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Unformatted text preview: 3.2 95 3
(c) yo:%24,aslongasﬁ26+6x/§. (d) limgﬂ00 to = ln(3/2), limgnoo yo 2 oo. 27.(a) Assuming that y is a constant, the ODE reduces to 0y 2 d. Hence the only
equilibrium solution is y = d/c. (b) Setting y = Y + d/c, substitution into the differential equation results in
aY” +bY’ +c(Y +d/c) = d.
The equation satisﬁed by Y is
aY”+bY’+cY=0. 1.
2t —3t 2
W(e2t, 6—3t/2) : 2:2t _e%e—:t/2 2. *get/2
3.
W(e—2t7te—2t) : _€2’e:t (1 156232—21: : e—4t
5.
W(et Sin t ’ 8t COS t) I et(sifit:1::os t) et(coest :in t)‘ : —e2t'
6. cos2 0 1 + cos 29 2 _
W(C°S 9"1 + COS 29) — l—2sin 0 cos 9 —2 sin 29 H. 7. Write the equation as y” + (3/t)y’ : 1. p(t) = 3/t is continuous for all t > 0.
Since 750 > 0, the IVP has a unique solution for all t > 0. 9. Write the equation as y” —— ﬁy’ + Wing : ﬂ. The coefﬁcients are not continuous at t = 0 and t = 4. Since to E (0 ,4) , the largest interval is 0 < t < 4. 10. The coefﬁcient 3111 t is discontinuous at t: 0. Since to > 0, the largest
interval of existence is 0 < t < oo . 11. Write the equation as y” —— ﬁy’ + %y = 0 . The coefﬁcients are discontin— uous at a: = 0 and x : 3. Since :80 E (0 ,3) , the largest interval is 0 < m < 3. 3.5 109 40. Note that cos t sin 15 : ésin 2t. Then 1— k cos t sin t : 1 — gsin 2t. Now if
0 < k < 2, then gsin 2t < Isin 2t and —% sin 2t > — lsin 2751. Hence k
1—kcostsint21—§sin2t>1—sin2t20. 41. The equation transforms into y” a 43/ + 43/ = 0. We obtain a double root 7" = 2.
The solution is y : 0162“7 + 02x62?” : 01621“ —— Cg 111756211” 2 01152 + Cgt21nt. 43. The equation transforms into y” — 7y’ / 2 + 5y/2 : O. The characteristic roots
are 7‘ = 17 5/2, so the solution is y : Clea: +0265m/2 : Clelnt +C2eSlHt/2 : Clt+62t5/2. 44. The equation transforms into y” + 2y’ + y = 0. We get a double root 7“ = ~1.
The solution is y 2 claim + Cgl‘STm = ale—h” + Cglnt€_1nt = alt—1 + 6275—1111t. 45. The equation transforms into y" — 3y’ + 9y/ 4 = 0. We obtain the double root
r = 3/2. The solution is y = 01631/2 —I— (area/2 = 01631“ “2 + Cgll’lte31nt/2 : clt3/2 + 02t3/2 Int. 46. The equation transforms into y” + 43/ + 13y = 0. The characteristic roots are
7“ = —2 :: 31'. The solution is y = cleﬁh‘ cos(33:) + 026’233 sin(3x) = (211572 cos(31nt) + 0275—2 sin(31nt). 2. The characteristic equation for the homogeneous problem is 7‘2 + 27“ + 5 = 0,
with complex roots 7" = —1 :: 21‘. Hence yc(t) : ale—t cos 2t + Cg€_t sin 2t. Since
the function g(t) = 3 sin 2t is not proportional to the solutions of the homogeneous
equation, set Y = Acos 215 + B sin 2t. Substitution into the given ODE, and com—
paring the coefﬁcients7 results in the system of equations B — 4A 2 3 and A + 4B : 0. Hence Y = —% cos 275 + % sin 275. The general solution is y(t) : yc(t) + Y. 3. The characteristic equation for the homogeneous problem is T2 — 27“ — 3 2 0,
with roots 7" : —1, 3. Hence yc(t) = ale—t + Cge3t. Note that the assignment Y : Ate—t is not sufﬁcient to match the coefﬁcients. Try Y : Ate‘t + Bt2e_t.
Substitution into the differential equation, and comparing the coefﬁcients, results
in the system of equations —4A+ ZB : 0 and —8B : —3. This implies that
Y = rite—t + 315264 . The general solution is y(t) : yc(t) + Y. 110 Chapter 3. Second Order Linear Equations 5. The characteristic equation for the homogeneous problem is 7‘2 + 9 = 0, with
complex roots 7" : :32'. Hence yc(t) = cl cos 31‘ + 02 sin 3t. To simplify the anal—
ysis, set 91(t) = 6 and 92(75) 2 t263t. By inspection, we have Y1 = 2/3. Based
on the form of 92, set 1/72 : A6“ + BteBt + C7526“. Substitution into the differ—
ential equation, and comparing the coefﬁcients, results in the system of equations 18A+6B+2C=0,18B+12C:0,and18C:1.Hence
1 1 1
Y:_3t__t3t _t23t'
2 1626 276 +18 6 The general solution is y(t) = yc(t) + Y1 + Y2 . 7. The characteristic equation for the homogeneous problem is 27'2 + 37“ + 1 = 0,
with roots 7‘ = —1, —1/2. Hence yc(t) = Cle‘t + (326—t/2. To simplify the analysis,
set 91(75) 2 t2 and 92(t) = 3sin t. Based on the form of 91, set Y1 : A + Bt + Ct2.
Substitution into the differential equation, and comparing the coefﬁcients, results
in the system of equations A + 3B + 40 = 0 , B + 60 = 0, and C : 1 . Hence we
obtain Y1 = 14 — 675 + t2. On the other hand, set Y2 = D cos t + E sin t. After
substitution into the ODE, we ﬁnd that D = —9/10 and E = —3/10. The general
solution is y(t) : yc(t) + Y1 + Y2. 9. The characteristic equation for the homogeneous problem is 7‘2 + wg : 0, with 1
complex roots 7" : :: we Hence yc(t) : cl cos we t + 02 sin wet. Since w 75 we ,
set Y = A cos wt + B sin wt. Substitution into the ODE and comparing the co
efﬁcients results in the system of equations (w?) — w2)A : 1 and (w?) — w2)B : 0.
Hence cos wt . The general solution is y(t) = yc(t) + Y. 10. From Problem 9, yc(t) is known. Since cos wet is a solution of the homoge—
neous problem, set Y : At cos wet + Bt sin wet. Substitution into the given ODE and comparing the coefﬁcients results in A 2 O and B : 201m . Hence the general
t 2WD solution is y(t) = 01 cos wet + Cg sin wet + sin wet .
12. The characteristic equation for the homogeneous problem is r2 — r — 2 : 0,
with roots 7’ : ~1 , 2 . Hence yc(t) : 01 e_t + 02 62‘. Based on the form of the right
hand side, that is, cosh(2t) = (62‘ + eg2‘)/2, set Y : At 6” + 36’”. Substitution
into the given ODE and comparing the coefﬁcients results in A = 1/6 and B = 1/8 .
Hence the general solution is y(t) = clc’t + 02 e” + tram/6 + e—Qt/8. 14. The characteristic equation for the homogeneous problem is r2 + 4 : 0, with
roots 7" = :: 2i. Hence yc(t) : cl cos 2t + 02 sin 2t. Set Y1 : A —— Bt + C752. Com
paring the coefﬁcients of the respective terms, we ﬁnd that A = —1/8, B : 0,
C : 1/4. Now set Y2 2 Det, and obtain D = 3/5. Hence the general solution is y(t) : (:1 cos 2t+cesin 2t—1/8+t2/4+3et/5. Invoking the initial conditions, we require that 19/40 — 61 = 0 and 3/5 + 2ce = 2.
Hence cl : —19/40 and Cg : 7/10. 115 3.6 9"
C» l 36. We have (D2 + 2D + 1)y 2: (D + 1) (D + 1)y. Let u = (D + Dy, and consider
(it) 2 2t 6% + 064. We therefore the ODE u’ + u = 26%. The general solution is u
t. The general solution of the have the first order equation u
latter differential equation is t
y(t) : 64/ [Zr + clldr + 026t :
t 0 ’+u:2te_t +016 e_t(t2 + 6175 + C2), and consider the equa—
2COS 2t+c. The
The general D2 +2D)y=D(D+2)y. Let u: (D+2)y,
ect integration results in u(t) : 3t —
’+2y:3t—2cos 2t+c. 37. We have (
tion u’ z 3 + 4sin 2t. Dir problem is reduced to solving the ODE y
solution of this ﬁrst order differential equation is
t
1105) = 621: / 62T [3T — 2cos 2T + 0] dr + 026’” z
750 (cos 2t + sin 2t) + 01 + Cge’zt. #3t_1
“2 2 ation is 3160;) = 01 eZt + 6263? The functions y1(t) : em and y2 (t) = 63‘ form a fundamental set of solutions. The Wronskian
5t. Using the method of variation of parameters, of these functions is W(y1, y2) = e
the particular solution is given by Y(t) : u1y1(t) + u; (t) yg (t), in which 1. The solution of the homogeneous equ 3t t
U1 (t : — “’6 dt : 26’t
2t 2 t
U205) = (142;) dt = ~e'2t
t t_ Hence the particular solution is Y(t) = 2 neous equation is yc(t) 016% + Cgt6_t. The func—
‘t form a fundamental set of solutions. The Wron—
‘Qt. Using the method of variation of (t) 31105) + U2 (75) {92 (t). in 3. The solution of the homoge
tions 3/105) : 6* and 14206) : te skian of these functions is W(y1,y2) = 8
parameters, the particular solution is given by Y(t) = ul which t t
t _
/e(%)a=—wﬂ Hence the particular solution is Y(t) = #3t26Tt/2 + 31526”t : 31526t/2.
75/2 form a fundamental set of solutions. = et. First write the equation in : (at/2 and mm : t6
functions is W(y1,y2) 4. The functions y1(t)
The Wronskian of these 116 Chapter 3. Second Order Linear Equations standard form, so that g(t) = 4et/2. Using the method of variation of parameters7
the particular solution is given by Y(t) = U10?) y1(t) + U2 y2(t), in which tat/2 Kiel/2
u1(t) : — Wdt : —2t2
t/2 4 75/2
u2(t) : / %(:))dt : 4t Hence the particular solution is Y(t) : —2t26t/2 + 4t2et/2 : 2t26t/2 . 6. The solution of the homogeneous equation is yaw) = C1 cos 313+ 02 sin 3t. The
two functions y1(t) : cos 3t and y2(t) = sin 3t form a fundamental set of solu
tions7 with W(y1, 3/2) : 3. The particular solution is given by Y(t) = u1(t) y1(t) +
u2 (t) y2(t), in which ' 9‘ 2375
u1(t):—/%2dt=~csc3t 3t 9 a 232:
71,203) = /%::c—)dt = ln Isec 3t + tan 375! Hence the particular solution is Y(t) : —1 + (sin 3t) 111 [sec 3t + tan 3tf. The gen—
eral solution is given by y(t) : cl cos 31‘ + ()2 sin 3t + (sin 323) 111 [sec 3t + tan 3t ~ 1 . 7. The functions y1(t) 2 e‘2t and y2(t) = t€_2t form a fundamental set of so—
lutions. The Wronskian of these functions is W(y1,y2) : 6—4t. The particular
solution is given by Y(t) : u1(t) yl(t) + u2(t) y2(t), in which t 72t —2 72t
u1(t) : ~/wdt= —lnt WU)
—2t 72 —2t
6 (t e )
7 t : _dt : —l t
Hence the particular solution is Y(t) : ~6T2t ln )5 — 6—2t. Since the second term is a solution of the homogeneous equation, the general solution is given by y(t) : 616—2t + Cgte’2t — (3’2t In if. 8. The solution of the homogeneous equation is yc(t) = c1 cos 215+ 62 sin 2t. The
two functions y1(t) : cos 2t and y2(t) = sin 2t form a fundamental set of solu
tions, with W(y1, yg) : 2. The particular solution is given by YO) : U1(t) 31105) +
U2(t) y2(t), in which mm _ _ / sin 27312:? 2t)dt : 3W2 2t 3 2t 3
W): / 2 Z in an a 118 Chapter 3. Second Order Linear Equations The general solution of the differential equation is given by l 1 t
y(t) = 01 cos 2t + 02 sin 213+ i/ 9(5) sin(2t — 25)ds. 13. Note ﬁrst that p(t) : 0 . q(t) = —2/t2 and g(t) : (3252 — 1)/t2. The functions
y1(t) and yg (t) are solutions of the homogeneous equation7 veriﬁed by substitution. The Wronskian of these two functions is W(y1,y2) : —3. Using the method of
variation of parameters7 the particular solution is Y(t) = u1(t) y1(t) + u2(t) y2(t), in which 1 2
15— (3t —1) _
152(3252—1)
t _ dt= t3 3 l
u2() / t2 / t/g Therefore Y(t) = 1/6 + t21n t — t2/3 + 1/3. Hence the general solution is
y(t) = 01752 + C2t_1 + t2 ln 75—0— 1/2. 15. Observe that g(t) : is”. The functions y1(t) and y2(t) are a fundamental set
of solutions. The Wronskian of these two functions is W(y1,y2) : tet. Using the
1' method of variation of parameters, the particular solution is Y(t) = 11,103) in + U205) yﬂt)7 in which 2f
u1(t) = 7/ egaj)dt = —€2t/2 62:
u2(t) 2 /(1—+Wt/)((:)—)dt= tet Therefore Y(t) = —(l +t)€2t/2 +13th 2 —62t/2 + ten/2. 16. Observe that g(t) : 2(1 — t) 8_t. Direct substitution of y1 2 6t and y2(t) : t
veriﬁes that they are solutions of the homogeneous equation. The Wronskian of the
two solutions is W(y1,y2) = (1 — t) at. Using the method of variation of parameters, the particular solution is Y(t) : U1(t) y1(t) + 1L2(t) yg (t), in which 2t 1 a , rt
u1(t) : — / (—mdt : te’2t + e—2t/2 W(t)
; _ 2(1_t) _ —L'
1112(07/ WU) dt——26 Therefore Y(t) = teﬁt + 6775/2 — 2t e—t : —te’t + eat/2. 17. Note that g(:z:) : ln 3:. The functions y1(:r) = :62 and y2(a:) : $2111 :3 are solu
tions of the homogeneous equation, as veriﬁed by substitution. The Wronskian of
the solutions is W(y1,y2) : 2203. Using the method of variation of parameters. the
particular solution is Y(l‘) = um?) mm) + U201“) 212(93), 142 Chapter 4. Higher Order Linear Equations 10. In polar form, 2(cos 7r/3 +i sin 7r/3) : 2 (am/3+2m”, in which m is any integer.
Thus [2(cos 7r/3 +z' sin 7r/3)]1/2 = 21/2 (am/6+7”? With m = 0, one square root  is given by 21/2 ei’r/6 : (x/3_+ With m : 1, the other root is given by 21/2 €i77r/6 : (_\/§ _ 11. The characteristic equation is r3 — 7‘2 r + 1 = 0. The roots are r = 71,1,1 .
One root is repeated, hence the general solution is y = 01 e_t + Cget —7 Cgtet. 13. The characteristic equation is r3 e 27'2 — 'r + 2 = 0, with roots 7“ = —1,1,2. The roots are real and distinct, so the general solution is y : ole—i —— Cget + 0362‘. 14. The characteristic equation can be written as r2 (7“2 — 47“ + 4) = 0. The roots
are 7" : 0, 0, 2, 2. There are two repeated roots, and hence the general solution is given by y : 01 + 0275 + 0362‘ + C4t€2t. 15. The characteristic equation is r6 + 1 : 0. The roots are given by r = (—1)1/6,
that is, the six sixth roots of 1 . They are 6—”i/6+m”i/3, m = 0,1, . . . ,5. Explic
itly, 7": (\/3_—i)/2, Mrs—Hm, 2‘, —i, (—\/§+i)/2, (—ﬁ—iw. Hence the general solution is given by
y = reﬁt/2 [01 cos (t/2) + 62 sin (t/2)] + C3 cos t+
+04 sin t+ tZ"/§t/2 [05 cos (t/2) + CG sin (if/2)]. 16. The characteristic equation can be written as (“r2 — 1)(r2 — 4) = 0. The roots
are given by 7" — 1, 2 . The roots are real and distinct, hence the general solution is y = 016% + 0261t + (336—2t + 046”. 17. The characteristic equation can be written as (r2 — 1)3 : 0. The roots are
given by 7’ = :: 1 , each with multiplicity three. Hence the general solution is y : cle’t + 021:6“ + c3t2e_t + C4€t + 05156t + 06t2et. 18. The characteristic equation can be written as r2(r4 e 1) = 0. The roots are
given by 7" = 0, 0, * 1, "i . The general solution is y : 01 +62t+036—t +C4et +05cos t+c6sin t. 19. The characteristic equation can be written as r(r4 — 373 + 3r2 — 37' + 2) : 0.
Examining the coefﬁcients, it follows that r4—3r3+3r2—3r+2= (re1)(r—2)(r2+1). Hence the roots are 7" : 0,1,2,ii. The general solution of the ODE is given by
y : c1 + Cget + 036% + C4 cos t + c5 sin 76. 20. The characteristic equation can be written as r(r3 e 8) = 0, with roots 7" = 0,
262mm/3, m = 0, 1, 2. That is, 7‘ = 0, 2, —1 g: Hence the general solution is
y : 01 + 0262’t + 6": [C3 cos x/Z’Tt + C4 sinx/Ht. ...
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