# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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In each of Problems 27 and 28 use the method of reduction of order (Problem 26) to solve the given differential equation.

27. (2 - t)Σ" + (2t - 3)y" - ty + y = 0, t < 2; y1(t) = e

28. t2(t + 3)Σ" - 3t(t + 2)Σ + 6(1 + t)y - 6y = 0, t > 0; yx(t) = t2, y2(t) = t3

4.2 Homogeneous Equations with Constant Coefficients

Consider the nth order linear homogeneous differential equation

L [y] = a0 y(n) + a! y(n-1) + + an-X Σ + any = 0, (1)

where a0, a1,..., an are real constants. From our knowledge of second order linear equations with constant coefficients it is natural to anticipate that y = ert is a solution ofEq. (1) for suitable values of r. Indeed,

L [ert] = ert(a0rn + a1rn-1 + + an-1r + an) = ert Z(r) (2)

for all r, where

Z(r) = a0rn + axrn-1 + + an-1r + an. (3)

For those values of r for which Z(r) = 0, it follows that L[ert] = 0 and y = ert is a

solution ofEq. (1). The polynomial Z (r) is called the characteristic polynomial, and

the equation Z(r) = 0 is the characteristic equation of the differential equation (1).

4.2 Homogeneous Equations with Constant Coefficients

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A polynomial of degree n has n zeros,1 say rv r2,..., rn, some of which may be equal; hence we can write the characteristic polynomial in the form

Z(r) = a0(r - r1)(r - r2) ¦¦¦(r - rn). (4)

Real and Unequal Roots. If the roots of the characteristic equation are real and no

two are equal, then we have n distinct solutions ΊΓ1*, ef2t,..., ernl of Eq. (1). If these functions are linearly independent, then the general solution of Eq. (1) is

y = c1er1t + c2er2l + ¦¦¦ + cnernl. (5)

One way to establish the linear independence of er1 (, er2t,..., ernt is to evaluate their

Wronskian determinant. Another way is outlined in Problem 40.

EXAMPLE 1

Find the general solution of

/" + /'- 7f- Σ + 6y = 0. (6)

Also find the solution that satisfies the initial conditions

y(0) = 1, y (0) = 0, /(0) = -2, Σ"(0) = -1 (7)

and plot its graph.

Assuming that y = ert, we must determine r by solving the polynomial equation

r4 + r3 - 7r2 - r + 6 = 0. (8)

The roots of this equation are r1 = 1, r2 = 1, r3 = 2, and r4 = 3. Therefore the

general solution of Eq. (6) is

y = c1et + c2e~l + c3e2t + c4e-3t. (9)

The initial conditions (7) require that cv ..., c4 satisfy the four equations

q + c2 + c3 + c4 = 1,

q - c2 + 2c3 - 3c4 = 0,

cj + c2 + 4c3 + 9c4 = -2, q - c2 + 8c3 - 27c4 = -1.

(10)

By solving this system of four linear algebraic equations, we find that

cl = 11/8, c2 = 5/12, c3 = -2/3, c4 = -1/8.

Therefore the solution of the initial value problem is

y = Δ e* + 12 e-t - 3 e2t - 8 e-3t. (11)

The graph of the solution is shown in Figure 4.2.1.

1An important question in mathematics for more than 200 years was whether every polynomial equation has

at least one root. The affirmative answer to this question, the fundamental theorem of algebra, was given by

Carl Friedrich Gauss in his doctoral dissertation in 1799, although his proof does not meet modern standards of rigor. Several other proofs have been discovered since, including three by Gauss himself. Today, students often meet the fundamental theorem of algebra in a first course on complex variables, where it can be established as a

consequence of some of the basic properties of complex analytic functions.

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Chapter 4. Higher Order Linear Equations

FIGURE 4.2.1 Solution of the initial value problem of Example 1.

As Example 1 illustrates, the procedure for solving an nth order linear differential equation with constant coefficients depends on finding the roots of a corresponding nth degree polynomial equation. If initial conditions are prescribed, then a system of n linear algebraic equations must be solved to determine the proper values of the constants c1, , c. While each of these tasks becomes much more complicated as n increases, they can often be handled without difficulty with a calculator or computer.

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