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Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
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8. Follow the steps outlined in Problem 7. Solve the first
3 1 '
D.E. for x2 to obtain x2 = x1 - x1. Substitute this
// /
into the second D.E. to obtain x1 - x1 - 2x1 = 0, which
2t -t
has the solution x1 = c1e + c2e . Differentiating this and substituting into the above equation for x2 yields x2
1 2t -t
= -c,e + 2c2e . The I.C. then give
2 1 2
11 ci + c2 = 3 and c1 + 2c2 = , which yield
11 2 11 2t 2 -t
c = , c2 = -. Thus x1 = e - e and
1 3 2 3 1 3 3
11 2t 4 -t
x2 = e - e . Note that for large t, the second
2 6 3
11 2t
term in each solution vanishes and we have x1 = e and
13
11 2t
x2 = e , so that x1 = 2x2. This says that the graph 6
will be asymptotic to the line x1 = 2x2 for large t.
4 ' 5
9. Solving the first D.E. for x2 gives x2 = x-, - -x1,
2 2 3 1 3 1
which substituted into the second D.E. yields
f t/2 2t
x1 - 2.5x1 + x1 = 0. Thus x1 = c^ + c2e and
x2 = -c1et/2 + c2e2t. Using the I.C. yields c1 = -3/2 and c2 = -1/3. For large t, x1 = (-1/2)e2t and x2 = (-1/2)e2t and thus the graph is asymptotic to x1 = x2 in the third quadrant. The graph is shown on the right.
122
Section 7.1
12.
9.
12. Solving the first D.E. for x2 gives x2 = x-, + xn and
2 2 2 1 4 1
substitution into the second D.E. gives
///17 -t/2
x1 = 0. Thus x1 = e (c1cos2t + c2sin2t) and
x0
-t/2
e (c2cos2t-c1sin2t). The I.C. yields c1 = -2 and
2.
14. If a12 ^ 0, then solve the first equation for x2,
obtaining x2 = [x^ - a11x1 - g1(t)]/a12. Upon substituting
this expression into the second equation, we have a second order linear O.D.E. for x1. One I.C. is
x1(0) = x1. The second I.C. is
x2(0) = [x1(0) - a11x1(0) - g1(0)]/a12 = x2. Solving for
f f 0 0 x1(0) gives x1(0) = a12x2 + a11x1 + g1(0). These results
hold when a11, ...,a22 are functions of t as long as the
derivatives exist and a12(t) and a21(t) are not both zero
on the interval. The initial conditions will involve a11(0) and a12(0).
19. Let us number the nodes 1,2, and 3 clockwise beginning with the top right node in Figure 7.1.4. Also let I1,
I2, I3, and I4 denote the currents through the resistor
1
R = 1, the inductor L = 1, the capacitor C = , and the
2
resistor R = 2, respectively. Let V1, V2, V3, and V4 be
the corresponding voltage drops. Kirchhoff's first law applied to nodes 1 and 2, respectively, gives
i) I1
I
0 and (ii) I2
I
I
0. Kirchhoff's
second law applied to each loop gives (iii) V1 + V2 + V3 = 0 and (iv) V3
V4 = 0.
The current-
c
2
Section 7.1
123
voltage relation through each circuit element yields four more equations: (v) V1 = I4, (vi) I2 = v2,
(vii) (1/2)V3 = I3 and (viii) V4 = 2I4. We thus have a
system of eight equations in eight unknowns, and we wish to eliminate all of the variables except I2 and V3 from
this system of equations. For example, we can use Eqs.(i) and (iv) to eliminate I1 and V4 in Eqs.(v) and
(viii). Then use the new Eqs.(v) and (viii) to eliminate V1 and I4 in Eqs.(ii) and (iii). Finally, use the new
Eqs. (ii) and (iii) in Eqs.(vi) and (vii) to obtain
i2 = - I2 - V3, V3 = 2I2 - V3. These equations are
identical (when subscripts on the remaining variables are dropped) to the equations given in the text.
21a. Note that the amount of water in each tank remains
constant. Thus Q1(t)/30 and Q2(t)/20 represent oz./gal of salt in each tank. As in Example 1 of Section 2.3, we assume the mixture in each tank is well stirred. Then, for the first tank we have dQ4 Qi(t) Q2(t)
= 1.5 - 3----------- + 1.5-------, where the first term on
dt 30 20
the right represents the amount of salt per minute
entering the mixture from an external source, the second
term represents the loss of salt per minute going to Tank
2 and the third term represents the gain of salt per
minute entering from Tank 2. Similarly, we have
dQ2 Qi(t) Q2(t)
= 3 + 3----------- - 4------- for Tank 2.
dt 30 20
21b. Solve the second equation for Q1(t) to obtain Q1(t) =
10Q2 +2Q2 - 30. Substitution into the first equation
1 9
then yields 10Q2 + 3Q2 + Q2 = . The steady state 2 2 8 2 2 E
solution for this is Q2 = 8(9/2) = 36. Substituting this
E
value into the equation for Q1 yields Q-l = 72 - 30 = 42.
21c. Substitute Q4 = x4 + 42 and Q2 = x2 + 36 into the equations found in part a.
124
Section 7.2
Section 7.2, Page 355
1a. 2A =
6 4 -2
so that
' 2+4 -4-2 0 + 3 " ' 6 -6 3 N
2A + B = 6-1 4 + 5 -2 + 0 = 5 9 -2
v-4 + 6 2 + 1 6+2, v 2 3 CO
1c. Using Eq.(9) and following Example 1 we have
AB
4 + 2 + 0
12 - 2 - 6
-8 - 1 + 18
-2 - 10 + 0 -6 + 10 - 1 4 + 5 + 3
(AB)C = A(BC) =
7 -11
11 20 17
3 -12
3 + 0 + 0
9 + 0 - 2
-6 + 0 + 6
which yields the correct answer.
' 6 -5 -7 " ' 5 3 3 N
6. AB = 1 9 1 and BC = -1 7 3
v-1 -2 8, V 2 3 -2 ,
so that
In problems 10 through 19 the method of row reduction, as illustrated in Example 2, can be used to find the inverse matrix or else to show that none exists. We start with the original matrix augmented by the indentity matrix, describe a suitable sequence of elementary row operations, and show the result of applying these operations.
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