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

**Download**(direct link)

**:**

**92**> 93 94 95 96 97 98 .. 609 >> Next

The characteristic equation is

r2 + 4r + 4 = (r + 2)2 = 0,

so rJ = r2 = — 2. Therefore one solution ofEq. (5) is y() = e—2t. To find the general solution ofEq. (5) we need a second solution that is not a multiple of y1. This second solution can be found in several ways (see Problems 20 through 22); here we use a method originated by D’Alembert5 in the eighteenth century. Recall that since y1(') is a solution ofEq. (1), so is cy1(') for any constant c. The basic idea is to generalize this observation by replacing c by a function v(') and then trying to determine v(') so that the product v(') yfi) is a solution ofEq. (1).

To carry out this program we substitute y = v(')yJ(') inEq. (1) and use the resulting equation to find v('). Starting with

y = v(t)/jC ) = v(t )e—2t, (6)

we have

/ = v'(' )e—2t — 2v(' )e—2t (7)

and

y" = v”{' )e—2t — 4v'(' )e—2t + 4v(t )e—2t. (8)

By substituting the expressions in Eqs. (6), (7), and (8) in Eq. (5) and collecting terms, we obtain

[v"(') — 4v'(') + 4v(') + 4v'(') — 8v(') + 4v(' )]e—2t = 0, which simplifies to

v"(') = 0. (9)

5Jean d’Alembert (1717-1783), a French mathematician, was a contemporary of Euler and Daniel Bernoulli, and is known primarily for his work in mechanics and differential equations. D’Alembert’s principle in mechanics and d’Alembert’s paradox in hydrodynamics are named for him, and the wave equation first appeared in his paper on vibrating strings in 1747. In his later years he devoted himselfprimarily to philosophy and to his duties as science editor of Diderot’s Encyclopedie.

162

Chapter 3. Second Order Linear Equations

Therefore

and

v'(') = c1 v(') = c1' + c2,

(10)

where c1 and c2 are arbitrary constants. Finally, substituting for v(t) in Eq. (6), we obtain

y = cj'e 2t + c2e 2t.

(11)

The second term on the right side of Eq. (11) corresponds to the original solution y1(') = exp(—2'), but the first term arises from a second solution, namely y2(') = ' exp(—2'). These two solutions are obviously not proportional, but we can verify that they are linearly independent by calculating their Wronskian:

W (y1; y2)(') =

2t

'e

2t

2e

2t

(1 2t)e

2t

= e—4t — 2'e—4t + 2'e—4t = e—4t = 0.

Therefore

Ó³ (') = e'

2t

y2(') = 'e

2t

(12)

form a fundamental set of solutions ofEq. (5), and the general solution of that equation is given by Eq. (11). Note that both y1 (') and y2(') tend to zero as' ^ to; consequently, all solutions ofEq. (5) behave in this way. The graph of a typical solution is shown in Figure 3.5.1.

The procedure used in Example 1 can be extended to a general equation whose characteristic equation has repeated roots. That is, we assume that the coefficients in Eq. (1) satisfy b2 4ac = 0, in which case

Ó³ (') = e—bt/2a

is a solution. Then we assume that

y = v(t) Ó³(') = v(' )e—bt/2a (13)

e

3.5 Repeated Roots; Reduction of Order

163

EXAMPLE

2

and substitute in Eq. (1) to determine v(t). We have

and

y' = v'it)e- bt/2a - bv(t)e- bt/2a 2a

b b2

y" = v'\t )e-bt/2a - - v'it )e-bt/2a + ^ vit )e-bt/2a.

a 4a2

(14)

(15)

Then, by substituting in Eq. (1), we obtain

b b2 Ã , b 1

v"(t) - -v'(t) + -1 v(t) b v'(t) - --- v(t)

a 4a2 + 2a

+ cv(t) I e-bt/2a = 0. (16)

Canceling the factor exp(-bt/2a), which is nonzero, and rearranging the remaining terms, we find that

b2 b2

av (t) + (-b + b)v (t) + I 4- - — + cl v(t) = 0.

(17)

The term involving v' (t)is obviously zero. Further, the coefficient of v(t) is c - (b2/4a), which is also zero because b2 - 4ac = 0 in the problem that we are considering. Thus, just as in Example 1, Eq. (17) reduces to

therefore,

Hence, from Eq. (13), we have

v"(t) = 0; v(t) = c1t + c2.

y = c1te-bt/2a + c2e-bt/2a. Thus y is a linear combination of the two solutions

y1(t) = e-bt/2a, y2(t) = te-bt/2a

The Wronskian of these two solutions is

-bt/2a te~bt/'2a

bt/2a

W(yv y2)(t) =

e b

------e

2a

1 - ^ e-bt/2a 2a

— e-bt/a.

(18)

(19)

(20)

Since W(yv y2)(t) is never zero, the solutions y1 and y2 given by Eq. (19) are a fundamental set of solutions. Further, Eq. (18) is the general solution ofEq. (1) when the roots of the characteristic equation are equal. In other words, in this case, there is one exponential solution corresponding to the repeated root, while a second solution is obtained by multiplying the exponential solution by t.

**92**> 93 94 95 96 97 98 .. 609 >> Next