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

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228

Chapter 4. Higher Order Linear Equations

EXAMPLE

1

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

u'm (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 <

!>m (t) m= 1 Jt0

giS.WM 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

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

f — / — Ó + 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)e‘

e

Factoring et from each of the first two columns and e-t 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) =

(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 =

W3(t) =

e

te1 0

(t + 1)et 0

(t + 2)et 1

e

te1

(t + 1)el

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

= 2,

e

Y(t) = el

L

g(s)(-1 - 2s)

4es

ds + te1

L

g(s)(2) 4es

ds + e

j

g(s )e'¦ 4es

2s

ds

1

4j' {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. /" + Ó = tan t, 0 < t < n 2. Ó" — Ó = t

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

5. Ó" - Ó' + Ó — Ó = e-t sin t 6. yv + 2Ó' + 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.

7. Ó" — Ó' + Ó — y = sec t, —æ/2 < t < æ/2

8. Ó" — Ó = csc t, 0 < t < æ

In each of Problems 9 through 12 find the solution of the given initial value problem. Then plot a graph of the solution.

9. Ó" + Ó = sec t, ó(0) = 2, Ó(0) = 1, Ó'(0) = —2

10. Ó4 + 2Ó' + ó = sin t, ó(0) = 2, /(0) = 0, /(0) = -1, /"(0) = 1

11. Ó" — Ó' + Ó — Ó = sec t, y(0) = 2, Ó(0) = — 1, Ó'(0) = 1

12. Ó" — Ó = csc t, ó (æ/2) = 2, Ó (æ/2) = 1, Ó'(æ/2) = — 1

13. Given that x, x2, and 1 /x are solutions of the homogeneous equation corresponding to

x3Ó" + x2Ó - 2xy + 2y = 2x4, x > 0,

determine a particular solution.

230

Chapter 4. Higher Order Linear Equations

14. Find a formula involving integrals for a particular solution of the differential equation

/" — y" + y — y = g(t)¦

15. Find a formula involving integrals for a particular solution of the differential equation

yv — y = g(t).

Hint: The functions sin t, cos t, sinh t, and cosh t form a fundamental set of solutions of the homogeneous equation.

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