BOYCE Resolvidos
BOYCE Resolvidos
BOYCE Resolvidos
ACT
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Capítulo 1
CHAPTER 1 1
dV
16. The D.E. expressing the evaporation is = - aS, a > 0.
dt
2/3
4 3 Ê 3 ˆ
= pr and S = = 4pÁÁ ˜˜
2
Now V 4pr , so S V2/3.
3 Ë 4p ¯
dv
Thus = - kV2/3, for k > 0.
dt
21.
22.
24.
dy
1b. dy/dt = -2y+5 can be rewritten as = -2dt. Thus
y-5/2
lnΩ y-5/2 Ω = -2t+c1, or y-5/2 = ce-2t. y(0) = y0 yields
c = y0 - 5/2, so y = 5/2 + (y0-5/2)e-2t.
If y0 > 5/2, the solution starts above the equilibrium
solution and decreases exponentially and approaches 5/2
as tƕ. Conversely, if y < 5/2, the solution starts
below 5/2 and grows exponentially and approaches 5/2 from
below as tƕ. 3
dy/dt
4a. Rewrite Eq.(ii) as = a and thus lnΩ y Ω = at+c1; or
y
y = ceat.
dy dy1
4b. If y = y1(t) + k, then = . Substituting both
dt dt
dy1
these into Eq.(i) we get = a(y1+k) - b. Since
dt
dy1
= ay1, this leaves ak - b = 0 and thus k = b/a.
dt
Hence y = y1(t) + b/a is the solution to Eq(i).
dQ dQ/dt
10a. = -rQ yields = -r, or lnΩ Q Ω = -rt + c1. Thus
dt Q
Q = ce-rt and Q(0) = 100 yields c = 100. Hence Q = 100e-rt.
Setting t = 1, we have 82.04 = 100e-r, which yields
r = .1980/wk or r = .02828/day.
dQ/dt -1
13a. Rewrite the D.E. as = , thus upon integrating and
Q-CV CR
simplifying we get Q = De-t/CR + CV. Q(0) = 0 Þ D = -CV and
thus Q(t) = CV(1 - e-t/CR).
dQ Q
13c. In this case R + = 0, Q(t1) = CV. The solution of this
dt C
-t /CR
D.E. is Q(t) = Ee-t/CR, so Q(t1) = Ee 1 = CV, or
t1/CR t1/CR -t/CR -(t-t1)/CR
E = CVe . Thus Q(t) = CVe e = CVe .
4 Section 1.3
r
19. Differentiating t twice and substituting into the D.E.
2 r-2 r-1 r 2 r
yields t [r(r-1)t ] + 4t[rt ] + 2t = [r +3r+2]t . If
r
y = t is to be a solution of the D.E., then the last term
2
must be zero for all t and thus r + 3r + 2 = 0.
22. The D.E. is second order since there are second partial 5
derivatives of u(x,y) appearing. The D.E. is nonlinear
due to the product of u(x,y) times ux(or uy).
CHAPTER 2
dt
4. m(t) = exp(Ú ) = elnt = t, so (ty)¢ = 3tcos2t, and
t
integration by parts yields the general solution.
4tdt
m(t) = et . m(t) = exp(Ú ) = (1+t2) 2.
2
7. 8.
1+t2
2dt
15. m(t) = exp(Ú ) = t2 and y = t2/4 - t/3 + 1/2 + c/t2.
t
Setting t = 1 and y = 1/2 we have c = 1/12.
CHAPTER 2
dt
4. m(t) = exp(Ú ) = elnt = t, so (ty)¢ = 3tcos2t, and
t
integration by parts yields the general solution.
4tdt
m(t) = et . m(t) = exp(Ú ) = (1+t2) 2.
2
7. 8.
1+t2
2dt
15. m(t) = exp(Ú ) = t2 and y = t2/4 - t/3 + 1/2 + c/t2.
t
Setting t = 1 and y = 1/2 we have c = 1/12.
21a. 24a.
Ú
2dt
24b. m(t) = exp = t2, so (t2y)¢ = sint and
t
-cost c p
y(t) = + 2 . Setting t = - yields
2 2
t t
4c ap 2 ap 2/4 - cost
= a or c = and hence y(t) = , which
p2 4 t2
is unbounded as t Æ 0 unless ap 2/4 = 1 or a0 = 4/p 2.
1 - cost
24c. For a = 4/p 2 y(t) = . To find the limit as
t2
t Æ 0 L’Hopital’s Rule must be used:
sint cost 1
limy(t) = lim = lim = .
tÆ0 tÆ0 2t tÆ0 2 2
dt
37. y(t) = A(t)exp(-Ú ) = A(t)/t.
t
3y + y2 = x3 - x + c.
x2
-y-1 = 2x + + c. y(0) = 1 yields c = -1 and thus
2
-1 2
y = = . This gives
2 - 4x - x2
2
x
+ 2x - 1
2
dy 8 + 4x
= , so the minimum value is attained at
dx (2-4x-x2) 2
x = -2. Note that the solution is defined for
-2 - 6 < x < -2 + 6 (by finding the zeros of the
denominator) and has vertical asymptotes at the end
points of the interval.
4
for c and solving for y yields y(t) = .
1 + 7e-2t /3
2
cy+d
29. Separating variables yields dy = dx. If a π 0 and
ay+b
c ad-bc
ay+b π 0 then dx = ( + )dy. Integration then
a a(ay+b)
yields the desired answer.
dy dv
30c. If v = y/x then y = vx and = v + x and thus the
dx dx
dv v-4
D.E. becomes v + x = . Subtracting v from both
dx 1-v
dv v2-4
sides yields x = .
dx 1-v
1-v 1
30d. The last equation in (c) separates into dv =
dx. To
2 x
v -4
integrate the left side use partial fractions to write
1-v A B
= + , which yields A = -1/4 and B = -3/4.
v-4 v-2 v+2
1 3
Integration then gives - ln|v-2| - ln|v+2| = ln|x| - k, or
4 4
ln|x4||v-2||v+2|3 = 4k after manipulations using
properties of the ln function.
2 - v 1
dv = dx. Applying a partial fraction
(v+3)(v-1) x
decomposition to the left side we obtain
1 1 5 1 dx
[ - ]dv = , and upon integrating both sides
4 v-1 4 v+3 x
1 5
we find that ln|v-1| - ln|v+3| = ln|x| + c.
4 4
Substituting for v and performing some algebraic
manipulations we get the solution in the implicit form
|y-x| = c|y+3x|5. v = 1 and v = -3 yield y = x and
y = -3x, respectively, as solutions also.
1. Note that q(0) = 200 gm, where q(t) represents the amount of
dye present at any time.
4. Salt flows into the tank at the rate of (1)(3) lb/min. and
it flows out of the tank at the rate of (2) lb/min. since
the volume of water in the tank at any time t is 200 +
(1)(t) gallons (due to the fact that water flows into the
tank faster than it flows out). Thus the I.V.P. is dQ/dt =
3 - Q(t), Q(0) = 100.
same numerical equation for and hence the doubling time has
not changed.
dv 1
22. The I.V.P. in this case is m = - v - mg, v(0) = 20,
dt 30
where the positive direction is measured upward.
dv
24a. The I.V.P. is m = mg - .75v, v(0) = 0 and v is
dt
measured positively downward. Since m = 180/32, the D.E.
dv 2
becomes = 32 - v and thus v(t) = 240(1-e-2t/15) so
dt 15
that v(10) = 176.7 ft/sec.
dv
24c. After the parachute opens the I.V.P. is m = mg-12v,
dt
v(0) = 176.7, which has the solution
v(t) = 161.7e-32t/15 + 15 and where t = 0 now represents
the time the parachute opens. Letting tƕ yields the
limiting velocity.
mg mg -kt/m
26b.From part (a) v(t) = - + [v0 + ]e . As k Æ 0
k k
this has the indeterminant form of -• + •. Thus rewrite
v(t) as v(t) = [-mg + (v0k + mg)e-kt/m ]/k which has the
indeterminant form of 0/0, as k Æ 0 and hence
L’Hopital’s Rule may be applied with k as the variable.
dx
31b. From part a) = v = ucosA and hence
dt
x(t) = (ucosA)t + d1. Since x(0) = 0, we have d1 = 0 and
dy
x(t) = (ucosA)t. Likewise = -gt + usinA and
dt
therefore y(t) = -gt2/2 + (usinA)t + d2. Since y(0) = h
we have d2 = h and y(t) = -gt2/2 + (usinA)t + h.
31d. Let tw be the time the ball reaches the wall. Then
L
x(tw) = L = (ucosA)tw and thus tw = . For the ball
ucosA
to clear the wall y(tw) ≥ H and thus (setting
L
tw = , g = 32 and h = 3 in y) we get
ucosA
-16L2
+ LtanA + 3 ≥ H.
u2cos2A
-161.98 sinA
31e. Setting L = 350 and H = 10 we get + 350 ≥ 7
2 cosA
cos A
or 7cos A - 350cosAsinA + 161.98 £ 0. This can be solved
2
-16(350) 2
yields + 350tanA = 7, where we have chosen the
u2cos2A
equality sign since we want to just clear the wall.
1,960,000
Solving for u2 we get u2 = . Now u will
175sin2A-7cos2A
have a minimum when the denominator has a maximum. Thus
350cos2A + 7sin2A = 0, or tan2A = -50, which yields
A = .7954 rad. and u = 106.89 ft./sec.
17. From the direction field and the given D.E. it is noted
that for t > 0 and y < 0 that y¢ < 0, so y Æ -• for
y0 < 0. Likewise, for 0 < y0 < 3, y¢ > 0 and y¢ Æ 0 as
Section 2.4 19
-t+[t2+4(1-t)] 1/2
22a. For y1 = 1-t, y¢1 = -1 =
2
-t+[(t-2) 2] 1/2
=
2
-t+Ωt-2Ω
= = -1 if
2
(t-2) ≥ 0, by the definition of absolute value. Setting
t = 2 in y1 we get y1(2) = -1, as required.
27a. For n = 0,1, the D.E. is linear and Eqs.(3) and (4)
apply.
dv dy dy 1 n dv
27b. Let v = y1-n then = (1-n)y-n so = y ,
dt dt dt 1-n dt
which makes sense when n π 0,1. Substituting into the
n
y dv
D.E. yields + p(t)y = q(t)yn or
1-n dt
20 Section 2.4
dv dy dy 1 3 dv
28. n = 3 so v = y-2 and = -2y-3 or = - y .
dt dt dt 2 dt
Substituting this into the D.E. gives
1 dv 2 1 3
- y3 + y = y . Simplifying and setting
2 dt t t2
y-2 = v then gives the linear D.E.
4 2 1
v’ - v = - , where m(t) = and
t t2 t4
2 2+5ct5
v(t) = ct4 + = . Thus y = ±[5t/(2+5ct5)] 1/2.
5t 5t
dv dv
29. n = 2 so v = y-1 and = -y2 . Thus the D.E.
dt dt
dv dv
becomes -y2 - ry = -ky2 or + rv = k. Hence
dt dt
m(t) = ert and v = k/r + ce-rt. Setting v = 1/y then
yields the solution.
Ï e-2t 0 £ t £ 1
y’= ÔÔÌ
ÔÔ (1-e2)e-2t
Ó 1 < t.
Evaluating the two parts of y’ at t0 = 1 we see that they
are different, and hence y’ is not continuous at t0 = 1.
Section 2.5 21
dy
3. The critical points are found by setting equal to
dt
zero. Thus y = 0,1,2 are the critical points. The graph
of y(y-1)(y-2) is positive for O < y < 1 and 2 < y and
negative for 1 < y < 2. Thus y(t) is increasing
dy
( > 0) for 0 < y < 1 and 2 < y and decreasing
dt
dy
( < 0) for 1 < y < 2. Therefore 0 and 2 are unstable
dt
critical points while 1 is an asymptotically stable
critical point.
dy dy
6. is zero only when arctany is zero (ie, y = 0). > 0
dt dt
dy
for y < 0 and < 0 for y > 0. Thus y = 0 is an
dt
asymptotically stable critical point.
dy
7c. Separate variables to get = kt. Integration
(1-y) 2
1 1 kt + c - 1
yields = kt + c, or y = 1 - = .
1-y kt + c kt + c
c-1
Setting t = 0 and y(0) = y0 yields y0 = or
c
1 (1-y0)kt + y0
c = . Hence y(t) = . Note that for
1-y0 (1-y0)kt + 1
y0 < 1 y Æ (1-y0)k/(1-y0)k = 1 as t Æ •. For y0 > 1
notice that the denominator will have a zero for some
value of t, depending on the values chosen for y0 and k.
Thus the solution has a discontinuity at that point.
dy
9. Setting = 0 we find y = 0, ± 1 are the critical
dt
dy dy
points. Since > 0 for y < -1 and y > 1 while < 0
dt dt
for -1 < y < 1 we may conclude that y = -1 is
asymptotically stable, y = 0 is semistable, and y = 1 is
unstable.
22 Section 2.5
dy du
17a. If u = ln(y/K) then y = Keu and = Keu so that the
dt dt
D.E. becomes du/dt = -ru.
dy
21a. Set = 0 and solve for y using the quadratic formula.
dt
21d. If h > rK/4 there are no critical points (see part a) and
dy
< 0 for all t.
dt
1 dx x dn
24a. If z = x/n then dz/dt = - . Use of
n dt n2 dt
Equations (i) and (ii) then gives the I.V.P. (iii).
ndz
24b. Separate variables to get = - bdt. Using
z(1-nz)
dz dz
partial fractions this becomes + = -bdt.
z 1-z
Integration and solving for z yields the answer.
dx
26b. x = p is a semistable critical point and since > 0,
dt
x(t) is an increasing function. Thus for x(0) = 0, x(t)
approaches p as t Æ •. To solve the D.E., separate
variables and integrate.
3b. Use the same formula as in Problem 3a, except now h = .05
and n = 0,1...7. Notice that only results for n = 1,3,5
and 7 are needed to compare with part a.
3c. Again, use the same formula as above with h = .025 and
n = 0,1...15. Notice that only results for n = 3,7,11 and
15 are needed to compare with parts a and b.
15c. There are two factors that explain the large differences.
From the limit, the slope of y, y¢, becomes very “large”
for values of y near -1.155. Also, the slope changes
sign at y = -1.155. Thus for part a,
y(1.7) @ y7 = -1.178, which is close to -1.155 and the
slope y¢ here is large and positive, creating the large
change in y8 @ y(1.8). For part b, y(1.65) @ -1.125,
resulting in a large negative slope, which yields
y(1.70) @ -3.133. The slope at this point is now positive
and the remainder of the solutions “grow” to -3.098 for
the approcimation to y(1.8).
16. For the four step sizes given, the approximte values for
y(.8) are 3.5078, 4.2013, 4.8004 and 5.3428. Thus, since
these changes are still rather “large”, it is hard to
give an estimate other than y(.8) is at least 5.3428. By
using h = .005, .0025 and .001, we find further
approximate values of y(.8) to be 5.576, 5.707 and 5.790.
Thus a better estimate now is for y(.8) to be between 5.8
and 6. No reliable estimate is obtainable for y(1),
which is consistent with the direction field of Prob.9.
Section 2.8 29
s2 t2 t3
Ú
t
f 3(t) = - (1-s + )ds = -t + - . ;
0 2 2 23
2 3
t2 t3 t4
f (t) = -Ú
t s s
4 (1 - s + - )ds = -t + - + .
0 2 3! 2 3! 4!
n
 (-1) ktk
Based upon these we hypothesize that: f n(t) =
k!
k=1
and use mathematical induction to verify this form for
f n(t). Using Eq.(7) again we have:
n
 (-1)
k k+1
Ú
t t
f n+1(t) = - [ 1 + f n(s) ] ds = -t -
0 (k+1)!
k=1
n n+1
 Â
(-1) k+1tk+1 (-1) iti
= = , where i = k+1. Since this
(k+1)! i!
k=0 i=1
is the same form for f n+1(t) as derived from f n(t) above,
we have verified by mathematical induction that f n(t) is
as given.
•
 (-1) ktk
4c. From part a, let f(t) = lim f n(t) =
nƕ k!
k=1
2 3
t t
- + ... . = -t +
2 3!
Since this is a power series, recall from calculus that:
•
Â
at aktk a2t2 a3t3
e = = 1 + at + + + ... . If we let
k! 2 3!
k=0
t2 t3
a = -1, then we have e-t = 1 - t + - + ... = 1 +f(t).
2 3!
Hence f(t) = e-t -1.
7. As in Prob.4,
 t2k-1
f n(t) = and use mathematical induction
1.3.5...(2k-1)
k=1
to verify this form for f n(t). Using Eq.(7) again we
have:
n
Ú Â
t s2k
f n+1(t) = ( + 1)ds
0 1.3.5...(2k-1)
k = 1
n
 1.3.5t...(2k+1) +
2k+1
= t
k=1
n
 1.3.5t...(2k+1)
2k+1
=
k=0
n+1
 1.3.5t...(2i-1),
2i-1
= where i = k+1. Since this is
i=1
the same form for f n+1(t) as derived from f n(t) above, we
have verified by mathematical induction that f n(t) is as
given.
x3 x5
11. Recall that sinx = x - + + O(x7). Thus, for
3! 5!
t2 t4 t6
f 2(t) = t - + - + O(t7) we have
2! 4! 6!
t2 3
(t - )
t2 t4 t6 2! t5
sin[ f 2(t) ] = (t - + - ) - + + O(t7).
2! 4! 6! 3! 5!
1
y2 = .5y1 + 6 = (.5) 2y0 + 6(1 + ) and
2
1 1
y3 = .5y2 + 6 = (.5) 3y0 + 6(1 + + ). In general, then
2 4
1 1
yn = (.5) ny0 + 6(1 + + ... + )
2 2n-1
1 - (1/2) n
= (.5) ny0 + 6( )
1 - 1/2
= (.5) ny0 + 12 - (.5) n12
= (.5) n(y0-12) + 12. Mathematical induction can now be
used to prove that this is the correct solution.
r-1
14. Substituting un = + vn into Eq.(21) we get
r
r-1 r-1 r-1
+ vn+1 = r( + vn)(1 - - vn) or
r r r
r-1 1
vn+1 = - + (r-1 + rvn)( - vn)
r r
1-r r-1
= + - (r-1)vn + vn- rv2n = (2-r)vn - rv2n.
r r
15a. For u0 = .2 we have u1 = 3.2u0(1-u0) = .512 and
u2 = 3.2u1(1-u1) = .7995392. Likewise u3 = .51288406,
u4 = .7994688, u5 = .51301899, u6 = .7994576 and
u7 = .5130404. Continuing in this fashion,
u14 = u16 = .79945549 and u15 = u17 = .51304451.
1. Linear 2. Homogeneous
5. Exact 6. Linear
2 dy dy
7. Letting u = x yields = 2x and thus
dx du
du
- 2yu = 2y3 which is linear in u(y).
dy
34 Miscellaneous Problems
8. Linear 9. Exact
CHAPTER 3
rt
10. Substituting y = e in the D.E.
we obtain the characteristic
equation r2 + 4r + 3 = 0, which
has the roots r1 = -1, r2 = -3.
-t -3t
Thus y = c1e + c2e and
-t -3t
y¢ = -c1e - 3c2e .
Substituting t = 0 we then have
c1 + c2 = 2 and -c1 - 3c2 = -1,
yielding c1 = 5/2 and
5 -t 1 -3t
c2 = -1/2. Thus y = e - e
2 2
and hence y Æ 0 as t Æ •.
-t 2t
21. The general solution is y = c1e + c2e . Using the I.C.
we obtain c1 + c2 = a and -c1 + 2c2 = 2, so adding the two
equations we find 3c2 = a + 2. If y is to approach zero
as t Æ •, c2 must be zero. Thus a = -2.
-2t
25c. From part (a), if b = 2 then y(t) = e and the solution
simply decays to zero. For b > 2, the solution becomes
unbounded negatively, and again there is no minimum
point.
-t
27. The second solution must decay faster than e , so choose
-2t -3t
e or e etc. as the second solution. Then proceed as
in Problem 17.
Ω cost sint Ω
W(cost,sint) = Ω Ω 2 2
2. Ω Ω = cos t + sin t = 1.
Ω
Ω -sint cost Ω
Ω
Ωx xex Ω
4. Ω
W(x,xex) = Ω Ω x 2 x x 2 x
Ω = xe + x e - xe = x e .
Ω
Ω 1 e + xe Ω
x x Ω
1 -1/2 1
14. For y = t1/2, y¢ = t and y≤ = - t-3/2. Thus
2 4
1 1
yy≤ + (y¢) 2 = - t-1 + t-1 = 0. Similarly y = 1 is also
4 4
a solution. If y = c1(1) + c2t1/2 is substituted in the
D.E. you will get -c1c2/4t3/2, which is zero only if
c1 = 0 or c2 = 0. Thus the linear combination of two
solutions is not, in general, a solution. Theorem 3.2.2
is not contradicted however, since the D.E. is not
linear.
Ωt g Ω 1
18. W(f,g) = Ω Ω = tg¢ - g = t2et, or g¢ - g = tet. This
Ω 1 g¢ Ω t
1 1 ¢ 1
has an integrating factor of and thus g - g = et
t t 2
t
1
or ( g) ¢ = et. Integrating and multiplying by t we
t
obtain g(t) = tet + ct.
21. From Section 3.1, et and e-2t are two solutions, and
since W(et,e-2t) π 0 they form a fundamental set of
solutions. To find the fundamental set specified by
Theorem 3.2.5, let y(t) = c1et + c2e-2t, where c1 and c2
satisfy
c1 + c2 = 1 and c1 - 2c2 = 0 for y1. Solving, we find
2 t 1 -2t
y1 = e + e . Likewise, c1 and c2 satisfy
3 3
c1 + c2 = 0 and c1 - 2c2 = 1 for y2, so that
1 t 1 -2t
y2 = e - e .
3 3
40 Section 3.2
30. We have P(x) = x, Q(x) = - cosx, and R(x) = sinx and the
condition for exactness is satisfied. Also, from Problem
27, f(x) = Q(x) - P’(x) = - cosx -1, so the D.E. becomes
(xy’)’ - [(1 + cosx)y]’ = 0. Hence
xy’ - (1 + cosx)y = c1. This is a first order linear
D.E. and the integrating factor (after dividing by x) is
m(x) = exp[-Ú x-1(1 + cosx)dx]. The general solution is
Út
x
y = [m(x)] -1[c1 m(t)dt + c2].
-1
x0
Ω t t-1 Ω
W(t,t ) = Ω Ω
Ω = -2/t π 0.
-1
6. Ω
Ω
Ω 1 -t-2 Ω
Ω
-(t+2)
W(t) = cexp[-Ú dt] = ct2et.
t
28. On 0 < t < 1, f(t) = t3 and g(t) = t3. Hence there are
nonzero constants, c1 = 1 and c2 = -1, such that
c1f(t) + c2g(t) = 0 for each t in (0,1). On -1 < t < 0,
f(t) = -t3 and g(t) = t3; thus c1 = c2 = 1 defines
constants such that c1f(t) + c2g(t) = 0 for each t in
(-1,0). Thus f and g are linearly dependent on
0 < t < 1 and on -1 < t < 0. We will show that f(t) and
g(t) are linearly independent on -1 < t < 1 by
demonstrating that it is impossible to find constants c1
and c2, not both zero, such that c1f(t) + c2g(t) = 0 for
all t in (-1,1). Assume that there are two such nonzero
constants and choose two points t0 and t1 in -1 < t < 1
such that t0 < 0 and t1 > 0. Then -c1t30 + c2t30 = 0 and
c1t31+ c2t31 = 0. These equations have a nontrivial solution
for c1 and c2 only if the determinant of coefficients is
zero. But the determinant of coefficients is -2t30t31 π 0
for t0 and t1 as specified. Hence f(t) and g(t) are
linearly independent on -1 < t < 1.
1-i)
5. Recall that 21-i = eln(2 = e(1-i)ln2
2
14. The characteristic equation is 9r + 9r - 4, which has
the real roots -4/3 and 1/3. Thus the solution has the
same form as in Section 3.1, y(t) = c1et/3 + c2e-4t/3.
1 23
and u¢ (0) = c1 + c2 = 0. Solving for c2 we find
6 6
23 2 23
u(t) = et/6(2cos t - sin t).
6 23 6
23b. To estimate the first time that Ω u(t) Ω = 10 plot the graph
of u(t) as found in part (a). Use this estimate in an
appropriate computer software program to find t = 10.7598.
a+2
25b. y(1) = e-1(2cos 5 + sin 5 ) = 0 and hence
5
2 5
a = -2 - = 1.50878.
tan 5
a+2
25c. For y(t) = 0 we must have 2cos 5t + sin 5 t = 0 or
5
-2 5
tan 5t = . For the given a (actually, for a > -2)
a+2
2 5
this yields 5 t = p - arctan since arctan x < 0 when
a+2
x < 0.
2 5
25d. From part (c) arctan Æ 0 as a Æ •, so t Æ p/ 5 .
a+2
d
31. [elt(cosmt + isinmt)] = lelt(cosmt + isinmt)
dt
+ elt(-msinmt + imcosmt) = lelt(cosmt + isinmt)
+ imelt(isinmt + cosmt) = elt(l+im) (cosmt + isinmt).
d rt
Setting r = l+im we then have e = rert.
dt
2
35. We use the result of Problem 34. Note that q(t) = e-t > 0
for -• < t < •. Next, we find that (q¢ + 2pq)/q3/2 = 0.
Hence the D.E. can be transformed into an equation with
2
constant coefficients by letting x = u(t) = Úe-t /2dt.
Substituting x = u(t) in the differential equation found in
part (b) of Problem 34 we obtain, after dividing by the
coefficient of d2y/dx2, the D.E. d2y/dx2 - y = 0. Hence the
general solution of the original D.E. is
2
y = c1cosx + c2sinx, x = Ú e-t /2dt.
1 -t/2 5 -t/2
17b. From part (a), y¢(t) = - (1 + 5t/2)e + e = 0, when
2 2
1 5t 5 8 -4/5
- - + = 0, or t0 = and y0 = 5e .
2 4 2 5
48 Section 3.5
1 1
17c. From part (a), - + c2 = b or c2 = b + and
2 2
1
y(t) = [1 + (b + )t]e-t/2.
2
1 1 1
17d. From part (c), y¢(t) = - [1 + (b+ )t]e-t/2 + (b+ )e-t/2 = 0
2 2 2
4b
which yields tM = and
2b+1
2b+1 . 4b
yM = (1 + )e-2b/(2b+1) = (1 + 2b)e-2b/(2b+1).
2 2b+1
Ú
t
is W(y1,y2) = c exp[- p(r)dr]. Hence
t0
exp[-Ú
t
(y2/y1)¢ = cy1-2 p(r)dr]. Integrating and setting
t0
c = 1 (since a solution y2 can be multiplied by any
constant) and taking the constant of integration to be
zero we obtain
Ú
s
exp[- p(r)dr]
Ú
t s0
y2(t) = y1(t) ds.
t0 [y1(s)] 2
since y1 = sin2(t2).
-t -t 2 -t
6. Since yc = c1e + c2te we must assume Y = At e , so
-t 2 -t -t -t 2 -t
that Y¢ = 2Ate - At e and y≤ = 2Ae - 4Ate + At e .
2 -t
Substituting in the D.E. gives (At -4At+2A)e +
2 -t 2 -t -t
2(-At +2At)e + At e = 2e . Notice that all terms on
2
the left involving t and t add to zero and we are left
-t -t 2 -t
with 2A = 2, or A = 1. Hence y = c1e + c2te + t e .
rt
11. First solve the homogeneous D.E. Substituting y = e
-t/2
gives r2 + r + 4 = 0. Hence yc = e [c1cos( 15 t/2) +
t -t
c2sin( 15 t/2)]. We replace sinht by (e - e )/2 and
t -t t -t
then assume Y(t) = Ae + Be . Since neither e nor e
are solutions of the homogeneous equation, there is no
need to modify our assumption for Y. Substituting in the
t -t t -t
D.E., we obtain 6Ae + 4Be = e - e . Hence, A = 1/6
and B = -1/4. The general solution is
-t/2 t -t
y = e [c1cos( 15 t/2) + c2sin( 15 t/2)] + e /6 - e /4.
[For this problem we could also have found a particular
solution as a linear combination of sinht and cosht:
Y(t) = Acosht + Bsinht. Substituting this in the D.E.
gives (5A + B)cosht + (A + 5B)sinht = 2sinht. The
solution is A = -1/12 and B = 5/12. A simple calculation
t -t
shows that -(1/12)cosht + (5/12)sinht = e /6 - e /4.]
-2t t
13. yc = c1e + c2e so for the particular solution we
assume Y = At + B. Since neither At or B are solutions of
the homogeneous equation it is not necessary to modify
the original assumption. Substituting Y in the D.E. we
obtain 0 + A -2(At+B) = 2t or -2A = 2 and A-2B = 0.
-2t t
Solving for A and B we obtain y = c1e + c2e - t - 1/2
as the general solution. y(0) = 0 Æ c1 + c2 - 1/2 = 0
and y¢(0) = 1 Æ -2c1 + c2 - 1 = 1, which yield c1 = -1/2
t -2t
and c2 = 1. Thus y = e - (1/2)e - t - 1/2.
-3t
19a. The solution of the homogeneous D.E. is yc = c1e + c2.
After inspection of the nonhomogeneous term, for 2t4 we
must assume a fourth order polynominial, for t2e-3t we
must assume a quadratic polynomial times the exponential,
and for sin3t we must assume Csin3t + Dcos3t. Thus
4 3 2 2 -3t
Y(t) = (A0t +A1t +A2t +A3t+A4) + (B0t +B1t+B2)e +Csin3t+Dcos3t.
-3t
However,since e and a constant are solutions of the
homogeneous D.E., we must multiply the coefficient of e-3t
and the polynomial by t. The correct form is
Y(t) = t(A0t4 + A1t3 + A2t2 + A3t + A4) +
t(B0t2 + B1t + B2)e-3t + Csin3t + Dcos3t.
-t
22a. The solution of the homgeneous D.E. is yc = e [c1cost +
c2sint]. After inspection of the nonhomogeneous term, we
-t 2 -t 2
assume Y(t) = Ae + (B0t + B1t + B2)e cost + (C0t + C1t
-t -t -t
+ C2)e sint. Since e cost and e sint are solutions of
the homogeneous D.E., it is necessary to multiply both
the last two terms by t. Hence the correct form is
-t 2 -t
Y(t) = Ae + t(B0t + B1t + B2)e cost +
2 -t
t(C0t + C1t + C2)e sint.
Ï t, 0 £ t £ p
y = f(t) = ÔÌÔ
ÔÓ -(p/2)cost - (1 + p/2)sint + (p/2)ep-t, t > p.
2t
33. From Problem 32 we write the D.E. as (D-4)(D+1)y = 3e .
2t
Thus let (D+1)y = u and then (D-4)u = 3e . This last
2t
equation is the same as du/dt - 4u = 3e , which may be
-4t
solved by multiplying both sides by e and integrating
2t 4t
(see section 2.1). This yields u = (-3/2)e + Ce .
Substituting this form of u into (D+1)y = u we obtain
2t 4t t
dy/dt + y = (-3/2)e + Ce . Again, multiplying by e
2t 4t -t
and integrating gives y = (-1/2)e + C1e + C2e , where
C1 = C/5.
= Ú [e3(t-s)- e2(t-s)]g(s)ds.
The complete solution is then obtained by adding
c1e3t + c2e2t to Y(t).
14. That t and tet are solutions of the homogeneous D.E. can
be verified by direction substitution. Thus we assume
Y = tu1(t) + tetu2(t). Following the pattern of earlier
problems we find tu¢1(t) + tetu¢2(t) = 0 and
56 Section 3.7
t -y1(t)y2(s)g(s)ds t y2(t)y1(s)g(s)ds
= t Ú
0 W(y1,y2)(s) Ú
+ t
0 W(y1,y2)(s)
t [y1(s)y2(t) - y1(t)y2(s)]g(s)ds
= t Ú
0
y1(s)y¢2(s) - y¢1(s)y2(s)
. To show that
∂G y1(s)y¢2(t) - y¢1(t)y2(s)
and = g(s). Likewise
∂t W(y1,y2)(s)
∂G(t,t) ∂2G
Ú
t
Y≤ = + (t,s)ds
∂t t 0 ∂t2
y1(s)y≤2(t) - y≤1(t)y2(s)
Ú
t
= g(t) + ds.
W(y1,y2)(s)
t0
Ú
t
Y(t) = g(s)ds
t0
me2ls
Ú
t
= m-1 el(t-s)[cosms sinmt - cosmt sinms]g(s)ds
t0
Ú
t
= m-1 el(t-s)[sinm(t-s)]g(s)ds.
t0
so
u = Acos14t + Bsin14t. The I.C. are u(0) = 0 Æ A = 0
and u¢(0) = 10 cm/sec Æ B = 10/14 = 5/7. Hence
u(t) = (5/7)sin14t, which first reaches equilibrium when
14t = p, or t = p/14.
2pg
23. From Problem 21: D = = Tdg/2m. Substituting the
m(2m)
(1/2) (3)
known values we find g = = 5 lb sec/ft.
.3
2k
24. From Eq.(13) w 20 = so P = 2p/ 2k/3 = p Æ k = 6.
3
Thus u(t) = c1cos2t + c2sin2t and u(0) = 2 Æ c1 = 2 and
v
u¢(0) = v Æ c2 = v/2. Hence u(t) = 2cos2t + sin2t =
2
60 Section 3.8
v2 v2
4+ cos(2t-g). Thus 4+ = 3 and v = ±2 5 .
4 4
11a. For this problem the mass m = 8/32 lb-sec2/ft and the
spring constant k = 8/(1/2) = 16 lb/ft, so the D.E. is
0.25u≤ + 0.25u¢ + 16u = 4cos2t where u is measured in ft
and t in sec. To determine the steady state response we
need only compute a particular solution of the
nonhomogeneous D.E. since the solutions of the
homogeneous D.E. decay to zero as t Æ •. We assume
u(t) = Acos2t + Bsin2t, and substitute in the D.E.:
- Acos2t - Bsin2t + (1/2)(-Asin2t + Bcos2t) + 16(Acos2t +
Bsin2t) = 4cos2t. Hence 15A + (1/2)B = 4 and
-(1/2)A + 15B = 0, from which we obtain A = 240/901 and
B = 8/901. The steady state response is
u(t) = (240cos2t + 8sin2t)/901.
62 Section 3.9
24.
CHAPTER 4
13. That et, e-t, and e-2t are solutions can be verified by
direct substitution. Computing the Wronskian we obtain,
Ω et e-t e-2t Ω Ω 1 1 1Ω
Ω
Ω Ω
Ω Ω
Ω et Ω 1 -1 -2 Ω
Ω = -6e-2t
W(et,e-t,e-2t)=Ω -2e-2t Ω
-2t
-e-t Ω = e Ω Ω
Ω Ω Ω Ω
Ω Ω Ω
Ω 1 1 4Ω Ω
Ω et e-t 4e-2t Ω
CHAPTER 4
13. That et, e-t, and e-2t are solutions can be verified by
direct substitution. Computing the Wronskian we obtain,
Ω et e-t e-2t Ω Ω 1 1 1Ω
Ω
Ω Ω
Ω Ω
Ω et Ω 1 -1 -2 Ω
Ω = -6e-2t
W(et,e-t,e-2t)=Ω -2e-2t Ω
-2t
-e-t Ω = e Ω Ω
Ω Ω Ω Ω
Ω Ω Ω
Ω 1 1 4Ω Ω
Ω et e-t 4e-2t Ω
W = ce Ú
- 2dt
= ce-2t.
Ú
Using Abel’s identity, W(t) = cexp(- p1(t)dt) = cet.
Section 4.4 71
Ú
1 t e ds 1 1 1 1
u1 = , u2 = - t - ln(cost) and u3 = - t + ln(cost)
2 t 0 coss 2 2 2 2
which, when substituted into the assumed form for Y, yields
the desired solution.
Ú Ú
1 t t t
Y(t) = {e e-sg(s)ds + cost (sins - coss)g(s)ds
2 t0 t0
Ú
t
-sint (sins + coss)g(s)ds}.
t0
This can be written in the form
Ú
t
Y(t) = (1/2) (et-s + costsins - costcoss
t0
-sintsins - sintcoss)g(s)ds.
If we use the trigonometric identities sin(A-B) =
sinAcosB - cosAsinB and cos(A-B) = cosAcosB + sinAsinB,
we obtain the desired result. Note: Eqs.(11) and (12) of
this section give the same result, but it is not
recommended to memorize these equations.
Ú Ú
t 1 t
tetu2 = -tet t se-sg(s)ds = - t 2tse(t-s)g(s)ds, and
0 2 0
likewise for u1 and u3. If g(t) = t-2et then g(s) = es/s2
and the integration is accomplished using the power rule.
Note that terms involving t0 become part of the
complimentary solution.
Capítulo 5
73
CHAPTER 5
12. For this problem f(x) = x2. Hence f′(x) = 2x, f″(x) = 2,
and f(n)(x) = 0 for n > 2. Then f(-1) = 1, f′(-1) = -2,
f″(-1) = 2 and x2 = 1 - 2(x+1) + 2(x+1) 2/2! = 1 - 2(x+1) +
(x+1) 2. Since the series terminates after a finite
number of terms, it converges for all x. Thus ρ = ∞.
19. Set m = n-1 on the right hand side of the equation. Then
n = m+1 and when n = 1, m = 0. Thus the right hand side
•
becomes  a (x-1) m
m+1
, which is the same as the left hand
m=0
side when m is replaced by n.
 nanx n
+ Âa x n
n
= Â (n+1)a x . n
n
• •
25. Â
m=2
m(m-1)amxm-2 + x  ka x
k=1
k
k-1
=
• •
 (n+2)(n+1)an+2xn +  ka x
k=1
k
k
=
n=0
•
 [(n+2)(n+1)a
n=0
n+2 + nan]xn. In the first case we have
∑ nanx n-1
= ∑ (m+1)am+1xm. Substituting this into the
n=1 m=0
given equation and letting m = n again, we obtain:
∞ ∞
∑ (n+1)an+1 x + 2 n
∑ anxn = 0, or
n=0 n=0
∞
∑ [(n+1)an+1 + 2an]xn = 0.
n=0
Hence an+1 = -2an/(n+1) for n = 0,1,2,3,... . Thus
a1 = -2a0, a2 = -2a1/2 = 2 a0/2, a3 = -2a2/3 = -23a0/2.3 =
2
∑ anx =n
∑ (-1) 2 n n n
a0x /n! = a0 ∑ (-2x) n/n! = a0e-2x.
n=0 n=0 n=0
∞ ∞
2. y = ∑ n
anx ; y′ = ∑ nanxn-1 and since we must multiply
n=0 n=1
y′ by x in the D.E. we do not shift the index; and
∞ ∞
y″ = ∑ n(n-1)anx n-2
= ∑ (n+2)(n+1)an+2xn. Substituting
n=2 n=0
in the D.E., we obtain
∞ ∞ ∞
∑ (n+2)(n+1)an+2x - n
∑ nanx - n
∑ anxn = 0. In order to
n=0 n=1 n=0
have the starting point the same in all three summations,
we let n = 0 in the first and third terms to obtain the
following
∞
(2.1 a2 - a0)x + 0
∑ [(n+2)(n+1)a n+2 - (n+1)an]xn = 0.
n=1
Thus an+2 = an/(n+2) for n = 1,2,3,... . Note that the
recurrence relation is also correct for n = 0. We show
how to calculate the odd a’s:
76 Section 5.2
∑ ∑
x2m 2mm! x2m+1
y = a0 + a1 .
m=0 2mm! m=0
(2m+1)!
∞ ∞ ∞
3. y = ∑ a (x-1)
n
n
; y′ = ∑ na (x-1) n
n-1
= ∑ (n+1)a n+1(x−1)
n
,
n=0 n=1 n=0
and
∞ ∞
y″ = ∑ n(n-1)an(x-1) n-2
= ∑ (n+2)(n+1)an+2(x-1) n.
n=2 n=0
Substituting in the D.E. and setting x = 1 + (x-1) we
obtain
∞ ∞ ∞
∑ (n+2)(n+1)an+2(x-1) n
- ∑ (n+1)a n+1(x-1)
n
- ∑ nan(x-1) n
n=0 n=0 n=1
∞
- ∑ an(x-1) n = 0,
n=0
where the third term comes from:
∞ ∞
5. y = ∑ n
anx ; y′ = ∑ nanx n-1
; and y″ = ∑ n(n-1)anxn-2.
n=0 n=1 n=2
Substituting in the D.E. and shifting the index in both
summations for y″ gives
∞ ∞ ∞
∑ (n+2)(n+1)an+2x - n
∑ (n+1)n an+1x + n
∑ anxn =
n=0 n=1 n=0
∞
8. If y = ∑ an(x-1) n then
n=1
∞ ∞
xy = [1+(x-1)]y = ∑ an(x-1) n
+ ∑ an(x-1) n+1,
n=1 n=1
∞
∞ ∞
= ∑ n(n-1)an(x-1) n-2
+ ∑ n(n-1)an(x-1) n-1.
n=1 n=1
14. You will need to rewrite x+1 as 3 + (x-2) in order to
multiply x+1 times y′ as a power series about x0 = 2.
∑ (n+2)(n+1)an+2t + n
∑ (n-1)an-1t + n
∑ an-2tn
n=0 n=2 n=2
∞
+ ∑ 2an-1tn = 0,
n=1
∞
23. 26.
2
6. The zeros of P(x) = x - 2x - 3 are x = -1 and x = 3. For
x0 = 4, x0 = -4, and x0 = 0 the distance to the nearest
zero of P(x) is 1,3, and 1, respectively. Thus a lower
bound for the radius of convergence for series solutions
in powers of (x-4), (x+4), and x is ρ = 1, ρ = 3, and
ρ = 1, respectively.
2
9f. Since P(x) = x + 2 has zeros at x = ± 2 i, the lower
bound for the radius of convergence of the series
solution about x0 = 0 is ρ = 2.
∞ ∞
y″ = ∑ n(n-1)a x n
n-2
. Substituting in the D.E., shifting
n=2
indices of summation, and collecting coefficients of like
powers of x yields the equation
(2.1.a + α a )x + [3.2.a + (α -1)a ]x
2 0 2 1
2 0 3 1
Section 5.3 81
+ ∑ [(n+2)(n+1)a n+2
2
+ (α -n )an]x
2 n
= 0.
n=2
Hence the recurrence relation is
2 2
an+2 = (n -α )an/(n+2)(n+1), n = 0, 1,2,... . For the
first solution we choose a1 = 0. We find that
a2 = -α a0/2.1, a3 = 0, a4 = (2 -α )a2/4.3 = -(2 -α )α a0/4!
2 2 2 2 2 2
2 2 2 2 2
..., a2m = -[(2m-2) - α ]... (2 -α )α a0/(2m)!,
2 2 2 2
α 2 (2 -α )α 4
and a2m+1 = 0, so y1(x) = 1 - 2!x - x - ...
4!
2 2 2 2 2
[(2m-2) -α ]...(2 -α )α 2m
x -
- ...,
(2m)!
where we have set a0 = 1. For the second solution we take
a0 = 0 and a1 = 1 in the recurrence relation to obtain
the desired solution.
2 2 2
10b. If α is an even integer 2k then (2m-2) - α = (2m-2) -
2
4k = 0. Thus when m = k+1 all terms in the series for
2k
y1(x) are zero after the x term. A similar argument
shows that if α = 2k+1 then all terms in y2(x) are zero
2k+1
after the x .
n=2
3 4 5
+ 20a5x + 30a6x + 42a7x + ...
3 5 2 3 4
+ (x-x /3!+x /5!−...)(a0+a1x+a2x +a3x +a4x +...)
2 3
= 2a2 + (6a3+a0)x + (12a4+a1)x + (20a5+a2-a0/6)x +
4 5
(30a6+ a3-a1/6)x + (42a7+a4+a0/5!)x + ... = 0. Hence
a2 = 0, a3 = -a0/6, a4 = -a1/12, a5 = a0/120,
a6 = (a1+a0)/180, a7 = -a0/7! + a1/504, ... . We set
a0 = 1 and a1 = 0 and obtain
3 5 6
y1(x) = (1 - x /6 + x /120 + x /180 + ...). Next we set
a0 = 0 and a1 = 1 and obtain
4 6 7
y2(x) = (x - x /12 + x /180 + x /504 +...). Since
82 Section 5.3
x 2 3
18. We know that e = 1 + x + x /2! + X /3! + ..., and
x2 2 4 6
therefore e = 1 + x + x /2! + x /3! + ... . Hence, if
n
y = Σanx , we have
2 2 4 2
a1 + 2a2x + 3a3x + ... = (1+x + x /2+...)(a0+a1x+a2x +...)
2
= a0 + a1x + (a0+a2)x + ...:
Thus, a1 = a0 2a2 = a1 and 3a3 = a0 + a2, which yield the
desired solution.
∞
20. Substituting y = ∑a x n
n
into the D.E. we obtain
n=2
∞ ∞
∑ na xn
n-1
- ∑a x n
n
= x .
2
Shifting indices in the summation
n=2 n=2
∞
yields ∑ [(n+1)a n+1 - an]x
n
= x .
2
Equating coefficients of
n=2
both sides then gives: a1 - a0 = 0, 2a2 - a1 = 0, 3a3 - a2 = 1
and (n+1)an+1 = an for n = 3,4,... . Thus a1 = a0,
a2 = a1/2 = a0/2, a3 = 1/3 + a2/3 = 1/3 + a0/2.3,
a4 = a3/4 = 1/3.4 + a0/2.3.4, ..., an = an-1/n = 2/n! + a0/n! and
hence
2 n 3 4 n
x x x x x
y(x) = a0(1 + x + +...+ ...) + 2( + +...+ +...).
2! n! 3! 4! n!
x
Using the power series for e , the first and second sums
x x 2
can be rewritten as a0e + 2(e − 1 − x − x /2).
∞
22. Substituting y = ∑a x n
n
into the Legendre equation,
n=2
shifting indices, and collecting coefficients of like
powers of x yields
[2.1.a2 + α(α+1)a0]x + {3.2.a3 - [2.1 - α(α+1)]a1}x +
0 1
∞
∑ {(n+2)(n+1)a n+2 - [n(n+1) - α(α+1)]an}x
n
= 0. Thus
n=2
a2 = -α(α+1)a0/2!, a3 = [2.1 - α(α+1)]a1/3! =
-(α-1)(α+2)a1/3! and the recurrence relation is
Section 5.4 83
2 2
28. Since [(1-x )y’]’ = (1-x )y” - 2xy’, the Legendre
Equation, from Problem 22, can be written as shown.
Thus, carrying out the steps indicated yields the two
equations:
P [(1-x )P′ ]′ = -n(n+1)P P
2
m n n m
∫
1
which is zero. Thus P (x)Pm(x)dx = 0 for n ≠ m.
-1 n
2
lim x p(x) = lim 1-x = 1, lim x q(x) = 0 and thus x = 0
x→0 x→0 x→0
is a regular singular point.
(x-nπ) 2
= [(x-nπ)+nπ][±1 ±
± ...], which
6
converges about x0 = nπ and thus (x-nπ)p(x) is analytic
at x= nπ. Similarly (x+nπ)p(x) and (x±nπ) 2q(x) are
analytic and thus x0 = ±nπ are regular singular points.
∞
19. Substituting y = ∑a x n
n
into the D.E. yields
n=0
∞ ∞ ∞
2∑ n(n-1)a x n
n-1
+ 3 ∑ na x n
n-1
+ ∑a x n
n+1
= 0. The last sum
n=2 n=1 n=0
∞
becomes ∑a n-2x
n-1
by replacing n+1 by n-1, the first term
n=2
of the middle sum is 3a1, and thus we have
∞
2
f(ξ) = [2P(1/ξ)/ξ - Q(1/ξ)/ξ ]/P(1/ξ) = 2/ξ - 1/ξ = 1/ξ
and g(ξ) = R(1/ξ)/ξ4P(1/ξ) = −4/ξ2. Thus the point at
infinity is a singular point. Since both ξf(ξ) and
ξ2g(ξ) are analytic at ξ = 0, the point at infinity is a
regular singular point.
4. If y = xr then r(r-1) + 3r + 5 = 0. So r2 + 2r + 5 = 0
and r = (-2 ± 4-20 )/2 = -1 ± 2i. Thus the general
solution of the D.E. is
y = c1x-1cos(2ln|x|) + c2x-1sin(2ln|x|), x ≠ 0.
2
16. We have F(r) = r(r-1) + 3r + 5 = r + 2r + 5 = 0. Thus
Section 5.5 87
-1
r1,r2 = -1 ± 2i and y = x [c1cos(2lnx) + c2sin(2lnx)].
-2
Then y(1) = c1 = 1 and y’ = -x [cos(2lnx) + c2sin(2lnx)]
-1
+ x [-sin(2lnx)2/x + c2cos(2lnx)2/x] so that
y’(1) = -1-2c2 = -1, or c2 = 0.
21a. The real part of the root must be positive so, from
Eq.(5), α < 0. Also β > 0, since the (α-1) 2-4β term
must be less than α-1.
2z 2z
u(z) = c1e + c2ze + (1/4)z + 1/4. Hence
2 2
y(x) = c1x + c2x lnx + (1/4)lnx + 1/4.
∑ ∑ ∑
n+r n+r 1 2
(n+r)(n+r-1)anx + (n+r)anx + (x - ) anxn+r = 0.
9 n=0
n=0 n=0
∞ ∞ ∞
Note that x 2
∑ anx n+r
= ∑ anx n+r+2
= ∑ an-2xn+r. Thus we
n=0 n=0 n=2
1 1
have [r(r-1) + r - ]a0xr + [(r+1)r + (r+1) - ]a xr+1 +
9 9 1
∞
∑ 1
{[(n+r)(n+r-1) + (n+r) - ]a + an-2} xn+r = 0. From
9 n
n=2
the first term, the indicial equation is r2 - 1/9 = 0
with roots r1 = 1/3 and r2 = - 1/3. For either value of
r it is necessary to take a1 = 0 in order that the
coefficient of xr+1 be zero. The recurrence relation is
[(n+r) 2 - 1/9]an = -an-2. For r = 1/3 we have
-an-2 an-2
an = = - , n = 2,3,4,... .
1 2 1 2 2
(n + ) -( ) (n + )n
3 3 3
Section 5.6 89
∑ (-1) m x
y1(x) = x1/3[1 + ( ) 2m].
1 1 1 2
m=1 m! (1 + )(2 + )...(m + )
3 3 3
Since r2 = - 1/3 ≠ r1 and r1 - r2 = 2/3 is not an integer,
we can calculate a second series solution corresponding
to r = - 1/3. The recurrence relation is
n(n-2/3)an = - an-2, which yields the desired solution
following the steps just outlined. Note that a1 = 0, as
in the first solution, and thus all the odd coefficients
are zero.
∑ an+1(r+n+1)(r+n)x n+r
+ ∑ an+1(r+n+1)x n+r
- ∑ anxn+r = 0,
n=-1 n=-1 n=0
90 Section 5.6
∑ (n+r)(n+r-1)ant n+r
+ 2 ∑ (n+r+1)(n+r)an+1tn+r
n=0 n=-1
∞ ∞
n=0 n=-1
- α(α+1) ∑ antn+r = 0, or
n=0
∞
[2r(r-1) + 2r]a0 t
r-1
+ ∑ {2(n+r+1) 2an+1
n=0
+ [(n+r)(n+r+1) - α(α+1)]an}tn+r = 0.
The indicial equation is 2r2 = 0 so r = 0 is a double
root. Thus there will be only one series solution of the
Section 5.6 91
∑ (n+r)(n+r-1)anx n+r
+ ∑ (n+r)anx n+r
+ ∑ an-2xn+r = 0,
n=0 n=0 n=2
or
[r(r-1)+r]a0xr + [(1+r)r+1+r]a1xr+1
∞
∑
+ [(n+r) 2an + an-2]xn+r = 0. The indicial equation
n=2
2
is r = 0 so r = 0 is a double root. It is necessary to
take a1 = 0 in order that the coefficient of xr+1 be zero.
The recurrence relation in n2an = -an-2, n = 2,3,... .
Since a1 = 0 it follows that a3 = a5 = a7 = ... = 0. For
the even coefficients we let n = 2m, m = 1,2,... . Then
a2m = -a2m-2/22m2 so a2 = -a0/22.1 2, a4 = a0/22.2 2.1 2.2 2,... ,
and a2m = (-1) ma0/22m(m!) 2. Thus one solution of the Bessel
92 Section 5.6
∫ ∫
dx 1
y2(x) = J0(x) = J0(x) [ + x + ...]dx
2 x
x[J (x)] 0
x2
= J0(x)[lnx + + ...], and it is clear that y2(x)
x
will contain a logarithmic term.
x -1
∑ bnxn.
n=0
-(1+x) 2
9. For this D.E., p(x) = and q(x) = and thus
2 x(1-x)
x (1-x)
x = 0, -1 are singular points. Since xp(x) is not
analytic at x = 0, x = 0 is not a regular singular point.
1+x 2(1-x)
Looking at (x-1)p(x) = and (x-1) 2q(x) = we
2 x
x
see that x = 1 is a regular singular point and that
p0 = 2 and q0 = 0.
sinx cosx
17a. We have p(x) = and q(x) = - , so that x = 0 is
2
x x2
a singular point. Note that xp(x) = (sinx)/x → 1 = p0
as x → 0 and x2q(x) = -cosx → -1 = q0 as x → 0. In
order to assert that x = 0 is a regular singular point we
must demonstrate that xp(x) and x2q(x), with xp(x) = 1 at
x = 0 and x2q(x) = -1 at x = 0, have convergent power
series (are analytic) about x = 0. We know that cosx is
analytic so we need only consider (sinx)/x. Now
∞
(sinx)/x = ∑ n
(-1) x2n/(2n+1)! and hence is analytic.
n=0
Thus we may conclude that x = 0 is a regular singular
point.
2 4 2 3 4
...) - (1 - x /2! + x /4! - ...)(a0x + a1x + a2x + a3x +
5 2
a4x + ...) = 0. Collecting terms, (2a1 + 2a1 - a1)x +
(6a2 + 3a2 - a0/6 - a2 + a0/2)x3 + (12a3 + 4a3 - 2a1/6 - a3+
a1/2)x4 + (20a4 + 5a4 - 3a2/6 + a0/120 - a4 + a2/2 -
a0/24)x5 + ... = 0. Simplifying, 3a1x2 + (8a2 + a0/3)x3 +
(15a3 + a1/6)x4 + (24a4 - a0/30)x5 + ... = 0. Thus, a1 = 0,
a2 = -a0/4!, a3 = 0, a4 = a0/6!,... . Hence
y1(x) = x - x3/4! + x5/6! + ... where we have set a0 = 1.
From Eq. (24) the second solution has the form
∞
n=1
1
= ay1(x)lnx + + c1 + c2x + c3x2 + c4x3 + …, so
x
y′2 = ay′1lnx + ay1x−1 − x−2 + c2 + 2c3x + 3c4x2 + …, and
y″2 = ay″1 lnx + 2ay′1x−1 − ay1x−2 + 2x−3 + 2c3 + 3c4x + ….
When these are substituted in the given D.E. the terms
including lnx will appear as
a[x2y″1 + (sinx)y′1 − (cosx)y1], which is zero since y1 is
a solution. For the remainder of the terms, use
y1 = x − x3/24 + x5/720 and the cosx and sinx series as
shown earlier to obtain
−c1 + (2/3+2a)x + (3c3+c1/2)x2 + (4/45+c2/3+8c4)x3 +…= 0.
These yield c1 = 0, a = −1/3, c3 = 0, and
c4 = −c2/24 − 1/90. We may take c2 = 0, since this term
will simply generate y1(x) over again. Thus
1 1 3
y2(x) = − y1(x)lnx + x−1 − x . If a computer algebra
3 90
system is used, then additional terms in each series may
be obtained without much additional effort. The next
terms, in each case, are shown here:
x3 x5 43x7
y1(x) = x − + − + … and
24 720 1451520
1 1 x4 41x6
y2(x) = − y1(x)lnx + [1− + − …].
3 x 90 120960
18. We first write the D.E. in the standard form as given for
Theorem 5.7.1 except that we are expanding in powers of
(x-1) rather than powers of x:
96 Section 5.7
20b. Substituting y = x ∑
r
anxn in x3y″ + αxy′ + βy = 0 gives
n=0
∞ ∞ ∞
∑ (n+r)(n+r-1)a x n
n+r+1
+ α ∑ (n+r)a x n
n+r
+ β ∑ anxn+r = 0.
n=0 n=0 n=0
β n(n-1)an-1
For = -1, then, an = , which is zero for
α αn
n = 1 and thus y(x) = x is the solution. Similarly for
β (n-1)(n-2)
= 0, an = and again for n = 1 a1 = 0 and
α αn
y(x) = 1 is the solution. Continuing in this fashion, we
see that the series solution will terminate for β/α any
positive integer as well as 0 and -1. For other values
β β
(n- -1)(n- -2)
an 2 α
of β/α, we have = , which approaches
an-1 αn
∞ as n → ∞ and thus the ratio test yields a zero radius
of convergence.
∞
∑ (n+r)(n+r-1)anx
n+r
+ α ∑ (n+r)anx
n+r+1-s
n=0 n=0
98 Section 5.8
+ β ∑ anxn+r+2-t = 0.
n=0
If s = 2 and t = 2 the first term in each of the three
series is r(r-1)a0xr, αra0xr-1, and βa0xr, respectively.
Thus we must have αra0 = 0 which requires r = 0. Hence
there is at most one solution of the assumed form.
Next
cn(-1) =
d
[(r+1)an(r)] = (-1) n
d
[
(r+1)
] ,
dr dr F(r+1) ... F(r+n)
r=-1 r=-1
where we again have set a0 = 1. Observe that
(r+1)/F(r+1)... F(r+n)=1/[(r+2) 2(r+3) 2...(r+n) 2(r+n+1)]=1/Gn(r).
1 -1/2
12. If ξ = αxβ, then dy/dx = x f + x1/2f′αβxβ-1 where f′
2
denotes df/dξ. Find d2y/dx2 in a similar fashion and use
algebra to show that f satisfies the D.E.
ξ2f″ + ξf′ + [ξ2 - υ2]f = 0.
14. First we verify that J0(λ jx) satisfies the D.E. We know
that J0(t) is a solution of the Bessel equation of order
zero:
t2J″0(t) + tJ′0(t) + t2J0(t) = 0 or
CHAPTER 6
•.
-1
2. Note that lim+ (t-1) =
tÆ1
-t2 -st M
Ú
2 M -st
= lim [
e |0 + e tdt]
M Æ • s s 0
Ú
2 1 1 M -st
= lim [- te-st|M0 + e dt]
sM Æ • s s 0
2 1 -st M 2
= lim - e |0 = .
2M Æ •
s s s3
6. That f(t) = cosat satisfies the hypotheses of Theorem
6.1.2 can be verified by recalling that |cosat| £ 1 for
• -st
all t. To determine £{cosat} = Ú0
e cosatdt we
• -st
must integrate by parts twice to get 0 Ú e cosatdt =
-1 -st -2 -st
lim [(-s e cos at + as e sinat)|M0
M Æ •
Úe
M
- (a2/s2) -st
cos at dt]. Evaluating the first two
0
terms, letting M Æ •, and adding the third term to both
• -st
sides, we obtain [1 + a2/s2] Ú
0
e cos at dt = 1/s, s > 0.
2 2
Division by [1 + a /s ] and simplification yields the
desired solution.
Ú
A
21. The integral (t2 + 1) -1dt can be evaluated in terms of
0
the arctan function and then Eq. (3) can be used. To
illustrate Theorem 6.1.1, however, consider that
•
Ú
1 1
< for t ≥ 1 and, from Example 3, 1 t-2dt
t2+1 t2
•
converges and hence Ú1
(t2 + 1) -1dt also converges.
Ú
1
(t2 + 1) -1dt is finite and hence does not affect the
0
•
convergence of Ú 0
(t2 + 1) -1dt at infinity.
Section 6.2 105
•
25. If we let u = f and dv = e-stdt then F(s) = Ú0
e-stf(t)dt
• -st
Ú
1 -st 1
= lim - e f(t)|M0 + e f¢(t)dt. Now use an
M Æ • s s 0
argument similar to that given to establish Theorem 6.1.2.
Ú•e x dx = s Ú•e x dx
1 n
27b. From part a, £{tn} = -x n -x n-1
sn+1 0 n+1 0
Ú
1 •
27c. From part a, £{t-1/2} = -x -1/2
e x dx. Let x = y2, then
s 0
s Ú
2 • -y2
2dy = x-1/2dx and thus £{t-1/2} = e dy.
0
4 2!
2. We have = 2
and thus the inverse Laplace
3
(s-1) (s-1) 2+1
transform is 2t2et, using line 11.
3s 3s 9/5 6/5
4. We have = = + using partial
2 (s-3)(s+2) s-3 s+2
s -s-6
fractions. Thus (9/5)e3t + (6/5)e-2t is the inverse
transform, from line 2.
11. Take the Laplace transform of the D.E., using Eq.(1) and
Eq.(2), to get
s2Y(s) - sy(0) - y¢(0) - [sY(s) - y(0)] - 6Y(s) = 0.
Using the I.C. and solving for Y(s) we obtain
s-2
Y(s) = . Following the pattern of Eq.(12) we have
s2-s-6
s-2 a b a(s-3)+b(s+2)
= + = . Equating like
2 s+2 s-3 (s+2)(s-3)
s -s-6
powers in the numerators we find a+b = 1 and
-3a + 2b = -2. Thus a = 4/5 and b = 1/5 and
4+5 1/5
Y(s) = + , which yields the desired solution
s+2 s-3
using Table 6.2.1.
(w 2-5)s s
= (w 2-4) -1[ + ].
s +w s +4
2 2 2
• -st p
Ú Ú Ú (e
M
e-st dt + lim -st
e f(t)dt = )(0)dt
0 0 M Æ •p
p
= Ú e-stdt = (1 - e-ps)/s.
0
Hence, the Laplace transform Y(s) of the solution is
given by Y(s) = s/(s2+4) + (1 - e-ps)/s(s2+4).
•
= Â [(-1)
n=0
n
/(2n+1)!L{t2n}
•
= Â [(-1)
n=0
n
(2n)!/(2n+1)!]s-(2n+1)
•
= Â [(-1)
n=0
n
/(2n+1)]s-(2n+1), which converges for s > 1.
Ï 1 ¸
38b. Since £-1ÔÔÌ s-r ÔÔý = erkt, the result follows.
ÔÔ kÔ Ô
Ó þ
• •
= (1/s) + Â
k=1
Â(-e
(-1) k e-ks/s = [
k=0
) ]/s.
-s k
We recognize
•
the last infinite series as the geometric series, Âar ,
k=0
k
Úe
T
-rs -sT
a = f(r)dr and u = e . The geometric series
0
converges to a/(1-u) for |u| < 1 and consequently we
obtain
Úe
T
F(s) = (1 - e-sT) -1 -rs
f(r)dr, s > 0.
o
Úe
2
-st
Problem 28 gives us £{f(t)} = f(t)dt/(1-e-2s).
0
Calculating the integral we have
Úe Úe Úe
2 1 2
-st -st -st
f(t)dt = dt - dt
0 0 1
= (1-e-s)/s + (e -e )/s -2s -s
= (e-2s-2e-s+1)/s
= (1-e-s) 2/s. Since the denominator of
£{f(t)}, 1 - e-2s, may be written as (1-e-s)(1+e-s) we
obtain the desired answer.
as+b cs+d
fraction expansion of the quotient is + ,
2 2
s +1 s +s+5/4
where
a+c = 0, a+b+d = 0, (5/4)a+b+c = 0 and (5/4)b+d = 1 by
equating coefficients. Solving for the constants yields
the desired solution.
1 s+1/4
H(s) = ( - ), then
s s2+s/4+1
3 7 7 3 7
h(t) = 1 - e-t/8(cos t+
sin t) and
8 21 8
u(t) = ku3/2(t)h(t-3/2) - ku5/2(t)h(t-5/2).
16c. In all cases the plot will be zero for 0 £ t < 3/2. For
3/2 £ t < 5/2 the plot will be the system response
114 Section 6.4
19b. Taking the Laplace transform of the D.E. and using the I.C. we
•
 (-1)
1 k -pks
have Y(s) = [1 + 2 e ], since
2
s(s +1) k=1
e-pks 1 1 s
£{upk(t)} = . Since = - , we then obtain
s 2 s 2
s(s +1) s +1
•
 (-1)
1 k -pks
Y(s) = [1 + 2e ]. Using partial
2
s(s +.1s+1) n=1
fractions we have
1 1 s+.1
H(s) = = -
2 s 2
s(s +.1s+1) s +.1s+1
1 s+.05 .05
= - - , where
s 2 2 2 2
(s+.05) +b (s+.05) +b
2 2
b = [1-(.05) ] = .9975. Now let
-1 -.05t .05 -.05t
h(t) = £ {H(s)} = 1 - e cosbt - e sinbt. Hence,
b
•
20b. From the graph of part a, A @ 12.5 and the frequency is 2p.
1 1
Y(s) = + e-3ps[ ]. Using partial
(s2+1)(s2+2s+3) s2+2s+3
fractions or a computer algebra system we obtain
1 1 1 -t 1
y(t) = sint - cost + e cos 2 t + u3p(t)h(t-3p),
4 4 4 2
where h(t) = e-tsin 2 t.
13a. From Eq. (22) y(t) will complete one cycle when
15 (t-5)/4 = 2p or T = t - 5 = 8p/ 15 , which is
consistent with the plot in Fig. 6.5.3. Since an impulse
causes a discontinuity in the first derivative, we need
to find the value of y¢ at t = 5 and t = 5 + T. From Eq.
(22) we have, for t ≥ 5,
-1 15 1 15
y¢ = e-(t-5)/4[ sin (t-5) + cos (t-5)]. Thus
2 15 4 2 4
1 1 -T/4
y¢(5) = and y¢(5+T) = e . Since the original
2 2
impulse, d(t-5), caused a discontinuity in y¢ of 1/2, we
must choose the impulse at t = 5 + T to be -e-T/4, which
is equal and opposite to y¢ at 5 + T.
20 20
e-ks
17b. We have (s2+1)Y(s) = Â e-kps so that Y(s) = Â s2+1
k=1 k=1
20
15 15
 Â
2 e-(2k-1)p
(s +1)Y(s) = e-(2k-1)p so that Y(s) = .
2
k=1 k=1 s +1
15
Úe
t
d(t-p)sin(t-t)dt.
-(t-t)
y = We know that the
0
integration variable is always less than t (the upper
limit) and thus for t < p we have t < p and thus
d(t-p) = 0. Hence y = 0 for t < p. For t > p utilize
Eq.(16).
Ú f(t-t)(g*h)(t)dt
t
f*(g*h) = 0
t
= Ú f(t-t)[Ú g(t-h)h(h)dh]dt
t
0 0
= Ú [Ú f(t-t)g(t-h)dt]h(h)(dh).
t t
0 h
The last double integral is obtained from the previous
line by interchanging the order of the h and t
integrations. Making the change of variable w = t - h
on the inside integral yields
Ú [Ú f(t-h-w)g(w)dw]h(h)dh
t t-h
f*(g*h) =
0 0
= Ú (f*g)(t-h)h(h)dh = (f*g)*h.
t
0
Ú g(t-t)h(t)dt.
t
f(t) = Using Table 6.2.1, we have
0
G(s) = £{g(t)} = 2/s3 and H(s) = £{h(t)} = s/(s2+4).
Hence, by Theorem 6.6.1, £{f(t)} = F(s) = G(s)H(s) =
2/s2(s2+4).
Section 6.6 119
Ú (t-t)
t
3
f(t) = (1/6) sintdt.
0
13. We take the Laplace transform of the D.E. and apply the
I.C.: (s2 + 2s + 2)Y(s) = a/(s2 + a 2). Solving for Y(s),
we have Y(s) = [a/(s2+a 2)][(s+1) 2 + 1] -1, where the second
factor has been written in a convenient way by completing
the square. Thus Y(s) is seen to be the product of the
transforms of sinat and e-tsint respectively. Hence,
Úe
t
-(t-t)
according to Theorem 6.6.1, y = sin(t-t)sinatdt.
0
17. Taking the Laplace transform, using the I.C. and solving,
we have Y(s) = (s+3)/(s+1)(s+2) + s/(s2+a 2)(s+1)(s+2).
As in Problem 15, there are several correct ways the
second term can be treated in order to use the
convolution integral. In order to obtain the desired
answer, write the second term as
s a b
( + ) and solve for a and b.
2 2 s+1 s+2
s +a
CHAPTER 7
12. 9.
1 ′ 1
12. Solving the first D.E. for x2 gives x2 = x + x and
2 1 4 1
substitution into the second D.E. gives
17
x″1 + x′1 + x = 0. Thus x1 = e-t/2(c1cos2t + c2sin2t) and
4 1
x2 = e-t/2(c2cos2t-c1sin2t). The I.C. yields c1 = -2 and
c2 = 2.
2 -4 0
1a. 2A = 6 4 -2 so that
-4 2 6
2+4 -4-2 0+3 6 -6 3
2A + B = 6-1 4+5 -2+0 = 5 9 -2
-4+6 2+1 6+2 2 3 8
6 -5 -7 5 3 3
6. AB = 1 9 1 and BC = -1 7 3 so that
-1 -2 8 2 3 -2
7 -11 -3
(AB)C = A(BC) = 11 20 17 .
-4 3 -12
1 4 . 1 0
.
0 1 . 2/11 1/11
Add (-4) times the second row to the first row.
1 0 . 3/11 -4/11
.
0 1 . 2/11 1/11
1 2 0 . -5 3 0
.
rows. 0 1 0 .-3 3 -1
.
0 0 1 . 2 -1 0
Add (-2) times the second row to the first row.
1 0 0 . 1 -3 2
.
0 1 0 .-3 3 -1
.
0
0 1 . 2 -1 0
The desired answer appears on the right side of this
augmented matrix.
4 8
22. x′ = 2e2t = e2t; and
2 4
3 -2 3 -2 4 2t 12-4 2t 8
x = e = e = e2t.
2 -2 2 -2 2 8-4 4
1 2 -1 . 1
.
2. The augmented matrix is 2 1 1 . 1 . Row reduction then
.
1 -1 2 . 1
1 2 -1 . 1
.
yields 0 -3 3 . -1 .
.
0 0 0 . 1
1 0 1 . 0
.
augmented matrix is 1 1 0 . 0
.
0 1 1 . 0
1 0 1 . 0
.
Row reduction yields 0 1 -1 . 0 .
.
0 0 2 . 0
1 -1 -2 -3 . 0
.
2 0 -1 0 . 0
.
2 3 1 -1 . 0
.
3 1 0 3 . 0
Add (-2) times the first row to the second, add (-2)
times the first row to the third, and add (-3) times the
first row to the fourth.
1 -1 -2 -3 . 0
.
0 2 3 6 . 0
.
0 5 5 5 . 0
.
0 4 6 12 . 0
Multiply the second row by (1/2) and then add (-5) times
the second row to the third and add (-4) times the second
row to the fourth.
1 -1 -2 -3 . 0
.
0 1 3/2 3 . 0
.
0 0 -5/2 -10. 0
.
0 0 0 0 . 0
et0 1 . 0
.
t0e t0 . 0
t0
Multiply the first row by (−t0) and add to the second row
et0 1 . 0
to obtain .
0 0 . 0
3 -1 x1 0
satisfy = , or 3x1 - x2 = 0. If we let
3 -1 x2 0
1
x1 = 1, then x2 = 3 and the eigenvector is x(1) = , or
3
any constant multiple of this vector. Similarly, the
eigenvector corresponding to λ 2 must satisfy
132 Section 7.3
1 x1
-1 0 1
= , or x1 - x2 = 0. Hence x(2) = , or
3 -3
x2 0 1
a multiple thereof.
−
18. Since a12 = a21, the given matrix is Hermitian and we
know in advance that its eigenvalues are real. To find
the eigenvalues and eigenvectors we must solve
1-λ i x1 0
= . The determinant of coefficients
-i 1-λ x2 0
1 x1
i 0
= , or x1 + ix2 = 0. Note that the second
-i 1 x2 0
equation -ix1 + x2 = 0 is a multiple of the first.
If x1 = 1, then x2 = i, and the eigenvector is
1
x(1) = . In a similar way we find that the
i
1
eigenvector associated with λ 2 is x(2) = .
-i
0
1 is the eigenvector corresponding to λ = 1 - 2i.
i
3-λ 2 4 x1 0
2 -λ 2 x2 = 0 . The determinant of
4 2 3-λ 0
x3
4 2 4 x1 0
2 1 2 x2 = 0 ; hence there is only the single
4 2 4 0
x3
-5 2 x1
4 0
2-8 2
x2 = 0 . Interchange the first and
4 2 -5 0
3
x
second rows and use row reduction to obtain the
equivalent system x1 - 4x2 + x3 = 0, 2x2 - x3 = 0. Since
there are two equations to satisfy only one variable can
be assigned an arbitrary value. If we let x2 = 1, then
2
= 1 .
(3)
x3 = 2 and x1 = 2, so we find that x
2
dW dx(1)
1
(2)
(1) dx1
(2)
(1) dx2 dx(1)
2
= [ x(2)
2 - x 2 ] + [ x 1 - 1 ].
x(2)
dt dt dt dt dt
The terms in the square brackets can now be recognized as
the respective determinants appearing in the desired
solution. A similar result was mentioned in Problem 20
of Section 4.1.
∫
W(t) = c exp [p11(t) + p22(t)]dt.
t t2
6a. W =
= 2t2 - t2 = t2.
1 2t
6b. Pick t = t0, then c1x(1)(t0) + c2x(2)(t0) = 0 implies
t0 t20 0
c1 + c2 = , which has a non-zero solution
1 2t0 0
t0 t20
for c1 and c2 if and only if = 2t20 - t20 = t20 = 0.
1 2t0
Thus x(1)(t) and x(2)(t) are linearly independent at each
point except t = 0. Thus they are linearly independent
on every interval.
consider
x1 t t2
x = c1x(1) + c2x(2), or = c1 + c2 .
x2 1 2t
x′1 1 2t
Taking the derivative we obtain = c1 + c2 .
x′2 0 2
Solving this last system for c1 and c2 we find
c1 = x′1 - tx′2 and c2 = x′2/2. Thus
x1 t x′2 t2
= (x′1 - tx′2) + , which yields
x2 1 2 2t
t2
x1 = tx′1 - x′2 and x2 = x′1. Writing this system in
2
t - t2/2
matrix form we have x = x′. Finding the
1 0
inverse of the matrix multiplying x′ yields the desired
solution.
4-r -3 ξ1 0
satisfy = . The determinant of the
8 -6-r ξ2 0
1−r i
9. The eigvalues are given by
= (1−r) + i =
2 2
−i 1−r
1 i ξ1
r(r−2) = 0. For r= 0 we have = 0 or
−i 1 ξ2
1
−iξ1 + ξ2 = 0 and thus is one eigenvector. Similarly
i
1
is the eigenvector for r = 2.
−i
1-r -1 4 ξ1 0
matrix satisfy 3 2-r -1
ξ2 = 0 . The determinant
2 1 -1-r ξ3 0
of coefficients set equal to zero reduces to
r3 - 2r2 - 5r + 6 = 0, so the eigenvalues are
r1 = 1, r2 = -2, and r3 = 3. The eigenvector
0 -1 4 ξ1 0
corresponding to r1 must satisfy 3 1 -1 ξ2 = 0 .
2 1 -2 ξ3 0
Using row reduction we obtain the equivalent system
ξ1 + ξ3 = 0, ξ2 - 4ξ3 = 0. Letting ξ1 = 1, it follows
that
1
ξ3 = -1 and ξ2 = -4, so ξ (1) = -4 . In a similar way the
-1
eigenvectors corresponding to r2 and r3 are found to be
1 1
ξ (2) = -1 and ξ (3) = 2 , respectively. Thus the
-1 1
general solution of the given D.E. is
1 1 1
t -2t
x = c1 -4 e + c2 -1 e + c3 2 e3t. Notice that the
-1 -1 1
“trajectories” of this solution would lie in the x1 x2 x3
three dimensional space.
1 1 . 1
matrix of this system is . and by row reduction
1 5 . 3
1 1 . 1
we obtain . . Thus c2 = 1/2 and c1 = 1/2.
0 1 .1/2
Substituting these values in the general solution gives
the solution of the I.V.P. As t → ∞, the solution
1 1 3t
becomes asymptotic to x = e , or x2 = 5x1.
2 5
20. Substituting x = ξ tr into the D.E. we obtain
2 -1
rξξtr = ξtr. For t ≠ 0 this equation can be
3 -2
2-r -1 ξ1 0
written as = . The eigenvalues and
3 -2-r ξ2 0
1
eigenvectors are r1 = 1, ξ(1) = and r2 = -1,
1
1
ξ(2) = . Substituting these in the assumed form we
3
1 1
obtain the general solution x = c1 t + c2 t-1.
1 3
25.
27.
Section 7.6 141
2−i −5 ξ1
is given by = 0 so that ξ1 = (2+i)ξ2 and
1 −2−i ξ2
thus one complex-valued solution is
2+i (−1+i)t
x(1)(t) = e . Finding the real and complex
1
parts of x(1) leads to the general solution
2cost − sint 2sint + cost
x = c1e−t + c2e−t . Setting
cost sint
1 2 1
t = 0 we find x(0) = = c1 + c2 , which is
1 1 0
2c1 + c2 = 1
equivalent to the system . Thus c1 = 1 and
c1 + 0 = 1
c2 = −1 and
2-r -5
15a. The eigenvalues satisfy
2
= r - 4 + 5α = 0, so
α -2-r
r1,r2 = ± 4-5α .
15c.
5/4−r 3/4
16a.
= r − 5r/2 + (25/16 − 3α/4) = 0, so
2
α 5/4−r
r1,2 = 5/4 ± 3α /2.
16c.
3-r α
18a. We have 2
= r + r - 12 + 6α = 0, so
-6 -4-r
r1,r2 = -1/2 ± 49-24α /2.
18c.
Section 7.6 145
23b.
29a. We have y′1 = x′1 = y2, y′3 = x′2 = y4, y′2 = -2y1 + y3, and
y′4 = y1 - 2y3. Thus
146 Section 7.7
0 1 0 0
−2
=
0 1 0
y′′ y.
0 0 0 1
1 0 −2 0
-3/4-r 1/2
2. The characteristic equation is given by
=
1/8 -3/4-r
r2 + 3r/2 + 1/2 = 0, so r = 1, 1/2. For r = 1 we have
1/4 1/2 ξ1 0 -2
= , and thus ξ (1) = . Likewise
1/8 1/4 ξ2 0 1
2 -2 -t 2
ξ(2) = and thus x(1)(t) = e and x(2)(t) = e-t/2. To
1 1 1
find the first column of Φ we choose c1 and c2 so that
1
c1x(1)(0) + c2x(2)(0) = , which yields -2c1 + 2c2 = 1 and
0
c1 + c2 = 0. Thus c1 = -1/4 and c2 = 1/4 and the first column
1/2e-t/2 + 1/2et
of Φ is . The second colunm of Φ is
1/4e-t/2 - 1/4e-t/2
0
determined by d1x(1)(0) + d2x(2)(0) = which yields d1 = d2 =
1
e-t/2 - e-t
1/2 and thus the second column of Φ is .
1/2e-t/2 + 1/2e-t
(1/5)e-3t + (4/5)e2t
. Similarly, the second column of
-(4/5)e-3t + (4/5)e2t
Φ is that linear combination of x(1)(t) and x(2)(t) that
0
satisfies the I.C. . Thus we must have
1
c1 + c2 = 0, -4c1 + c2 = 1; therefore c1 = -1/5 and
c2 = 1/5. Hence the second column of Φ (t) is
-(1/5)e-3t + (1/5)e2t
.
(4/5)e-3t + (1/5)e2t
0
column we have d1 x(1)(0)+ d2 x(2)(0)
+ d3 = 1 ,
x(3)(0)
0
which yields d1 = -1/3, d2 = 1/3 and d3 = 0 and thus
(-1/3et + 1/3e-2t, 4/3et - 1/3e-2t, 1/3et - 1/3e-2t) T is
the second column of Φ (t). Finally, for the third column
0
we have e1x (0) + e2x (0) + e3x (0) = 0 , which
(1) (2) (3)
1
gives e1 = 1/2, e2 = -1 and e3 = 1/2 and hence
(1/2et - e-2t + 1/2e3t, -2et + e-2t + e3t,
-1/2et + e-2t + 1/2e3t) T is the third column of Φ (t).
2 -4 η1 2
satisfy = , which verifies Eq.(16).
1 -2 η2 1
Solving these equations yields η1 - 2η2 = 1. If η2 = k,
where k is an arbitrary constant, then η1 = 1 + 2k.
Hence the second solution that we obtain is
2 1 + 2k t 2 1 2
x(2)(t) = tet + e = tet + et + k et.
1 k 1 0 1
The last term is a multiple of the first solution x(1)(t)
and may be neglected, that is, we may set k = 0. Thus
2 1
x(2)(t) = tet + et and the general solution is
1 0
x = c1x(1)(t) + c2x(2)(t). All solutions diverge to
infinity as t → ∞. The graph is shown on the right.
2 1 1 ξ1 0 -3
2 2 -1 ξ2 = 0 which yields ξ
(1) = 4 and
0 -1 2 ξ3 0 2
-3
x(1) = 4 e-t. The eigenvectors corresponding to the
2
-1 1 1 ξ1 0
double eigenvalue must satsify 2 -1 -1 ξ2 = 0 ,
0 -1 -1 ξ3 0
Section 7.8 151
0
which yields the single eigenvector ξ(2) = 1 and hence
-1
0
x(2)(t) = 1 e2t. The second solution corresponding to
-1
the double eigenvalue will have the form specified by
0
Eq.(13), which yields x (3) = 1 te2t + η e2t.
-1
Substituting this into the given system, or using
-1 1 1 η1 0
Eq.(16), we find that η satisfies 2 -1 -1 η2 = 1 .
0 -1 -1 η3 -1
Using row reduction we find that η1 = 1 and η2 + η3 = 1,
where either η2 or η3 is arbitrary. If we choose η2 = 0,
1 0 1
then η = 0 and thus x(3) = 1 te2t + 0 e2t. The
1 -1 1
general solution is then x = c1x(1) + c2x(2) + c3x(3).
2−r 3/2
9.
We have
= (r−1/2) = 0. For r = 1/2, the
2
−3/2
−1−r
3/2 3/2 ξ1 1
eigenvector is given by = 0, so ξ =
−3/2 −3/2 ξ2 −1
1 t/2
and e is one solution. For the second solution we
−1
1
have x = ξ tet/2 + η et/2, where (A − η = ξ , A being
I)η
2
the coefficient matrix for this problem. This last
equation reduces to 3η η1/2 + 3η η2/2 = 1 and
η1/2 − 3η
−3η η2/2 = −1. Choosing η2 = 0 yields η1 = 2/3
and hence
1 t/2 2/3 t/2 1 t/2 3
x = c1 e + c2 e + c2 te . x(0) =
−1 0 −1 −2
gives c1 + 2c2/3 = 3 and −c1 = −2, and hence c1 = 2,
c2 = 3/2. Substituting these into the above x yields the
solution.
152 Section 7.8
12.
1
. Hence one solution of the given D.E. is
1
1
x(1)(t) = t-3. By analogy with the scalar case
1
considered in Section 5.5 and Example 2 of this section,
we seek a second solution of the form x = η t-3lnt + ζ t-3.
Substituting this expression into the D.E. we find that η
and ζ satisfy the equations (A + 3I)η η = 0 and
1 -4
(A + 3I)ζζ = η , where A = and I is the identity
4 -7
1
matrix. Thus η = , from above, and ζ is found to be
1
0
. Thus a second solution is
-1/4
1 1 −3
x(2)(t) = t-3lnt + t .
1 -1/4
1 -1 -1 η1 0
equivalent to the system 0 1 1 η2 = 1 . If we
0 0 0 η3 0
1
choose η3 = 0, then η2 = 1 and η1 = 1, so η = 1 . Hence
0
a second solution of the D.E. is
0 1
2t.
x (t) = 1 te2t +
(2)
1 e
-1 0
1 -1 -1 ζ1 -1
equivalent system 0 1 1 ζ2 = 3 . If we let
0 0 0 ζ3 0
Section 7.8 155
2
ζ2 = 0, then ζ3 = 3 and ζ1 =2, so ζ = 0 and
3
0 1 2
x(3)(t) 2
= 1 (t /2)e2t + 1 te2t + 0 e2t.
-1 0 3
0 1 2
17f. T = 1 1 0 and using row operations on T and I, or a
-1 0 3
-3 3 2
computer algebra system, T-1 = 3 -2 -2 and thus
-1 1 1
2 1 0
T-1AT = 0 2 1 = J.
0 0 2
λ 1 λ 1 λ 2 2λ
19a. J2 = JJ = =
0 λ 0 λ 0 λ 2
λ 1 λ 2 2λ λ 3 3λ 2
J3 = JJ2 = =
0 λ 0 λ 2 0 λ 3
∞
∑ Jntn
exp(Jt) = I +
n!
n=1
λ ntn nλ n-1tn
∞
∑ n! n!
= I +
λ t
n n
n=1 0
n!
∞ ∞
λ n-1tn
∑ ∑
λ ntn
1 +
n! (n-1)!
=
n=1 n=1
∞
∑ λ ntn
0 1 +
n!
n=1
eλt teλt
= , since
0 eλt
∞ ∞
∑ ∑
λ n-1tn λ ntn
= t( 1 + ) = teλt.
(n-1)! n!
n=1 n=1
3 2t 1 -2t
x(c) = c1 e + c2 e . Writing the
1 - 3
1 0 -t
nonhomogeneous term as et + e we see that we
0 3
can assume x(p) = aet + be-t. Substituting this in the
D.E., we obtain
1 0 -t
aet - be-t = Aaet + Abe-t + et + e , where A
0 3
is the given coefficient matrix. All the terms involving
1 0
et must add to zero and thus we have Aa - a + = .
0 0
This is equivalent to the system
3 a2 = -1 and 3 a1 - 2a2 = 0, or a1 = -2/3 and
a2 = -1/ 3 . Likewise the terms involving e-t must add
0 0
to zero, which yields Ab + b + = . The solution
3 0
of this system is b1 = -1 and b2 = 2/ 3 . Substituting
these values for a and b into x(p) and adding x(p) to
x(c) yields the desired solution.
5cost 5sint
Ψ(t) = . The inverse
2cost + sint -cost + 2sint
matrix is
cost - 2sint
sint
5
Ψ −1(t) = , which may be found as
2cost + sint
-cost
5
cost - 2sint
sint
5 -cost
u′(t) = sint
2cost + sint
-cost
5
1 2 - 3cost2t + sin2t
= ,
5 -1 - cos2t - 3sin2t
after multiplying and using appropriate trigonometric
identities. Integration and multiplication by Ψ yields
the desired solution.
CHAPTER 8
y21 + 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
y22 + 2(.1)y2
y2 = .5050895 + .05 which is again quadratic
3 + (.1) 2
in y2, yielding y2 = .5111273.
n yn tn yn+1
0 1 0 1.06513
1 1.06513 .025 1.13303
2 1.13303 .050 1.20381
15. If y¢ = 1 - t + 4y then
y≤ = -1 + 4y¢ = -1 + 4(1-t+4y) = 3 - 4t + 16y. In
Eq.(12) we let yn, y¢n and y≤n denote the approximate
values of f(tn), f¢(tn), and f≤(tn), respectively.
Keeping the first three terms in the Taylor series we
have
yn+1 = yn + y¢nh + y≤n h2/2
= yn + (1 - tn + 4yn)h + (3 - 4tn + 16yn)h2/2. For n = 0,
t0 = 0 and y0 = 1 we have
(.1) 2
y1 = 1 + (1 - 0 + 4)(.1) + (3 - 0 + 16) = 1.595.
2
-
19. The local truncation error is en+1 = f≤(tn)h2/2. For this
problem f¢(t) = 5t - 3f 1/2(t) and thus
f≤(t) = 5 - (3/2)f -1/2f¢ = 19/2 - (15/2)tf -1/2.
Substituting this last expression into en+1 yields the
desired answer.
-
22d. Since y≤ = -5psin5pt, Eq.(21) gives en+1 = -(5p/2)sin(5ptn)h2.
5p 2 1
Thus Ωen+1Ω < h < .05, or h < @ .08.
2 50p
4. In this case y¢n = 2tn + e-tnyn and thus the improved Euler
formula is
[(2tn + e-tnyn) + 2tn+1 + e-tn+1(yn + hy¢n) ]h
yn+1 = yn + . For
2
n = 0, 1, 2 we get y1 = 1.05122, y2 = 1.10483 and y3 = 1.16072
for h = .05.
f[tn+h,yn+hf(tn,yn)] = f(tn,yn)+ft(tn,yn)h+fy(tn,yn)hf(tn,yn)
+ [ftt(x,h)h2+2fty(x,h)h2f(tn,yn)+fyy(x,h)h2f2(tn,yn)]/2!
f(t0,y0) + f(t1,y1)
y1 = y 0 + h
2
= 2 + [-3 2 + (.5 - 3 1.57574 )].05 = 1.62458.
14a.
7a. Using the predictor and corrector formulas (Eqs.6 and 10)
with fn = .5 - tn + 2yn and using the Runge-Kutta method
to calculate y1,y2 and y3, we obtain the following table
for h = .05, t0 = 0, y0 = 1:
168 Section 8.4
n yn fn yn+1
0 1 2.5
1 1.130171 2.710342
2 1.271403 2.942805 1.424859
3 1.424859 3.199718 1.591825
4 1.591825 3.483650 1.773722
5 1.773722 3.797444 1.972120
6 1.972120 4.144241 2.188755
7 2.188755 4.527510 2.425544
8 2.425544 4.951088 2.684607
9 2.684607 5.419214 2.968287
10 2.968287
n yn yn+1
0 1
1 1.130170833
2 1.271402571
3 1.424858497 1.591825573
4 1.591825573 1.773724801
5 1.773724801 1.972125968
6 1.972125968 2.188764173
7 2.188764173 2.425557376
8 2.425557376 2.684625416
9 2.684625416 2.968311063
10 2.968311063
Subtracting Eq. (i) from Eq. (ii) to get Eq. (v) (not
shown) and subtracting Eq. (ii) from Eq. (iii) to get Eq.
(vi) (not shown), then subtracting Eq. (v) from Eq. (vi)
yields yn+1 - 2yn + yn-1 = 2Ah2, which can be solved for
A. Thus B = fn+1 - 2A(tn+h) [from Eq. (iv)] and
C = yn - tnfn+1 + At2n + 2Atnh [from Eq. (ii)]. Using
these values for A, B and C in Eq. (iv) yields
yn+1 = (1/3)(4yn-yn-1+2hfn+1), which is Eq. (15).
1
variables. For f 1(t) we have dy = dt, or
1+ey
e-y
dy = dt. Integrating both sides yields
e-y+1
-ln(e-y+1) = t + c. Solving for y we find
y = ln[1/(c1e-t-1)]. Setting t = 0 and y = 0, we obtain
c1 = 2 and thus f 1(t) = ln[et/(2-et)]. As t Æ ln 2, we
see that f 1(t) Æ •. A similar analysis shows that
f 2(t) = ln[1/(c2-t)], where c2 = 1 when the I.C. are
used. Thus f 2(t) Æ • as t Æ 1 and thus we conclude
that f(t) Æ • for some t such that ln2 £ t £ 1.
Ê 2 ˆ
k 01 = ÁÁ ˜˜
Ë -3 ¯
Section 8.6 173
Ê 2+(-.15) ˆ Ê 1.85 ˆ
k 02 = ÁÁ ˜˜ = ÁÁ ˜˜
ËÁ .05-3(1.1)-(.05) 2(1.85) ¯˜ Ë -3.25463 ¯
Ê 2+(-.16273) ˆ Ê 1.83727 ˆ
k 03 = ÁÁ ˜˜ = ÁÁ ˜˜
ËÁ .05-3(1.0925)-(.05) (1.83727) ¯˜
2 Ë -3.23209 ¯
Ê 2+(-.32321) ˆ Ê 1.67679 ˆ
k 04 = ÁÁ ˜˜ = ÁÁ ˜˜
ËÁ .1-3(1.18373)-(.1) 2(1.67679) ¯˜ Ë -3.46796 ¯
and thus
x1 = 1+(.1/6)[2 + 2(1.85)+2(1.83727)+(1.67679)]=1.18419,
y1 = 2+(.1/6)[-3-2(3.25463)-2(3.23209)-3.46796]=1.67598,
Ê 1.67598 ˆ Ê 1.50251 ˆ
k 11 = ÁÁ ˜˜ k12 = ÁÁ ˜˜
Ë -3.46933 ¯ Ë -3.68777 ¯
Ê 1.49159 ˆ Ê 1.30983 ˆ
k 13 = ÁÁ ˜˜ k14 = ÁÁ ˜˜
Ë -3.66151 ¯ Ë -3.85244 ¯
and thus
x2=x1+(.1/6)[1.67598+2(1.50251)+2(1.49159)+1.30983]=1.33376
y2=y1-(.1/6)[3.46933+2(3.68777)+2(3.66151)+3.85244]=1.30897.
CHAPTER 9
1d.
Ω 1-r -4 Ω
4a. Again the eigenvalues are given by Ω Ω Ω
Ω =
Ω
Ω 4 -7-r Ω
Ω
2
r + 6r + 9 = 0 and thus r1 = r2 = -3. The eigenvectors
Ê 4 -4 ˆÊ x1 ˆ Ê 0ˆ
are solutions of ÁÁ ˜˜Á ˜ = ÁÁ ˜˜ and hence there is
Ë 4 -4 ¯ËÁ x2 ¯˜ Ë 0¯
Ê 1ˆ
just one eigenvector x = ÁÁ ˜˜.
Ë 1¯
4b. Since the eigenvalues are negative, (0,0) is an improper
node which is asymptotically stable. If we had found
that there were two independent eigenvectors then (0,0)
would have been a proper node, as indicated in Case 3a.
Section 9.1 175
4d.
7d.
10d.
18b. From Chapter 7, the solution is x(t) = c1x (1) + c2x (2)er2t,
which can be written in scalar form as
x1 = c1x(1) + c2x(2) and x2 = c1x(1) + c2x(2)
r 2t r 2t
1 1 e 2 2 e .
Assuming x1(2) π 0, the first equation can be solved for
c2er2t, which is then substituted into the second
equation to yield x2 = c1x(1) 2 2 /x1 ][x1-c1x1 ].
+ [x(2) (2) (1)
2
a + a a < 0 then Eq.(iv) is an ellipse. Using
22 21 12
2
a + a = 0 we have a = -a a and hence
11 22 22 11 22
-a a + a a < 0 or a11a22 - a a > 0, which is true by
11 22 21 12 21 12
Eqs.(ii). Thus Eq.(iv) is an ellipse under the conditions
of Eqs.(ii).
Êa a ˆ
d Ê xˆ 12 ˜˜Ê x ˆ
ÁÁ ˜˜ = ÁÁÁ
11
20. The given system can be written as
dt Ë y ¯ Á ˜˜ËÁÁ y ¯˜˜.
Ë a21 a
22 ¯
Thus the eigenvalues are given by
r2-(a +a )r + a a -a a = 0 and using the given
11 22 11 22 12 21
definitions we rewrite this as r2 - pr + q = 0 and thus
r1,2 = (p ± p2-4q )/2 = (p ± D )/2. The results are
now obtained using Table 9.1.1.
7b.
7c. For (0,0) since all trajectories leave this point, this
is an unstable node. For (0,2) and (1,0) since the
trajectories tend to these points, respectively, they are
asymptotically stable nodes. For (1/2,1/2), one
trajectory tends to (1/2,1/2) while all others tend to
infinity, so this is an unstable saddle point.
12b.
12c. Clearly ( 6 ,0) and (- 6 ,0) are spiral points, and are
asymptotically stable since the trajectories tend to each
point, respectively. (0,0) is a saddle point, which is
unstable, since the trajectories behave like the ones for
(1/2,1/2) in Problem 7.
dy dy/dt 8x
15a. = = , so 4xdx - ydy = 0 and thus 4x2 - y2 = c,
dx dx/dt 2y
which are hyperbolas for c π 0 and straight lines y = ±2x for
c = 0.
15b. 19b.
dy y-2xy
19a. = , so (y-2xy)dx + (x-y-x2)dy = 0, which is an
dx -x+y+x2
exact D.E. Therefore f(x,y) = xy - x2y + g(y) and hence
∂f
= x - x2 + g¢(y) = x - y - x2, so g¢(y) = -y and
∂y
g(y) = -y2/2. Thus 2x2y - 2xy + y2 = c (after multiplying by
-2) is the desired solution.
dy -sinx
21a. = , so ydy + sinxdx = 0 and thus y2/2 - cosx = c.
dx y
Section 9.3 181
21b.
24. Suppose that t1 > t0. Let s = t1 - t0. Since the system
is autonomous, the result of Problem 23, with s replaced
by -s shows that x = f 1(t+s) and y = y1(t+s) generates
the same trajectory (C1) as x = f 1(t) and y = y1(t). But
at t = t0 we have x = f 1(t0+s) = f 1(t1) = x0 and
y = y1(t0+s) = y1(t1) = y0. Thus the solution
x = f 1(t+s), y = y1(t+s) satisfies exactly the same
initial conditions as the solution x = f 0(t), y = y0(t)
which generates the trajectory C0. Hence C0 and C1 are
the same.
d Ê xˆ Ê1 0ˆ Ê xˆ Ê 2ˆ Ê1 0ˆ
ÁÁ ˜˜ = ÁÁ ˜˜ ÁÁ ˜˜ + Á y ˜ and thus A = ÁÁ ˜˜ and
dt Ë y¯ Ë1 1¯ Ë y¯ ËÁ 0 ¯˜ Ë1 1¯
Ê y2 ˆ Ê 0ˆ
g = Á ˜. Since g(0) = ÁÁ ˜˜ we conclude that (0,0) is a
ËÁ 0 ¯˜ Ë 0¯
critical point. Following the procedure of Example 1, we
let x = rcosq and y = rsinq and thus
2 2
r sin q
g1(x,y)/r = Æ 0 as r Æ 0 and thus the system is
r
2
almost linear. Since det(A-rI) = (r-1) , we find that
the eigenvalues are r = r = 1. Since the roots are
1 2
equal, we must determine whether there are one or two
eigenvectors to classify the type of critical point. The
Ê 0 0 ˆÊÁÁ x1 ˆ˜˜ Ê 0ˆ
eigenvectors are determined by ÁÁ ˜˜Á ˜ = ÁÁ ˜˜ and hence
Ë 1 0 ¯ËÁ x ¯˜ Ë 0¯
2
Section 9.3 183
Ê 0ˆ
there is only one eigenvector x = ÁÁ ˜˜. Thus the critical
Ë 1¯
point for the linear system is an unstable improper node.
From Table 9.3.1 we then conclude that the given system,
which is almost linear, has a critical point near (0,0)
which is either a node or spiral point (depending on how
the roots bifurcate) which is unstable.
6b, For the critical point (0,0) the D.E. is already in the
6c. form of an almost linear system; and the corresponding
linear system is du/dt = u, dv/dt = 3v which has the
eigenvalues r1 = 1 and r2 = 3. Thus the critical point
(0,0) is an unstable node. Each of the other three
critical points is dealt with in the same manner; we
consider only the critical point (-1,2). In order to
translate this critical point to the origin we set
x(t) = -1 + u(t), y(t) = 2 + v(t) and substitute in the
D.E. to obtain
du/dt = -1 + u - (-1+u) 2 - (-1+u)(2+v) = u + v - u2 - uv
and
dv/dt = 3(2+v) - (-1+u)(2+v) - 2(2+v) 2 = -2u - 4v - uv - 2v2.
Writing this in the form of Eq.(4) we find that
Ê 1 1 ˆ Ê u2 + uv ˆ
A = ÁÁ ˜˜ and g = - Á ˜ which is an almost
Ë -2 -4 ¯ ÁË uv + 2v2 ¯˜
linear system. The eigenvalues of the corresponding
linear system are r = (-3 ± 9 + 8 )/2 and hence the
critical point (-1,2), of the original system, is an
unstable saddle point.
2
10a. The critical points are solutions of x + x2 + y = 0 and
y(1-x) = 0, which yield (0,0) and (-1,0).
dv
= 2v - uv. Thus the corresponding linear system is
dt
u’ = -u and v’ = -2v.
Ê 1 0ˆ
10c. For (0,0) A = ÁÁ ˜˜ which has r = r = 1, so that
Ë 0 1¯ 1 2
d Ê xˆ Ê1
Ê xˆ Ê 0 ˆ
0ˆ
18a. The system is ÁÁ ˜˜ = ÁÁ ˜˜
ÁÁ ˜˜ + Á 3 ˜ and thus is
dt Ë y¯ Ë0
Ë y¯ -2 ¯
ËÁ x ¯˜
almost linear using the procedures outlined in the
earlier problems. The corresponding linear system has
the eigenvalues r1 = 1, r2 = -2 and thus (0,0) is an
unstable saddle point for both the linear and almost
linear systems.
22a.
23a.
Ú
0
T/4 = - L/2g (1/ cosq - cosa )dq.
a
dx d2x dy
28a. If = y, then = = -g(x) - c(x)y.
dt 2 dt
dt
d Ê uˆ Ê 1.5-x0-y0 -x0 ˆÊ u ˆ
3c. From Eq (5) we get ÁÁ ˜˜ = Á ˜ÁÁ ˜˜. For
dt Ë v ¯ ËÁ -1.125y0 2-2y0-1.125x0 ¯˜Ë v ¯
(0,0) we get u¢ = 1.5u and v¢ = 2v, so r = 3/2 and r = 2, and
thus (0,0) is an unstable node. For (0,2) we have u¢ = -.5u
and v¢ = -2.25u-2v, so r = -.5, -2 and thus (0,2) is an
asymptotically stable node. For (3,0) we get u¢ = -1.5u-3v
and v¢ = -1.375v, so r = -1.5, -1.375 and hence (3,0) is an
symptotically stable node. For (.8,1.1) we have
u¢ = -.4u -.8v and v¢ = -1.2375u - 1.1v which give
r = -1.80475, .30475 and thus (.8,1.1) is an unstable saddle
point.
Section 9.4 187
3e.
5e.
Ê uˆ Ê 1 ˆ -t Ê 1 ˆ 11t/4
ÁÁ ˜˜ = c1 ÁÁ ˜˜ e + c2 ÁÁ ˜˜ e
Ë v¯ Ë 0¯ Ë 15/2 ¯
and
Ê uˆ Ê 0 ˆ -5t/2 Ê 1 ˆ 11t/6
ÁÁ ˜˜ = c1 ÁÁ ˜˜ e + c2 ÁÁ ˜˜ e ,
Ë v¯ Ë 1¯ Ë 5/52 ¯
8b. The conditions e 1/s1 > e 2/a 2 and e 1/a 1 > e 2/s2 imply that
e 2s1 - e 1a 2 < 0 and e 1s2 - e 2a 1 > 0 so again one of the
coordinates of the fourth point in 8a. is negative and
hence a mixed state is not possible. An analysis similar
to that in part(a) shows that (0,0) and (0,e 2/s2) are
unstable while (e 1/s1,0) is stable. Hence the bluegill
(represented by x) survive in this case.
s1 a1 1 g1
9a. x¢ = e 1x(1 - x - y) = e 1x(1 - x - y)
e1 e1 B B
s2 a2 1 g2
y¢ = e 2y(1 - y - x) = e 2y(1 - y - x). The coexistence
e2 e2 R R
1 g1 g2 1
equilibrium point is given by x + y = 1 and x + y =
B B R R
1. Solving these (using determinants) yields
X = (B - g 1R)/(1 - g 1g 2) and Y = (R - g 2B)/(1-g 1g 2).
Ê u ˆ¢ Ê -2 -2 ˆÊ u ˆ
12b. For a = .75 the linear system is ÁÁ ˜˜ = ÁÁ ˜˜ÁÁ ˜˜, which
Ë v¯ Ë -1.5 -2 ¯Ë v ¯
has the characteristic equation r2 + 4r + 1 = 0 so that
r = -2 ± 3 . Thus the critical point is an asymptotically
Ê u ˆ¢ Ê -2 -2 ˆÊ u ˆ
stable node. For a = 1.25, we have ÁÁ ˜˜ = ÁÁ ˜˜ÁÁ ˜˜, so
Ë v¯ Ë -2.5 -2 ¯Ë v ¯
r2 + 4r -1 = 0 and r = -2 ± 5. Thus (2,2) is an unstable
saddle point.
Ê -1 -1 ˆ
A = ÁÁ ˜˜, which has the eigenvalues r1 = -1,
Ë 0 3/4 ¯
Ê 1ˆ Ê -4 ˆ
r2 = 3/4 and corresponding eigenvectors ÁÁ ˜˜ and ÁÁ ˜˜.
Ë 0¯ Ë 7¯
Thus (2,0) is also an unstable saddle point.
Ê 1 3ˆ
For Á 2 , 2 ˜ we let x = 1/2 + u and y = 3/2 + v in the
ËÁ ¯˜
du 1 1 dv 3
given equations, which yields = - u - v, = u
dt 4 4 dt 4
Ê 1 1ˆ
Á-4 -4 ˜
Á ˜˜, which has
as the linear portion. Thus A = ÁÁ ˜˜
Á 3
Á 0˜
Ë 4 ¯
Ê 1 3ˆ
eigenvalues r1,2 = (-1 ± 11 i)/8. Thus Á 2 , 2 ˜ is an
ÁË ¯˜
asymptotically stable spiral point since the eigenvalues
are complex with negative real part. Using
r1 = (-1 + 11 i)/8 we find that one eigenvector is
Ê -2 ˆ
Á ˜ and by Section 7.6 the second eigenvector is
ËÁ 1 + 11 i ¯˜
Ê -2 ˆ
Á ˜.
ËÁ 1 - 11 i ¯˜
3e.
3f. For (x,y) above the line x + y = 2 we see that x¢ < 0 and
thus x must remain bounded. For (x,y) to the right of
x = 1/2, y¢ > 0 so it appears that y could grow large
asymptotic to x = constant. However, this implies a
contradiction (x = constant implies x¢ = 0, but as y gets
larger, x¢ gets increasingly negative) and hence we
conclude y must remain bounded and hence (x,y) Æ
(1/2,3/2) as t Æ •, again assuming they start in the
first quadrant.
Section 9.5 193
Ú
x
if 0 < x < k then g(s)ds > 0,
0
Ú g(s)ds = - Ú g(s)ds
x 0
if -k < x < 0 then > 0.
0 x
Ú g(s)ds
x
Since V(0,0) = 0 it follows that V(x,y) = y2/2 +
0
is positive definite for -k < x < k, -• < y < •. Next,
. dx dy
we have V(x,y) = Vx + Vy = g(x)y + y[-g(x)] = 0.
dt dt
.
Since V(x,y) is never positive, we may conclude that it
is negative semidefinite and hence by Theorem 9.6.1 (0,0)
is at least a stable critical point.
7c. Since V(0,0) = 0, 1 - cosx > 0 for 0 < |x| < 2p and y2 >
0 for y π 0, it follows that V(x,y) is positive definite
in a neighborhood of the origin. Next Vx(x,y) = sinx,
Vy(x,y) = y, so
. .
V(x,y) = (sinx)(y) + y(-y - sinx) = -y2. Hence V is
negative semidefinite and (0,0) is a stable critical
point by Theorem 9.6.1.
Ú
x
V(x,y) = y2/2 + g(s)ds, which yields
0
.
V = g(x)y + y[-c(x)y - g(x)] = -y2c(x).
10c. Since a11a22 - a12a21 > 0 and a11 + a22 < 0, we see that
D < 0 and so A > 0. Using the expressions for A, B, and
C found in part (b) we obtain
(4AC-B2)D 2 = [a221+a222 + (a11a22-a12a21)][a211+a212 + (a11a22-a12a21)]
- (a12a22+a11a21) 2
= (a211+a212+a221+a222)(a11a22-a12a21) + (a211+a212)(a221+a222)
+ (a11a22-a12a21) 2- (a12a22+a11a21) 2
= (a211+a212+a221+a222)(a11a22-a12a21) + 2(a11a22-a12a21) 2.
Since a11a22 - a12a21 > 0 it follows that 4AC - B2 > 0.
.
If we choose e = M/8 we obtain V[x(r,q), y(r,q)] £ -r2/2
.
for 0 £ r < R. Hence V is negative definite in 0 £ r < R
and from Problem 10c V is positive definite and thus V is
a Liapunov function for the almost linear system.
Ê 3(1-x2c) 3 ˆ
linear coefficient matrix to be A = Á ˜, where xc
ËÁ -1/3 -.8/3 ¯˜
is the critical point. For xc = 1.1994 we obtain complex
conjugate eigenvalues with a negative real part, and
therefore k = 0 yields an asymptotically stable spiral
point. For xc = .80485 the eigenvalues are also complex
conjugates, but with positive real parts, so k = .5 yields
an unstable spiral point.
16b. Letting k = .1, .2, .3, .4 in the cubic equation of part (a)
and finding the corresponding eigenvalues from the matrix in
part (a), we find that the real part of the eigenvalues
change sign between k = .3 and k = .4. Continuing to
iterate in this fashion we find that for k = .3464 that the
real part of the eigenvalue is -.0002 while for k = .3465
the real part is .00005, which indicates k = .3465 is the
0
critical point for which the system changes from stable to
unstable.
16d. You must plot the solution for values of k slightly less
than k0, found in part (c), to determine whether a limit
cycle exists.
1c. Simply substitute r = 28 into the answers in parts (a) and (b).
x2 y2 (z-2r)
M+r: + + = 1, then S1 is contained
2 2
(M+r) (M+r) (M+r2)
in S2. Thus, if we choose c, in Eq.(ii), such that
c c .
> (M+r) 2 and > (M+r) 2, then V < 0 as the trajectory
r s
crosses V(x,y,z) = c. Note that this is a sufficient
condition and there may be many other “better” choices
using different techniques.
CHAPTER 10
0 -1 ≤ x < 0
7. To start, let n = 0, then f(x) = ; for n = 1,
1 0 ≤ x < 1
0 1 ≤ x < 2 0 3 ≤ x < 4
f(x) = ; and for n = 2, f(x) = . By
1 2 ≤ x < 3 1 4 ≤ x < 5
continuing in this fashion, and drawing a graph, it can be
seen that T = 2.
∫ ∫
L L
a0 = (l/L) (-x)dx = 0 and am = (l/L) (-x)cos(mπx/L)dx = 0,
-L -L
m = 1,2... (these can be shown by direct integration, or
∫
a
using the fact that g(x)dx = 0 when g(x) is an odd
-a
function). Finally,
∫
L
bm = (l/L) (-x)sin(mπx/L)dx
-L
L
∫
L
= (x/mπ)cos(mπx/L) - (l/mπ) cos(mπx/L)dx
-L -L
L
= (2Lcosmπ)/mπ − (L/m2π 2)sin(mπx/L) = 2L(−1) m/mπ
-L
Substituting these terms in the above Fourier series for
f(x) yields the desired answer.
∫
0
found to be a0 = (1/π) xdx = -π/2 since f(x) is zero on
-π
the interval [0,π]. Likewise
∫
0
an = (1/π) xcosnxdx = [1 - (-1) n]/n2π, using integration
-π
by parts and recalling that cosnπ = (-1) n. Thus an = 0
for n even and an = 2/n2π for n odd, which may be written
as a2n-1 = 2/(2n-1) 2π since 2n-1 is always an odd number.
∫
0
In a similar fashion bn = (1/π) xsinnxdx = (-1) n+1/n and
-π
thus the desired solution is obtained. Notice that in
this case both cosine and sine terms appear in the
Fourier series for the given f(x).
206 Section 10.2
15a. 21a.
1 2 x2 1 3 2
∫
4 a0 2
21b. a0 = dx = x = , so = and
2 −2 2 12 -2 3 2 3
1 2 x2
∫
nπx
an = cos dx
2 −2 2 2
1 2x2 nπx 8x nπx 16 nπx 2
= [ sin + 2 2 cos − 3 3 sin ]
4 nπ 2 n π 2 n π 2 -2
= (8/n2π 2)cos(nπ) = (−1) n8/n2π 2
where the second line for an is found by integration by
parts or a computer algebra system. Similarly,
1 2 x2
∫
nπx nπx
bn = sin dx = 0, since x2sin is an odd
2 −2 2 2 2
8 ∞ (−1) n
∑
2 nπx
function. Thus f(x) = + 2 cos .
3 π n 2 2
n=1
∑
2 8 m (−1) n nπx
21c. As in Eq. (27), we have sm(x) = + 2 cos
3 π n=1 n 2 2
∫ ∫ g(s)ds
a+T a
27a. First we have g(x)dx = by letting x = s + T
T 0
in the left integral. Now, if 0 ≤ a ≤ T, then from
elementary calculus we know that
∫ ∫ ∫ ∫ ∫
a+T T a+T T a
g(x)dx = g(x)dx + g(x)dx = g(x)dx + g(x)dx
a a T a 0
using the equality derived above. This last sum is
∫
T
g(x)dx and thus we have the desired result.
0
∫
π
yields a0 = (1/π) xdx = π/2;
0
∫
π
am = (1/π) xcosmxdx = (cosmπ - 1)/πm2 = 0 for m even and
0
= -2/πm2 for m odd; and
∫
π
bm = (1/π) xsinmxdx = -(πcosmπ)/mπ = (-1) m+1/m,
0
m = 1,2... . Substituting these values into Eq.(1) with
L = π yields the desired solution.
∫
1
yields a0 = (1-x2)dx = 4/3;
-1
208 Section 10.3
∫ ∫
1 1
an = (1-x2)cosnπxdx = (2/nπ) xsinnπxdx
-1 -1
∫
1
= (-2/n2π 2)[xcosnπx - 1
cosnπxdx]
-1 -1
= 4(-1) n+1/n2π 2; and
∫
1
bn = (1-x2)sinnπxdx = 0. Substituting these values
-1
into Eq.(1) gives the desired series.
∑[ 2cos(2k−1)x
π n
(−1) k+1sinkx
7b. en(x) = f(x) + − + ].
4 π(2k−1) 2 k
k=1
Using a computer algebra system, we find that for
n = 5, 10 and 20 the maximum error occurs at x = −π in
each case and is 1.6025, 1.5867 and 1.5787 respectively.
Note that the author’s n values are 10, 20 and 40, since
he has included the zero cosine coefficient terms and the
sine terms are all zero at x = −π.
∫ ∫
1 1
12a. a0 = (x-x3)dx = 0 and an = (x-x3)cosnπxdx = 0 since
-1 -1
(x-x3) and (x-x3)cosnπx are odd functions.
∫
1
bn = (x−x3)sinnπxdx
−1
x3 3x2 (n2π 2+6) (n2π 2+6)
= [ cosnπx− 2 2 sinnπx− xcosnπx+ sinnπx] 1−1
nπ n π n3π 3 n4π 4
12 ∞ (−1) n
∑
−12
= 3 3 cosnπ, so f(x) = − 3 sinnπx.
n π π n= n3
1
Section 10.3 209
∑ (-1)
k
12
12b. en(x) = f(x) + sinkπx. These errors will be
π3 k 3
k=1
much smaller than in the earlier problems due to the n3
factor in the denominator. Convergence is much more
rapid in this case.
∞
y′(0) = c2m - bm/2m + ∑ bnn/(m2-n2) = 0. Thus
n=1,n≠m
∞
c2 = bm/2m2 - ∑ bnn/m(m2-n2), which when substituted
n=1,n≠m
into the equation for y(t) yields the desired solution.
∫ ∫
π 2π
a0 = (1/π) dx - (1/π) dx = 0;
0 π
∫ cosnxdx - (1/π)∫
π 2π
an = (1/π) cosnxdx = 0; and
0 π
∫ sinnxdx - (1/π)∫
π 2π
bn = (1/π) sinnxdx = 0 for n even and
0 π
= 4/nπ for n odd. Thus
∞
f(t) = (4/π) ∑ sin(2n-1)t/(2n-1). Comparing this to the
n=1
forcing function of Problem 14 we see that bn of
Problem 14 has the specific values b2n = 0 and
b2n-1 = (4/π)/(2n-1) in this example. Substituting these
into the answer to Problem 14 yields the desired
solution. Note that we have asumed ω is not a positive
integer. Note also, that if the solution to Problem 14
is not available, the procedure for solving this problem
would be exactly the same as shown in Problem 14.
18a. We will assume f(x) is continuous for this part. For the
case where f(x) has jump discontinuities, a more detailed
proof can be developed, as shown in part b. From Eq.(3)
∫
1 L nπx
we have bn = f(x)sin dx. If we let u = f(x) and
L -L L
nπx -L nπx
dv = sin dx, then du = f′(x)dx and v = cos .
L nπ L
Thus
Section 10.4 211
∫
1 -L nπx L L L nπx
bn = [ f(x)cos -L + f′(x)cos dx]
L nπ L nπ -L L
∫
1 1 L nπx
= - [f(L)cosnπ - f(-L)cos(-nπ)] + f′(x)cos dx
nπ nπ -L L
∫
1 L nπx
= f′(x)cos dx, since f(L) = f(-L) and
nπ -L L
∫
1 L nπx
cos(-nπ) = cosnπ. Hence nbn = f′(x)cos dx, which
π -L L
exists for all n since f′(x) is piecewise continuous.
Thus nbn is bounded as n → ∞. Likewise, for an, we
∫
1 L nπx
obtain nan = - f′(x)sin dx and hence nan is also
π -L L
bounded as n → ∞.
∑
2 n nπxi n
n bn = [f(xi+)-f(xi-)]cos - [f(L-)-f(-L+)]cosnπ
π L π
i=1
m
n=1
7.
10.
14. For the cosine series we use the even extension of the
function given in Eq.(13) and hence
0 -2 ≤ x <-1
f(x) = on the interval -2 ≤ x < 0.
1+x -1 ≤ x < 0
However, we don’t really need this, as the coefficients
in this case are given by Eqs.(7), which just use the
original values for f(x) on 0 < x ≤ 2. Applying Eqs.(7)
we have L = 2 and thus
∫ ∫
1 2
a0 = (2/2) (1-x)dx + (2/2) 0dx = 1/2. Similarly,
0 1
∫ (1-x)cos(nπx/2)dx = 4[1-cos(nπ/2)]/n π
1 2 2
an = (2/2) and
0
bn = 0. Substituting these values in the Fourier series
yields the desired results.
For the sine series, we use Eqs.(8) with L = 2. Thus
an = 0 and
∫ ∫
1 1
a0 = dx = 1 and an = cos(nπx/2)dx = 2sin(nπ/2)/nπ.
0 0
Thus an = 0 for n even, an = 2/nπ for n = 1,5,9,...and
an = -2/nπ for n = 3,7,11,... . Hence we may write
a2n = 0 and a2n-1 = 2(-1) n+1/(2n-1)π, which when
substituted into the series gives the desired answer.
∫
π
L = π, that bn = (2/π) sinnxdx = 2[1-(-1) n]/nπ and thus
0
bn = 0 for n even and bn = 4/nπ for n odd.
∫ ∫
1 1
a0 = 2 xdx = 1, an = 2 xcos(2nπx)dx = 0 and
0 0
∫ xsin(2nπx)dx = -1/nπ.
1
bn = 2 [Note: We have used the
0
214 Section 10.4
∫ (2−x )sin
2 2 nπx
25a. bn = 2 dx
2 0 2
−2 nπx 8x nπx 16 nπx 2
= [ (2−x2)cos − 2 2 sin − 3 3 cos ]
nπ 2 n π 2 n π 2 0
4 16
= (1+cosnπ) + 3 3 (1−cosnπ) and thus
nπ n π
∞
∑ 4n2π 2(1+cosnπ) + 16(1−cosnπ) nπx
f(x) = ( )sin
n3π 3 2
n=1
25b.
25c.
∫
2 1 1
a0 = xdx =
2 0 2
Section 10.4 215
∫
2 1 nπx 2x nπx 4 nπx 1
an = xcos dx = [ sin + 2 2 cos ]
2 0 2 nπ 2 n π 2 0
2 nπ 4 nπ 4
= sin + 2 2 cos − 2 2 , so
nπ 2 n π 2 n π
∞
∑ 4cos(nπ/2) + 2nπsin(nπ/2)
1 − 4 nπx
g(x) = + cos .
4 n π 2 2 2
n=1
For the sine series (odd extension) we have
−2x
∫
2 1 nπx nπx 4 nπx 1
bn = xsin dx = [ cos + 2 2 sin ]
2 0 2 nπ 2 n π 2 0
−2 nπ 4 nπ
= cos + 2 2 sin , so
nπ 2 n π 2
∞
h(x) = ∑ 4sin(nπ/2) n− π2nπcos(nπ/2) sin nπx
2 2 2
.
n=1
28c.
28d. The maximum error does not approach zero in either case,
due to Gibb’s phenomenon. Note that the coefficients in
both series behave like 1/n as n → ∞ since there is an
n in the numerator.
∫ ∫ ∫ f(x)dx.
L 0 L
31. We have f(x)dx = f(x)dx + Now, if we let
-L -L 0
x = -y in the first integral on the right, then
∫ ∫ ∫ ∫
0 0 L L
f(x)dx = f(-y)(-dy) = f(-y)dy = - f(y)dy. Thus
-L L 0 0
∫ ∫ f(y)dy + ∫ f(x)dx
L L L
f(x)dx = - = 0.
-L 0 0
∫ f(t)dt we have
x
34. Since F(x) =
0
F(-x) = ∫ f(t)dt = -∫ f(-s)ds
-x x
by letting t = -s. If f
0 0
is an even function, f(-s) = f(s), we then have
∫
x
F(-x) = - f(s)ds = -F(x) from the original definition of
0
F. Thus F(x) is an odd function. The argument is
similar if f is odd.
∫
0
L
[f(x)] 2dx = ∫
0
L
[f(x) ∑ b sin(nπx/L)]dx
n
n=1
∞ ∞
= ∑ ∫ bn
0
L
f(x)sin(nπx/L)dx = (L/2) ∑b , 2
n by Eq.(8).
n=1 n=1
This result is identical to that of Problem 17 of Section
10.3 if we set an = 0, n = 0,1,2,... , since
∫ ∫
1 L 2 L
[f(x)] 2dx = [f(x)] 2dx . In a similar manner, it
L -L L 0
can be shown that
∞
(2/L) ∫0
L
[f(x)] 2dx = a20/2 + ∑a . 2
n
n=1
∫
2L
bn = (2/2L) f(x)sin(nπx/2L)dx, n = 1,2,... . The
0
∞
Fourier sine series for f is f(x) = ∑ b sin(nπx/2L).
n
n=1
∫
2L
39. From Problem 38 we have bn = (1/L) f(x)sin(nπx/2L)dx
0
10. Since the B.C. for this heat conduction problem are
u(0,t) = u(40,t) = 0, t > 0, the solution u(x,t) is given
by Eq.(19) with α 2 = 1 cm2/sec, L = 40 cm, and the
coefficients cn determined by Eq.(21) with the I.C.
u(x,0) = f(x) = x, 0 ≤ x ≤ 20; = 40 - x, 20 ≤ x ≤ 40.
∫ ∫
1 20 nπx 40 nπx
Thus cn = [ xsin dx + (40−x)sin dx]
20 0 40 20 40
160 nπ
= 2 2 sin . It follows that
n π 2
∞
∑ sin(nπ/2)
160 −n2π2t/1600sin nπx
u(x,t) = e .
π2 n 2 40
n=1
Section 10.5 219
15a.
15b.
15c.
π 2
1600 160
t = ln 2 = 451.60 sec.
π 2 π
18a. Since the B.C. for this heat conduction problem are
u(0,t) = u(20,t) = 0, t > 0, the solution u(x,t) is given by
Eq.(19) with L = 20 cm, and the coefficients cn determined by
Eq.(21) with the I.C. u(x,0) = f(x) = 100oC. Thus
∫
20
cn = (1/10) (100)sin(nπx/20)dx = -200[(-1) n-1]/nπ and hence
0
220 Section 10.5
∞
e−(2n−1) π α
∑
2 2 2t/400
400 (2n−1)πx
u(x,t) = sin
π 2n−1 20
n=1
19b. Using only one term in the series for u(x,t), we must
solve the equation 5 = (400/π)exp[-π 2(.86)t/400] for t.
Taking the logarithm of both sides and solving for t
yields t ≅ 400ln(80/π)/π 2(.86) = 152.56 sec.
9a. Since the B.C. are not homogeneous, we must first find
the steady state solution. Using Eqs.(9) and (10) we have
v″ = 0 with v(0) = 0 and v(20) = 60, which has the
solution v(x) = 3x. Thus the transient solution w(x,t)
satisfies the equations α 2wxx = wt, w(0,t) = 0,
w(20,t) = 0 and w(x,0) = 25 - 3x, which are obtained from
Eqs.(13) to (15). The solution of this problem is given
by Eq.(4) with the cn given by Eq.(6):
∫
1 20 nπx
cn = (25−3x)sin dx = (70cosnπ+50)/nπ, and thus
10 0 20
∞
∑ 70cosnπ + 50 −0.86n2π2t/400sin nπx
u(x,t) = 3x + e since
nπ 20
n=1
α 2 = .86 for aluminum.
9b. 9c.
20 −0.86π2t/400 π
e sin = .15, which yields t = 160.30 sec.
π 4
∫
L
c0 = (2/L) sin(πx/L)dx = 4/π and
0
= (2/L)∫ sin(πx/L)cos(nπx/L)dx
L
cn
0
= (1/L)∫ {sin[(n+1)πx/L] - sin[(n-1)πx/L]}dx
L
0
= (1/π){[1 - cos(n+1)π]/(n+1) - [1 - cos(n-1)π]/(n-1)}
= 0, n odd; = -4/(n2-1)π, n even. Thus
∞
u(x,t) = 2/π-(4/π) ∑ exp[-4n π α t/L ]cos(2nπx/L)/(4n -1)
2 2 2 2 2
n=1
where we are now summing over even terms by setting
n = 2n.
12c.
∫ ∫
2 30 1 10 25
c0 = f(x)dx =
25dx = and
30 0 15 5 3
∫ ∫
2 3 nπx 1 10 nπx 50 nπ nπ
cn = f(x)cos dx = 25cos dx = [sin - sin ].
30 0 30 15 5 30 nπ 3 6
14b.
14c.
∫
L
coefficients cn = (2/L) f(x)sin[(2n-1)πx/2L]dx.
0
∫ ∫
2 L/2 2x nπx L 2(L−x) nπx
cn = [ sin dx + sin dx]
L 0 L L L/2 L L
8 nπ
= 2 2 sin . Thus Eq. (20) becomes
n π 2
∞
∑ n1 sin nπ2 sin nπx
8 nπat
u(x,t) = cos .
π2 2 L L
n=1
Section 10.7 225
1b.
1c.
6a. The motion is governed by Eqs.(1), (3) and (31), and thus
the solution is given by Eq.(34) where the kn are given
by Eq.(36):
∫ ∫ ∫
2 L/4 4x nπx 3L/4 nπx L 4(L−x) nπx
kn = [ sin dx + sin dx + sin dx]
nπa 0 L L L/4 L 3L L L
8L nπ 3nπ
= 3 3 (sin + sin ). Substituting this in Eq.(34)
n π a 4 4
nπ 3nπ
∞ sin + sin
∑
8L 4 4 nπx nπat
yields u(x,t) = sin sin .
3 3 L L
aπ n
n=1
6b.
226 Section 10.7
6c.
∫
L
cn = (2/L) f(x)sin[(2n-1)πx/2L]dx, n = 1,2,... .
0
Substituting these values into the above series for
u(x,t) yields the desired solution.
∫
2 (L+2)/2 (2n−1)πx
cn = sin dx
L (L−2)/2 2L
−4 (2n−1)π(L+2) (2n−1)π(L−2)
= [cos( ) − cos( )]
(2n−1)π 4L 4L
8 (2n−1)π (2n−1)π
= sin sin using the
(2n−1)π 4 2L
Section 10.7 227
10b.
10c.
17b. Using the hint and the first equation obtained in part (a)
leads to φ(x) + ψ(x) = 2φ(x) + c = f(x) so
φ(x) = (1/2)f(x) - c/2 and ψ(x) = (1/2)f(x) + c/2. Hence
u(x,t) = φ(x - at) + ψ(x + at) = (1/2)[f(x - at) - c] +
(1/2)[f(x + at) + c] = (1/2)[f(x - at) + f(x + at)].
2 -1 - at < x < 1 - at
f(x + at) = .
0 otherwise
∫
-1 x
Integration then yields φ(x) - φ(x0) = g(ξ)dξ and
2a x 0
hence
∫
x
ψ(x) = (1/2a) g(ξ)dξ - φ(x0).
x0
∫ ∫
x-at x+at
= -(1/2a) g(ξ)dξ + φ(x0) + (1/2a) g(ξ)dξ - φ(x0)
x0 x0
∫ ∫
x+at x-at
= (1/2a)[ g(ξ)dξ - g(ξ)dξ]
x0 x0
= (1/2a)[∫ ∫
x+at x0
g(ξ)dξ + g(ξ)dξ]
x0 x-at
= (1/2a∫
x+at
g(ξ)dξ.
x-at
∫ g(x)sin(nπx/a)dx, n = 1,2,... .
a
cnsinh(nπb/a) = (2/a)
0
∫
a/2
cnsinh(nπb/a) = (2/a)[ xsin(nπx/a)dx +
0
∫
a
(a-x)sin(nπx/a)dx] = [4a sin(nπ/2)]/n2π 2, n = 1,2,...,
a/2
so cn = [4a sin(nπ/2)]/[n2π 2sinh(nπb/a)]. Substituting
these values for cn in the above series yields the
desired solution.
1c.
Section 10.8 231
1d.
∫ k(y)sin(nπy/b)dy.
b
u(0,y) = k(y) we obtain dn = (2/b)
0
Using the B.C. u(a,y) = f(y) we obtain
∫
b
cnsinh(nπa/b) + dncosh(nπa/b) = (2/b) f(y)sin(nπy/b)dy, which
0
can be solved for cn, since dn is already known. The second
problem, in this approach, would be uxx + uyy = 0, u(x,0) =
h(x), u(x,b) = g(x), u(0,y) = 0 and u(a,y) = 0. This has the
fundamental solutions
232 Section 10.8
∫
2
a-ncn = (1/π) f(θ)cosnθdθ, n = 0,1,2,... and
0
∫ f(θ)sinnθdθ, n = 1,2... .
2
a-nkn = (1/π)
0
4a2
∫
2 a nπx
8b. cn = x(a−x)sin dx = 3 3 (1−cosnπ)
a 0 a n π
∞
u(a,y) = ∑ c sinh[(2n-1)πa/2b]sin[(2n-1)πy/2b] = f(y).
n
n=1
By properly extending f as a periodic function of period
4b as in Problem 39, Section 10.4, we find that the
coefficients cn are given by
∫ f(y)sin[(2n-1)πy/2b]dy,
b
cnsinh[(2n-1)πa/2b] = (2/b)
0
n = 1,2,... .
Capítulo
11