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

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
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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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