# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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+ cnyn (0-

(2)

4.4 The Method of Variation of Parameters

227

The method of variation of parameters for determining a particular solution of Eq. (1) rests on the possibility of determining n functions u1, u2,, un such that Y(t) is of the form

Y (t) = u 1(t)y1(t) + u2(t)y2(t) + ? ? ? + un (t) yn (t). (3)

Since we have n functions to determine, we will have to specify n conditions. One of these is clearly that Y satisfy Eq. (1). The other n — 1 conditions are chosen so as to make the calculations as simple as possible. Since we can hardly expect a simplification in determining Y if we must solve high order differential equations for u 1,..., un, it is natural to impose conditions to suppress the terms that lead to higher derivatives of u1, . . . , un. From Eq. (3) we obtain

Y' = (u1 y1 + u2y2 + - ? ? + "Jn) + (u1 y1 + u2y2 + ? ? ? + unyn^ (4)

where we have omitted the independent variable t on which each function in Eq. (4) depends. Thus the first condition that we impose is that

u1 y1 + u2 y2 + ? ? ? + unyn = 0. (5)

Continuing this process in a similar manner through n — 1 derivatives of Y gives

Y(m) = u1 y(m) + u2y(m) + ? ? ? + uny(m), m = 0, 1, 2,..., n — 1, (6)

and the following n — 1 conditions on the functions u 17..., un:

u 1 y1m—1) + u’2y2m—1) + ? ? ? + u'Jr1 = 0, m = 1, 2,..., n — 1. (7)

The nth derivative of Y is

Y(n) = u y1n) + ? ? ? + ujnn)) + (ui y1n—1) + ? ? ? + u.nynn—l)). (8)

Finally, we impose the condition that Y must be a solution of Eq. (1). On substituting for the derivatives of Y from Eqs. (6) and (8), collecting terms, and making use of the fact that L[y ] = 0, i = 1, 2,..., n, we obtain

u1 y{n—1) + u2y2n—1:1 + ? ? ? + ujr1 = g. (9)

Equation (9), coupled with the n — 1 equations (7), gives n simultaneous linear non-homogeneous algebraic equations for ur1, iW2,..., u'n:

y1u 1 + y2u 2 + ? ? ? + ynu'n = ^

y1 u1 + y2u 2 + + yn u'n = 0,

/I11! + y^u2 + —+ ynu'n =0, (10)

y(r1]u'1 + ? ? ? + y^w» = g.

The system (10) is a linear algebraic system for the unknown quantities ur1,..., u’n. By solving this system and then integrating the resulting expressions, you can obtain the coefficients u1,..., un .A sufficient condition for the existence of a solution of the system of equations (10) is that the determinant of coefficients is nonzero for each value of t. However, the determinant of coefficients is precisely W(y1, y2,..., yn), and it is nowhere zero since y1,..., yn are linearly independent solutions of the homogeneous

228

Chapter 4. Higher Order Linear Equations

EXAMPLE

1

equation. Hence it is possible to determine iYl,..., u'n. Using Cramer’s rule, we find that the solution of the system of equations (10) is

Um (t) =

g(t) Wm(t) W(t) -

m = 1, 2,..., n.

(11)

Here W(t) = W(y1,y2,..., yn)(t) and Wm is the determinant obtained from W by replacing the mth column by the column (0, 0,..., 0, 1). With this notation a particular solution of Eq. (1) is given by

Y (t ) =

n p t

J^Ym (t)

m= 1 Jt0

gisWgl ds,

W(s)

(12)

where t0 is arbitrary. While the procedure is straightforward, the algebraic computations involved in determining Y( t) from Eq. (12) become more and more complicated as n increases. In some cases the calculations may be simplified to some extent by using Abel’s identity (Problem 20 of Section 4.1),

W (t) = W (yv ..., yn )(t) = c exp

lp1(t )

dt

The constant c can be determined by evaluating W at some convenient point.

Given that y() = el, y2(t) = tel, and y3 (t) = e 1 are solutions of the homogeneous equation corresponding to

/' - / - / + y = g(t ),

determine a particular solution of Eq. (13) in terms of an integral. We use Eq. (12). First, we have

(13)

W(t) = W(et, tet, e-t)(t) =

(t + 1)e' (t + 2)el

e

Factoring el from each of the first two columns and e 1 from the third column, we obtain

W (t) = e(

t1 t + 1 -1

t + 2 1

Then, by subtracting the first row from the second and third rows, we have

W(t) = e

1

-2

0

Finally, evaluating the latter determinant by minors associated with the first column, we find that

W(t) = 4e".

-t

te e

e

e

t

e

e

4.4 The Method of Variation of Parameters

229

PROBLEMS

Next,

W(t) =

te

(t + 1)et (t + 2) et

e

Using minors associated with the first column, we obtain

Wi(t) =

te1 (t + 1)et

e

= -2t- 1.

In a similar way

and

W2(0 =

W() =

e

te1 0

(t + 1)e' 0

(t + 2)et 1

e

tet

(t + 1)et

Substituting these results in Eq. (12), we have

= 2,

e

Y(t) = el

t

g(s)(-1 - 2s)

4es

ds + te1

t

g(s)(2) 4es

ds + e

tt

t

g(s )e'? 4es

2s

ds

1 fl

{et-s[-1 + 2(t - s)] + e~(t-s)} g(s) ds.

e

e

t

e

t

0

e

e

e

e

0

e

e

t

1

e

e

e

e

e

e

e

In each of Problems 1 through 6 use the method of variation of parameters to determine the general solution of the given differential equation.

1. /" + y = tan t, 0 < t < n 2. y — y = t

3. y — 2y" — y + 2y = e4t 4. y'' + y = sec f, —n/2 < t < n/2

5. /" - y" + y7 - y = e-t sin t 6. yv + 2y" + y = sin t

In each of Problems 7 and 8 find the general solution of the given differential equation. Leave your answer in terms of one or more integrals.

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