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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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4t
so
e yi = (1/5)e + (1/10)e + ci. For the second equation
the integrating factor is e (e-2ty2)' = (4/5)e-4t - (2/5)e
2t
so
Hence
-2t.
y2 = -(1/5)e-4t + (2/5)e
-t
y =
1/5
-1/5
e-2t +
1/10
2/5
et +
+ c2. Thus
Finally,
-3t
c1e-3t
c2e2t /
multiplying by T, we obtain
x = Ty = ' 0 ė e-2t + ' 1/2 1 e ' 11 e-3t + c2 ' 11
rt
+
c
H
v-1 , v 0 , V-4, v 1 ,
2t
The last two terms are the general solution of the corresponding homogeneous system, while the first two terms constitute a particular solution of the nonhomogeneous system.
e
Section 7.9
159
12. Since the coefficient matrix is the same as that of Problem 3, use the same procedure as done in that problem, including the ?-1 found there. In the interval n/2 < t < n sint > 0 and cost < 0; hence |sint| = sint, but |cost| = -cost.
14. To verify that the given vector is the general solution of the corresponding system, it is sufficient to substitute it into the D.E. Note also that the two terms in x(c) are linearly independent. If we seek a solution of the form x = ?(t)u(t) then we find that the equation corresponding to Eq.(26) is t?(t)u'(t) = g(t), where
?(t) =
It 1/tė 2
1-t2
and g(t) = . Thus
t V 2t -
3
t
u' = (1/t)?-1(t)g (t). Using a computer algebra system or row operations on ? and I, we find that
3
?
-1
t/2 t/2 2
and hence u1 =
2t
31 -  -  and 22t
-1 t2 -3 3t
u2 = ----- +  + t, which yields ui =  - --------------------- - lnt + ci
2 2 2t 2
1 t3 t2
and u2 = -  t +  +  + c2. 2 6 2
Multiplication of u by ?(t) yields the desired solution.
160
CHAPTER 8
Section 8.1, Page 427
In the following problems that ask for a large number of numerical calculations the first few steps are shown. It is then necessary to use these samples as a model to format a computer program or calculator to find the remaining values.
1a. The Euler formulas is yn+1 = yn + h(3 + tn - yn) for
n = 0,1,2,3... and with t0 = 0 and y0 = 1. Thus
y1 = 1 + .05(3 + 0 - 1) = 1.1
y2 = 1.1 + .05(3 + .05 - 1.1) = 1.1975 @ y(.1)
y3 = 1.1975 + .05(3 + .1 - 1.1975) = 1.29263
y4 = 1.29263 + .05(3 + .15 - 1.29263) = 1.38549 @ y(.2).
1c. The backward Euler formula is yn+i = yn + h(3 + tn+i - yn+i).
Solving this for yn+1 we find yn+1= [yn + h(3 + tn+1)]/(1+n).
1 + .05(3.05)
Thus y1 = ------------------ = 1.097619 and
1.05
1.097619 + .05(3.1)
y2 = ------------------------------ = 1.192971.
2 1.05
y0 + 2t0y0 (.5)2 + 0
5a. y1 = y0 + h-------------------------------------------------- - = .5 + .05- = .504167
3 + t20 3 + 0
(.504167)2 + 2(.05)(.504167)
y2 = .504167 + .05--------------------------------------- = .509239
3 + (.05) 2
y2 + 2(.05)y1
5c. y1 = .5 + .05----------- , which is a quadratic
3 + (.05) 2
equation in y1. Using the quadratic formula, or an equation solver, we obtain y1 = .5050895. Thus
Ó2 + 2(.1)Ó2
y2 = .5050895 + .05--------------- which is again quadratic
3 + (.1)2
in y2, yielding y2 = .5111273.
7a. For part a eighty steps must be taken, that is,
n = 0,1,...79 and for part b 160 steps must taken with
n = 0,1,...159. Thus use of a programmable calculator or
a computer is required.
7c. We have yn+1 = yn + h(.5-tn+1 + 2yn+1), which is linear in
yn + .5h -htn+1
yn+1 and thus we have yn+1 = -----------------------. Again, 80
1 - 2h
Section 8.1
161
steps are needed here and 160 steps in part d. In This case a spreadsheet is very useful. The first few, the middle three and last two lines are shown for h = .025:
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