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Unformatted text preview: 72 Section 4.1 CHAPTER 4
Section 4.1, Page 224
2. Writing the equation in standard form of Eq.(2), we
obtain y” + [(sint)/t]y" + (3/t)y = cost/t. The
functions p1(t) = sint/t, p3(t) = 3/t and g(t) = cost/t each have a discontinuity at t = 0. Hence Theorem 4.1.1
guarantees that a solution exists for t < 0 or for t > 0. 4. The equation is in standard form with pl(t), p2(t) and
lnt is II p3(t) being continuous for all t. However, g(t) defined and continuous only for t > 0. 2t—3 2t2+1 3t2+t
8. We have W(fl,f2,f3) = 2 4t 5t+1 = 0 for all t. 0 4 6 Thus by the extension of Theorem 3.3.1 (or by the
discussion following Eq.(8), the given functions are
linearly dependent. To find a linear relation we have c1(2t—3) + c2(2t2+1) + c3(3t2+t) = (2c2+3c3)t2 + (201+ca)t + (~3cl+c2) which is zero when
(2c2+3c3) = 0, 2cl+c3 = 0 and 3cl+c2 = 0. Solving, we find c:1 : 1, c2 = 3 and c3 = —2 and hence (2t—3)+3(2t2+1)—2(3t2+t) = o. 13. That et, e t, and e_2t are solutions can be verified by
direct substitution. For y = e'zt, y' = 2e"2t,
y" = 4e_2t, and y”’ = —8e—2t and thus
(—8e'2t) + 2(4e_2t) — (—2e_2t) — 2(e“2 that e—2t is a solution. Computing the Wronskian we t) = 0, verifying obtain,
at e—t e~2t I 1 l 1
W(et'e—t'e—2t) = at _e_t _Ze_2t = 921: 1 _1 “2 = _6el—2t
et e—t 49—2t 1 1 4 17. To show that the given Wronskian is zero, it is helpful,
in evaluating the Wronskian, to note that
(sinztf = Zaintcost = sin2t.
The result can also be obtained directly since 1
sinzt = (1 — cos2t)/2 = 13(5) + (—1/2)cosZt and hence 19a. 19b. 19c. 23. 25a. 25b. Section 4.4 73 sinzt is a linear combination of 5 and cosZt. Thus the
functions are linearly dependent and their Wronskian is zero.
If y 2 t“ then y' = ntn‘l, y" = n(n—1)t“'2,
y“‘1 = [n(n—1)(n—2)21t. so 12“” = [n(nl)(n—2)“'(2)(1H
The nth derivative of ert is y‘n’ — rnert.
If we let L[y} = y(4) — 5y” + 4y and if we use the result
of Prob. 19b, we have L[ert] = (r4 — 5r2 + 4)ert. Thus ert
will be a solution of the D.E. provided
(12—4)(r2—1) = 0. Solving for r, we obtain the four . t —t 2t ~2t .
solutions e , e , e and e . Since t — 2t — .
W(e , e t, e , e 2t) ¢ 0, the four functions form a fundamental set of solutions. Writing the equation in the form of Eq.(2), we have Zdt
_f_:w c
p1(t) = — and from Prob. 20, W — ce = —37.
t
On 0 < t < 1, f(t) = t3 and g(t) = t3. Hence there are
nonzero constants, c1 = 1 and c2 = 1, such that On —1 < t < 0, c1f(t) + czg(t) O for each t in (0,1). 1 defines f(t) —t3 and g(t) t3;
constants such that c1f(t) + c2g(t) (1,0). Thus f and g are linearly dependent on
0 < t < 1 and on 1 < t < 0. thus c1 = c2 0 for each t in We will show that f(t) and g(t) are linearly independent
on «1 < t < l by demonstrating that it is impossible to
find constants c1 and c2, not both zero, such that c1f(t) + czg(t) O for all t in (1,1).
two such nonzero constants and choose two points t0 and
t1 in ~l < t < 1 such that tO < 0 and t1 > 0. Assume that there are Then —cltg + cztg 0 and citi+ czti — 0. These equations
have a nontrivial solution for c1 and 02 only if the determinant of coefficients is zero. But the determinant
of coefficients is ~2tgti ¢ 0 for t0 and t1 as specified. Hence f(t) and g(t) are linearly independent on —1 < t < l. 74 Section 4 . 2 [ —t3 t3
25c. For —1 < t < O, W(f,g) = = O and for 0 s t < l,
—3t2 3t2
t3 t3
W(f,g) : = O. This shows that f and g cannot
3t2 3t2 27. be solutions of an equation y” + p(t)y'+ q(t)y = 0 with p
and q continuous on 1 < t < l. Differentiating st and substituting in the D.E. we verif}r that y = at is a solution: 2—t et + (2t—3)et — tet + et = 0.
) Now, as in Prob. 26, we let y = v(t)et. Differentiating
three times and substituting into the D.E. yields (2—t)etv”’ + (3—t)etv” = O. Dividing by (2—t)et and
letting w = v” we obtain the first order separable
t—3 1
equation w' = ——w = (—1 + —)w. Separating t and w,
t—2 tZ
integrating, and then solving for w yields
w = v" = c1(t—2)e_t. Integrating this twice then gives t _ —t _ t___ t
v—clte ~4—c2t+c3 so thaty—ve ~c1t+czte +c3e, which is the complete solution, since it contains the given y1(t) and three constants. Section 4.2, Page 231 2. If —1 + ix/E = R(cos6+isin9) = Re”, then Rcose = —1 and
RsinB = V . Thus R2 = (—1)2 + Mfg)2 = 4 and the angle 6
is given by RcosB = 2cosB = —l and Rsinﬂ = ZsinB = V . Hence cose = —l/2 and sinB = V3 [2 which has the solution 6 = 2313/3. The angle
3 is only determined up to an additive integer multiple
of 3 2n. 8 Writing (lui) in the form Re.l , we have RcosB = l and Raine = —l, which yield R = V2 and 9 = 41/4. Thus
(l—i) = V2 sum/4+2”) (where m is any integer) and hence
(1_i)1/2 = [21/2el(—ﬂ/4+2m)]1/2 = 21/4el(—J13/8+m2 We obtain the two square roots by setting m = 0,1. They are Elme—iﬂ/Band 21/4ei7n/8. Note that any other integer value of m gives one of these two values. Also note that
l—i could be written as l—i = V2 sum/4 + 2m.) 15. 23. 27. 29. Section 4.2 75 It We 100k for solutions of the form y = e Substituting in the D.E., we obtain the characteristic equation
3 r — 3r2 + 3r — l = O which has roots r = 1,1,1. Since
the roots are repeated, the general solution is y = clet + cztet + catzet. We look for solutions of the form y = ert Substituting in the D.E. we obtain the characteristic equation 5 in r + 1 = 0. The six roots of —1 = e are obtained by o,1,2,3,4,5 in (4)1”: ei‘“+2m”6. They are
= “[37 + i)/2, em” = i, aim/6 = (—ﬁ + i)/2,
= (—ﬁ — i)/2, 63””
eilml6 = (VF; — i)/2. Note that there are three pairs of
conjugate roots. The general solution is y = evgt/2[C1COS(t/2) + czsin(t/2)] setting m =
eix/é i7n 6
e / = —i, and + e—VEt/2[c3cos(t/2) + c4sin(t/2)] + 05cost + cﬁsint. The characteristic equation is r3 "5r2 + 3r + 1 = 0.
Using the procedure suggested following Eq. (12) we try, since a0 = an = 1, r = l as a root and find that indeed it is. Factoring out (r—l) we are then left with
rZ—ér—1=O, «I5.
The characteristic equation in this case is 12r4 + 31r3 + 75]:2 + 37r + 5 = O. which has the roots 2 1 Using an equation 1 1
solver we find I = —24 —r3, —1 : Zi. Thus
~t/4 st/E it . .
y = cle + c2e + e (c3cos2t +c4s1n2t). As in Prob. 23, it is possible to find the first two of these
roots without using an equation solver. Factoring then
reduces the characteristic equation to a quadratic, which
can be solved for the other two roots. The characteristic equation is r3 + r = 0 and hence r = 0, +i, vi are the roots and the general solution is
y(t) = c1 + czcost + eBaint. y(0) = 0 implies C1 + c2 = 0, y’(0) = l and y"(0) = 2 Use this last equation in the first to 1 implies c3 =
implies —c2 = 2.
find :21 = 2 and thus ytt) =
continues to oscillate about y = 2 as t —a an 2 — Zoost + sint, which 76 Section 4.2 29 A . n 30' i /\ 30. The general solution is given by Eq. (21). 31. The characteristic equation is r4 — 4r + 4r2 = O, which has the roots r = 0,0,2,2. Thus the general solution would normally be written y(t) = c1 + czt + c3e2t + c4te2t. However, in order to evaluate the c's when the initial
conditions are given at t = 1, it is advantageous to rewrite this as y(t) = c1+_c2t + 0592(t—1) + c6(t_l)62(t—1)I which also satisfies the given D.E. 34. 34. The characteristic equation is 4r3 + r + 5 = U, which has 1
roots —1,§ 2 i, where r1 = —1 can be found as in
—t t/2 .
Prob.23. Thus y(t) = cle + e (czcost + c3s1nt),
y’ﬁt) = —c1e_t + et’2[(c2/2 + c3)cost + ("£22 + c3/2)sint]
and
n "t t/Z _
y (t) = cle + e [(—3c2/4 + c3)cost + (—cz — 3c3/4)s1nt].
The I.C. then yield c1 + c2 = 2, —c1 + c2/2 + c3 = 1 and
c1 — 3c2/4 + c3 = —1. Solving these last three equations give c1 = 2/13, c2 = 24/13 and c3 = 3/13. 37. The approach for solving the D.E. would normally yield
Y(t) = Clcost + czsint + cset + cae‘t as the solution. Section 4.3 77 Since cosht = (et+e_t)/2 and sinht = (et—e_t)/2, y(t) can
be written as y(t) = clcost + czsint + c3cosht + c4sinht, where c3 and c4 can be written in terms of c5 and as. It is convenient to use cosht and sinht rather than et and e because the I.C. are given at t : 0. Since cosht and
sinht and all of their derivatives are either 0 or 1 at
t = 0, the algebra in satisfying the I.C. is greatly
simplified. _ :ﬁdt
38a.Since pl(t) = O, W = ce = c. 393.As in Sect.3.7, the force that the spring designated by
k1 exerts on mass n5 is 3u1. By an analysis similar to that shown in Sect.3.7, the middle spring exerts a force
of —2(ul—u2) on mass :111 and a force of —2(u2—u1) on mass n5. Thus Newton's Law gives mlul" = —3ul —2(ul—u2) and
m2u2" = ~—2{u2—u1), where u1 and u2 are measured from their equilibrium positions. Setting the masses equal to l and
rewriting each equation yields Eqs.(i). In all cases the
positive direction is taken in the direction shown in
Figure 4.2.4. 39b.The characteristic equation for (ii) is r4 + 7r2 +6 = 0,
or (r2+1)(r2+6) = 0. Thus the general solution of Eq,(ii)
is u1(t) = clcost + czsint + c3coswl6 t + c4sinx/6 t. 39c.From Eq.(iii) and u1 from 39b we have ul(0) = C1 + c3 = l
and u;(0) = 02 + V6504 = O. From Eq.(i) we have
u:(0) = 2u2(0) — 5u1(0) = —1 and u1”(0) = 2u;(0} — 5u1(0) = 0, hence —cl  6c3 = —l and —c2  6Vq;c4 we find c1 = 1 and c2 = c3 = c4 = 0, so that u1 = cost. 0. Solving the four equations for the ci The first of Eqs.(i) then gives 2u2 ul" + 5u1 = 4cost and thus u2 = Zoost. Section 4.3I Page 237 1. First solve the homogeneous D.E. The characteristic equation is r3 r2 — r + 1 = O, and the roots are r = —l, t l, 1; hence yE(t) = cle' + czet + catet. Using the 78 Section 4.3 superposition principle, we can write a solution as the sum of particular soluti ns corresponding
t to the D.E. y”—y’yHy = 2e and y”—y'—Y+y = 3. Our initial choice for a particular solution, Y1, of the particular is a solution of the
Thus, first equation is Ae‘t; but e—t
homogeneous equation so we multiply by t. Y1(t) =
Y2(t) = B, and there is no need to modify this choice. Ate—t. For the second equation we choose The constants are determined by substituting into the individual equations. We obtain A = 1/2, B = 3. Thus,
the general solution is y = cle‘t + czet + c3tet + (ts—t)/2 + 3. The characteristic equation is r4 — 4r2 = r2(r24) = 0, 21'. so yc(t) = c1 + cat + c3e_2t + c4e . For the particular solution correspnding to t2 we assume
Y1 = t2(At2 + Bt + C) and for the particular solution corresponding to et we assume Y2 = Det. Substituting Y1, in the D.E. yields —48A = l, B = 0 and 24A—8C = 0 and
substituting Y2 yields ~3D = 1. Solving for A, B, C and D gives the desired solution. The characteristic equation for the related homogeneous D.E. is r3 + 4r = 0 with roots r = 0, +2i, —2i. Hence
yc(t) = c1 + czcosZt + c3sin2t. The initial choice for Y(t) is At + B, but since B is a solution of the
homogeneous equation we must multiply by t and assume
Y(t) = t(At+B). A and s
are found by substituting
in the D.E., which gives 4
A = 1/8, B = 0, and thus E
the general solution is 3'
y(t)=c1+c2c052t+c3sin2t+(1/8)t .
Applying the I.C. we have 2
no) =o=>c1+c2=0.
y'(0) = 0 =9 2C3 = 0,
y”(0) = 1 =9 ~4c2 + 1/4 = 1,
which have the solution c1 = 3/16, c2 = —3/16, c3 = and * i 4 5 i
I D 1 2 0. For t = n, 2n... the graph will be tangent to t2/B and for large t the graph will be
approximated by t2/8. 13. 17. 20. 22a. Section 4.3 79 The characteristic equation for the homogeneous D.E. is 3  2r2 + r Hence the r = 0 with roots I = 0,1,1.
. . t
complementary solution is yc(t) = C1 + czet + c3te . We
I  I I HF J' 3
conSider the differential equations y — 2y” + y = t and y”  2y" + y' 2et separately. Our initial choice
for a particular solution, Y1, of the first equation is
Aot3 + Alt2 + Azt + A3; but since a constant is a solution of the homogeneous equation we must multiply by t. Thus Y1(t) t(Aot3 + A1122 + Azt + A3). t
Be , For the second equation
t but since both e and tet we first choose Y2(t) =
are solutions of the homogeneous equation, we multiply by
t2 to obtain Y2(t) = Btzet. Then Y(t) Y1(t) + Y2(t) by the superposition principle and y(t) yh(t) + Y(t). The characteristic equation is
4 3 r(r~l)(r2—l) 0, r — r
. . _ —t t t
solution is yb(t) m (21 + c2e + cae + c4te . so the complementary
The — r2 + r
superposition principle allows us to consider separately
the D.E. y‘4) — y — y" +y’ t2 + 4 and (4) tsint. For the first equation our Y S _ Y!!! __ y." + Yf
initial choice is Y1(t) Act2 + Alt + A2; but this must I" be multiplied by t since a constant is a solution of the
Hence Y1(t) t(A0t2 + Alt + A2). the second equation our initial choice that
(Bot + Bl)cost + + (Cot + C1)sint does not need to be For homogeneous D.E. Y2 —
modified. Hence Y{t} = t{Aot2 + Alt + A2} + (Bot + Bﬂcost + (Cot. + c1)sint.
(D—a)(D—b)f = (D—a)(Df—bf) = D2f — (a+b)Df + abf and
(D—b)(D—a)f = (D—b)(Df—af) = sz — (b+a)Df + baf. Since a+b = b+a and ab = be, we find the given equation holds
for any function f. The D.E. of Prob.13 can be written as
D(D—1)2y = t3 + Zet. Since 13“ annihilates t3 and (9—1) annihilates 2st, we have D5(D1)3y = 0, which corresponds to Eq.(ii} of Prob. 21. The solution of this equation is
ﬁt) = Alt4 + Azt3 + A3t2 + A4t + A5 + (131+;2 + 321. + B3)e_t. Since A5 and (th + :33)!t are solutions of the homogeneous equation related to the original D.E., they 80 22b. If y = te‘t Section 4.4 deleted and thus it" + 113:3
L d t may be
t + A3t2 + A4t + B1t2e_ . m
.i
. n then Dy = —te_t and Dzy = te—t  2e‘t,
so (D+l)2y = (D2+2D+l)y = 0 and thus (D+l)2 annihilates
t te_ . Likewise D21 annihilates 2c05t. Thus (D+l)2(DZ+1)
annihilates the right side of the D.E. of Prob.l4. + e—t 22e.D3(D2+1)2 annihilates the right side of the D.E.of Prob.l7. Section 4.4, Page 242 1. The complementary solution is ya = c1 + czcost + casint and thus we assume a particular solution of the form
Y = u1(t) + u2(t)cost + u3(t)sint. Differentiating and assnming Eq.(5), we obtain Y’ = —uzsint + u3cost and (a).
Continuing this process we obtain Y” = J J I .
u1 + ugcost + u3Sint = 0
—u2cost  u3sint,
and —u;sint + ugcost = 0 (b) a l' r 
and Y”' = uzsint — uacost ~ uzcost — ussint. Substituting Y and its derivatives, as given above, into
the D.E. we obtain the third equation: (C}«
Equations (a), (b) and (c) constitute Eqs.(10) of the
text for this problem and may be solved to give Thus I I _
~u2cost — u3s1nt = tant r t I I . 2
u1 = tant, u2 = —Sint, and u3 = Sin't/oost. sint — ln(sect + tant) u1 = —lncost, u2 cost and u3 = and substitution into Y above gives
Y —1ncost + 1 — (sint)ln(sect + taut), since sinzt + coszt = 1 Note that the constant
absorbed in C1 in yC above. 1 can be 1 by sect and use
sect, Replace tant in Eq.(c) of Prob.
Eqs.(a) and (b) as in Prob.1 to obtain u; = u; = —1 and u; = —sint/cost. Replace sect in Prob. 7 with e_tsint. 11. Section 4.4 81 Since et, cost and sint are solutions of the related
homogenous equation we have Y(t) = (10) then are t .
ule + uzcost + u3s1nt. Eqs. I t , J .
ule + uzcost + u3s1nt = 0 t .
ule — ugsint + ugcost = 0 u’let — ugcost — ugsint = sect.
Using Abel's identity, W(t) = cexp(:Iﬁl(t)dt) = Get.
1 1 0
Using the above equations, W(O) = 1 0 1 = 2, so
1 —1 0
c = 2 and W(t) = Zet. From Eq.(11), we have
0 cost sint
sect W1{t)
ua(t) = t , where W1 = 0 —sint cost = l
28 1 —cost sint
and thus
. 1 _ . .
u1(t) =§e tloost. LikeWise
r sect W2(t) 1
1.12 = —————e4——— = ———sect(cost — sint) and
2et 2
' sect W3(t) 1
u3 = —————————— = —asect(sint + cost). Thus
2et 2
1 eHSds 1 1 1 1
= — ———, u2 = —t — —1n(cost) and 113 = ——t + —ln(cost)
2 ocos(s) 2 2 2 2 which, when substituted into the assumed form for Y, yields
the desired solution. ' we may use the Since the D.E. is the same as in Prob. 7.
0. complete solution from that, with to = Thus
1 1
17(0) = 01 + '32 = 2; YWO) = C1 + C3 — E + E = —1 and
"(0) + 1 1 1 1
= C ' C —  + — = ,
Y 1 2 2 2 1a A computer algebra system
may be used to find the
respective derivatives.
Note that the solution is valid only for a
0 s t < 3, where we see ° 02 m u u 1 u 14 1a the vertical asymptote. 82 14. 16. Section 4.4 Since a fundamental set of solutions of the homogeneous
D.E. is yl = et, y2 = cost, Y3 = sint, a particular
solution is of the form Y(t) = etulg't) + (cost)u2(t) + (sint)u3(t). Differentiating and making the same assumptions that lead to Eqs.(10), we
obtain uaet + uécost + ugsint : G urlet  uésint + ugcost = G Git) Solving these equations using either determinants or by
(1/2)e‘tg(t). u'2=(1/2)(sint  cost)g(t),u’3 = ——(l/2)(sint + cost)g(t). Integrating these and substituting into Y yields
t 1 t —5
Eye]; 9 g(s)ds + cost
0 r t l f .
ule  uzcost — u3sint — . . . . 1'
elimination, we obtain u1 = t
Y(t) = ﬂ [sin(s) — cos(s)]g(s)ds
0 a t I
~sint£1[s1n(s) + cos(s)]g(s)ds}. Putting et, cost and sint inside the respective integrals yields
1: Y(t) = (1/2)]; [et—S + costsin(s) — costc0s(s) 
0 sintsin(s) — sintcos(s)]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. The characteristic equation has the repeated roots
r = 1,1,1 and thus the particular solution has the form Y = etu1(t) + tetu2(t) + tzetusrt).
making the same assumptions as in the earlier problems, Differentiating, and solving the three linear equations for u}, u;, and u;
yields ’ 2 t f at I _t
u1 =(1/2)t e g(t), u2 = —te g(t) and u3 = (l/2)e g(t).
Integrating and substituting into Y yields the desired solution. For instance
t 1 t tetu2 = —teiﬂ se‘sg(s)ds = —Et 2tse‘tﬁ)g(s)ds, and
0 D Section 4.4 83 t—Zet then 9(3) = es/sz, likewise for ud and u . If (t)
J. 3 g . . t. (t and the integration with repect and thus e 5)g(s) = ‘3
s
to s is accomplished using the power rule. Note that
terms involving t0 become part of the complimentary solution. ...
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