(x) is not zero. To prove linear independence, we use Theorem 3.1.3. The Wronskian of the functions is

The general solution of (3) on I is the linear combination

EXAMPLE 2 A Second Solution by Formula (5)

The function y1 = x2 is a solution of x2y″ − 3xy′ + 4y = 0. Find the general solution on the interval (0, ∞).

SOLUTION

From the standard form of the equation

we find from (5)

The general solution on the interval (0, ∞) is given by y = c1y1 + c2y2; that is, y = c1x2 + c2x2 ln x. ≡

REMARKS

  1. We have derived and illustrated how to use (5) because this formula appears again in the next section and in Section 5.2. We use (5) simply to save time in obtaining a desired result. Your instructor will tell you whether you should memorize (5) or whether you should know the first principles of reduction of order.
  2. The integral in (5) may be nonelementary. In this case we simply write the second solution in terms of an integral-defined function:

    where we assume that the integrand is continuous on the interval [x0, x]. See Problems 23 and 24 in Exercises 3.2.

3.2 Exercises Answers to selected odd-numbered problems begin on page ANS-5.

In Problems 1−18, the indicated function y1(x) is a solution of the given equation. Use reduction of order or formula (5), as instructed, to find a second solution y2(x).

  1. y″ − 4y′ + 4y = 0; y1 = e2x
  2. y″ + 2y′ + y = 0; y1 = xe−x
  3. y″ + 16y = 0; y1 = cos 4x
  4. y″ + 9y = 0; y1 = sin 3x
  5. y″ − y = 0; y1 = cosh x
  6. y″ − 25y = 0; y1 = e5x
  7. 9y″ − 12y′ + 4y = 0; y1 = e2x/3
  8. 6y″ + y′ − y = 0; y1 = ex/3
  9. x2y″ − 7xy′ + 16y = 0; y1 = x4
  10. x2y″ + 2xy′ − 6y = 0; y1 = x2
  11. xy″ + y′ = 0; y1 = ln x
  12. 4x2y″ + y = 0; y1 = x1/2 ln x
  13. x2y″ − xy′ + 2y = 0; y1 = x sin(ln x)
  14. x2y″ − 3xy′ + 5y = 0; y1 = x2 cos(ln x)
  15. (1 − 2x – x2)y″ + 2(1 + x)y′ − 2y = 0; y1 = x + 1
  16. (1 − x2)y″ + 2xy′ = 0; y1 = 1
  17. xy″ − (2x + 1)y′ + (x + 1)y = 0; y1 = ex
  18. y″ + 2(tan x)y′ − y = 0; y1 = sin x

In Problems 19–22, the indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find a second solution y2(x) of the homogeneous equation and a particular solution yp(x) of the given nonhomogeneous equation.

  1. y″ − 4y = 2; y1 = e−2x
  2. y″ + y′ = 1; y1 = 1
  3. y″ − 3y′ + 2y = 5e3x; y1 = ex
  4. y″ − 4y′ + 3y = x; y1 = ex

In Problems 23 and 24, the indicated function y1(x) is a solution of the given differential equation. Use formula (5) to find a second solution y2(x) expressed in terms of an integral-defined function. See (ii) in the Remarks at the end of this section.

  1. x2y″ + (x2 − x)y′ + (1 − x)y = 0; y1 = x
  2. 2xy″ − (2x + 1)y′ + y = 0; y1 = ex

Discussion Problems

    1. Give a convincing demonstration that the second-order equation ay″ + by′ + cy = 0, a, b, and c constants, always possesses at least one solution of the form y1 = , m1 a constant.
    2. Explain why the differential equation in part (a) must then have a second solution, either of the form y2 = or of the form y2 = x, m1 and m2 constants.
    3. Reexamine Problems 1–8. Can you explain why the statements in parts (a) and (b) above are not contradicted by the answers to Problems 3−5?
  2. Verify that y1(x) = x is a solution of xy″ − xy′ + y = 0. Use reduction of order to find a second solution y2(x) in the form of an infinite series. Conjecture an interval of definition for y2(x).

Computer Lab Assignment

    1. Verify that y1(x) = ex is a solution of

      xy″ − (x + 10)y′ + 10y = 0.

    2. Use (5) to find a second solution y2(x). Use a CAS to carry out the required integration.
    3. Explain, using Corollary (a) of Theorem 3.1.2, why the second solution can be written compactly as