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Unformatted text preview: CHAPTER Second Order Linear Equations 1. Let y r e”, so that y" : re” and y” :12 er". Direct substitution into the
differential equation yields (1'2 + 21' — 3k” 3 D. Canceling the exponentiah the
cl'iarncteristic equation is r2 + 21' — 3 = U . The roots of the equation are r = _3 , 1 . Hence the general solution is y = olet + age3‘. 2. Let y = e”. Substitution of the assumed solution results in the characteristic
+ ST + 2 = U. The roots of the equation are r = ~2, m1. Hence the will. equation 7‘2
gene a] solution is y z ole” + 628 4. Substitution. of the assumed solution 3] z 8"" results in the characteristic equation
212 w 31' + 1 = 0 . The roots of the equation are 1‘ = 1/2 , 1 . Hence the gene‘s} solution is y : clef/‘3 + age". 6. The characteristic equation is air: — Q = 0 7 with roots 7‘ = :i:3/2 . Therefore the
gene al solution is y : ole—3”2 + (gem/2.
8. The characteristic equation is r2 — 27' — 2 = U , with roots 1' 2 l, i ﬂ . Hence the general solution is y : clc(1"‘/§)" i cad“HE”. ' 9. Substitution of the assumed solution 1} = e” results in the characteristic equa—
tion 7'2 + r — 2 2 U. The roots ol? the equation are r = M? , 1. Hence the general solution is y = ole—2" + ege". Its derivative is y’ 2 w‘2c1e‘m + age": Based on the
91 Chapter 3. Second Order Linear Equations ﬁrst condition, y(0) = 1, we require that c1 4» Cg = 1 . In order to satisfy y’(0) = 17 I we ﬁnd that —2c1 + C2 5 1. Solving for the constants, cl = [} and Cg = 1 . Hence %
the speciﬁc solution is Mt) 2 e". 11. Substitution of the assumed solution 3; m e” results in the characteristic equa
tion 67'2 — 57' + 1 = U. The roots of the equation are 'r : 1/3,1/2. Hence the
general solution is y : clef‘f3 + cgei/ﬂ. Its derivative is y’ = clot/W3 + ege“/2/2 .
Based on the ﬁrst condition, 11(0) 2 1, we require that (:1 +ca = 4. In order to
satisﬁy the condition y’(0) = 1 , we ﬁnd that (:1/3 + (12/2 2 0. Solving for the con— stants, c1 x 12 and (:2 m —8. Hence the specific solution is 31(1‘.) = 12 (at/3 — Sci/2.
12. The characteristic equation is 7'2 + 37' z 0, with roots 1' = ~3, 0. Therefore
the general solution is y = c1 + ege‘3‘, with derivative y’ = «*3 egeml‘”. In order
to satisfy the initial conditions, we ﬁnd that C] + Cg : —_, and —3 Cg m 3_ Hence
the speciﬁc solution is y(t) :: —i — .54". 13. The characteristic equation is 'rﬂ + 57‘ + 3 : 07 with roots r «13
3i 9 _ 71,2 = m The general soiution is y = c1e(“5”‘/ﬁ)‘i/3 + ege(“5+ml‘/2, with derivative '11 : Talc “geimﬁi—mH/‘J. ; “5 ““ m (—5—mn/2 + ‘5 + Vl3 r
2 In order to satisfy the initial conditions, we require that —"—\/13 —" v13
c1+c3=1 and +c1+J—:—~cz=0. 14. The characteristic equation is at”: + 7‘ — £1 = D , with roots 1 33
 n =——:i:———.
’1“ a 4 Chapter 3. Second Order Linear Equations / 9"
0.: ll 43. P = 1, Q 5 3:32, R = 3:. Note that P” —» Q’ + R r— —5:r, and therefore the
differential equation is not exact. 45. P z .752, Q 2.1:, R. = _1. We have P” — Q’ + R z D. The equation is exact.
W'rite the equation as {:11'331’)’ ~— (n:y)’ = 0. After integration, we conclude that
9:23," —~ my 2 c. Divide both sides of the ODE by 1:2. The resulting equation is
linear, with integTa‘ting factor it x 1/13. Hence (y/a‘)’ = CHI—3. The solution is
Mr.) i aka"1 + egaz. £17. P = r1:2 , Q = :r, B, m 3:2 w I12. Hence the coefﬁcients are 2P” — Q = 39: and
P”  Q' + R = 3:2 + 1 ~19. The adjoint of the original differential equation is
given by may.” + 3.1; ,u’ + {:r2 + 1  ugh}, : U . :19. P = 1, Q I (l, R = —;r. Hence the coefficients are given by 213” — Q r— 0 and
P” — Q' + R t w—ra: . Therefore the adjoint of the original equation is a” — .1: h, r» O . 2" _ . _. a t . .
2. e 3‘ = ele '5’ n (:‘(cos d w “I. 5111 3). 3. eiTr = cos wii sin 7r = ——l.
4. cg“%i = e2(cos s —'i sin 1}) = —eﬂ’.i.
(3'. 7r"1+2i = e(”i"*'2i)1“ ‘“‘ = e'l“ Tre2 1“ "f 7 (cos(21n a) + i? sin(2 1n ﬁ))/7r. 8. The cl'iaracteristic equation is r2 — 2r + G = I}, with roots r L“ 1 i W5 . Hence
the general solution is y : cle" cos \/5—t + (33 e" sin \l5t. 9. The characteristic equation is r2 + 27' — 8 = 0, with roots r = —4 ,2 . The roots
are real and different, hence the general solution is y = ole—4" + Cg e”. 10. The characteristic equation is r2 + 21' + 2 ﬂ 0, with roots 1' r —1 :i: Hence
the general solution is y : cle‘t cos 1.? + r32 e" sin t. 12. The characteristic equation is 4?? + 9 = U, with roots r «1 ﬁg Hence the
general solution is y m (:1 cos(3t/2) + C2 sin(3t/2). 13. The characteristic equation is 7'2 + 27' + 1.25 = 0, with roots r = —1 i Hence the general solution is y : ole": cos(t/2) + C2 e‘i‘ shut/2). :izi. [JIM 15. The characteristic equation is 7'2 + r + 1.25 = O , with roots 7‘ = —
Hence the Teneral solution is 1' ﬁ c (ff/2 cos i. + Cg e'i/2 sin t.
s J 1 ii 3.3 99 MI.‘ 16. The characteristic equation is “1'2 + Air + 6.25 r» U , with roots '1' 7—“ _‘2 :i:
Hence the general solution is y : ole2" cos(3t/2) + c: ed" sin(3t/2). roots 7' = i‘li. Hence the 17. The characteristic equation is r2 +4 : D, with
y : c1 cos 2i; + c_: sin 2!. . Now *5" = —‘2c1 sin 21‘. + 2(12 cos 2t . = O, we require that c1 7— 0. In order to satisfy general solution is
The constants are c1 :0 and Based on the ﬁrst condition, y(0)
the condition 'y’(0) = 1, we ﬁnd that 20; :— 1.
(:2 : 1/2. Hence the speciﬁc solution is y(t) = 12 sin 2t. 19. The characteristic equation is 7'2 ~— ‘Z'r + 5 s 0, with roots r = 1 :1: 2i. Hence
the general solution is y = clet cos 2t + Cg e" sin 21:. Based on the initial condition
y(7r/‘2) = 0, we require that c1 7— D. It follows that y 3 c2 et sin 21,, and so the
ﬁrst derivative is y' = C2 et sin 2t + ‘20; ct cos 215. In order to satisfy the condition
11"(7r/2) :2 .‘2, we ﬁnd that —2E€7r/2C2 : ‘2. Hence we have (:2 r» —c‘”/'3 . Therefore 0 the speciﬁc solution is y(t) = —e / 5 \LJ ‘20. The characteristic equation is r2 + 1 r— 0, with roots 1' = ii. Hence the gen—
eral solution is y = {:1 cos '1: + Cg sin 15. Its derivative is y’ = —~c1 sin 15+ ca cos t. Based on the ﬁrst condition, Mir/3) = ‘2, we require that c1 + ﬂag = 4. In err
der to satisfy the condition y’(n/3) = —4 , we ﬁnd that —\/§ c1 + Cg = —~8. Solving
these for the constants, c1 t 1 + 2x/‘d and c: = » 2. Hence the speciﬁc solution
is a steady oscillation, given by 310;) = (1 + 2J3) cos t + — 2) sin t. ‘21. From Problem 15, the general solution is y = ole—"1’2 cos H C2 e“""2 sin t. In—
voking the ﬁrst initial condition, y(0) = 3 , which implies that c; z 3 . Substituting,
"/2 sin t, and so the ﬁrst derivative is it follows that y = 3(3’”2 cos ti— cg e'” :
' ii 3 . .
y’ = —3e"‘/J cos 1; —« Sta—""3 sin t + ca {5 .1 (31 _ _
Le 11"[2511113. I
0 ; , .4 ._ '1
V“ cos t — lJI'I Invoking the initial condition, y’m) = 1 , we ﬁnd tha —% + c; = 1 , and so cg = 3.04 Chapter 3. Second Order Linear Equations M 9°
4:. 45. p(t) = 3i. and q(t) : if" . We have :1; = ftdt = (32/2. Furthermore,
N) + 272mm) 3 (1 Haw
9 3/2 ' ‘ ’ '
~ {003)} The ratio is not constant, and therefore the equation cannot be transformed. 46. 11(13) 2 t — 1/1? and q(t) m t2 . We have .‘L‘ = ftdt 2152/2. Furthermore,
0’“) + QWMU) : 1
2 new“ The ratio is constant, and therefore the equation can be transformed. From Problem
43, the transformed equation is (12y dy
r. + — +1; =
(i‘r (in:
Based on the methods in this section, the characteristic equation is 1'2 + r + 1 = i],
with roots 7' 2 —% :i: . The general solution is y(;r) 2 alarm/2 cos Vim/2 + Cg 6““2 sin {gm/2. Since :1: = t2/2 , the solution in the original variable t is y(t) = eﬁl‘E/‘1[c1cos(\/§t2/4)+ Cg sin(\/§t2/4): . 2. The characteristic equation is 97'2 + 67' + 1 = O, with the double root 1‘ x —1/3.
The general solution is y(t) 7» cle‘l/3 + Cate—U3. 3. The cl'iaracteristic equation is 47'? — 417‘ w 3 = D , with roots '1' = —1/2, 3/2 . The
general solution is y(t) : c] 5‘”? + elem/2 4. The characteristic equation is 4175" + 12_1'+ 9 2 U, with double root r = “3/2.
The general solution is y(t) = (Cl + (:3 t)e“3"/2. 5. The characteristic equation is 1'2 u 21' + 10 = O, with complex roots 2" = 1 i 31"..
The general solution is Mt) = cle" cos 3t i— ege" sin 3t. 6. The characteristic equation is r2 — 67' + 9 = D, with the double root 1' =‘ 3 . The
general solution is y(t) = clc3t + eat em. 7. The cl'iaracteristic equation is 472 + 17‘1'44: 0, with roots 7‘ = —1/4, —4.
The general solution is Mt) = clc"’/‘I + ege‘“. 8. The characteristic equation is 167‘2 + 247‘ + Q = 0, with double root 1‘ 2 —3/4.
The general solution is 31(1) 2 r.'1e”“‘“‘/‘l + cgte —3.l./'1 106 Chapter 3. Second Order Linear Equations ((1) Given that y’(0) 2" b, we have —3/‘2 + (:3 = b, or Cg : bt— 3/2. Hence the
solution is y(t) = 6‘3"” + [h + %)te‘3‘/2. Since the second term dominates, the
long—term solution depends on the sign of the coefﬁcient b + 3/ 2. The critical value
is b = —3/2. 16. The characteristic roots are r1 = 7‘; x 1/2. Hence the general solution is given
by y(t) = cgei/2 + cat et/E. Invoking the initial conditions, we require that cl = 2,
and that l. + c; = b. The speciﬁc solution is Mt} = 2e”2 + (b ~1}t61/2' Since the second term dominates, the longmterni solution depends on the sign of the
coefficient {J — 1. The critical value is b = 1. 1.8.(a) The characteristic roots are n = r2 2 —2/3. Therefore the general solution
is given by y(t) = 8162” 3 + cate‘m/S. Invoking the initial conditions, we require
that cl = e, and that —2e/3+ c3 = —1. After solving for the coefﬁcients, the specific solution is y(t) = (1641/3 + — 1)te"”2"/3_ (13) Since the second term dominates, the long—term solution depends on the sign
of the coefﬁcient m i . The critical value is a. = 3/2. 20.(a) The characteristic equation is 7'2 + 2n '1' + a“) : ('1' + e.)2 ""—"“ D . (b) With Mt) = 20. , Abel’s Formula becomes LV{yl 7CUE) : 66— 2a (ll. : cenﬂut ' (c) y] m e‘“ is a solution. From part (1)),
“Eu? emﬂtyﬂt) +ee’""yg(t) =cc ‘, which can be written as resulting in
e‘” yg{t) = ct. 23. Set ya = t2 oft) . Substitution into the ODE results in
{1(th + zl'tu’ + 21)) — new + 2m) + 612% = 0. After collecting terms, we end lip with file” = 0. Hence aft) 2 (:1 + Cgt, and thus
ya“) 2 cit2 + Cgfiu. Setting (:1 = O and (:2 n 1, we obtain yg(t) = ﬁg. 24. Set 11305) = tv(t) . Substitution into the ODE results in t2(t’U” + 2u’) + now + e) — em = 0. 3.4 M 107 After collecting terms, we end up with tan” + =1t2'u’ "—r 0. This equation is linear
in the vru'iable m 2 e’. It follows that e’U) 2 ct'“1, and n05) :— c. {73 + c2 . Thus “'3 m“) 7— ent—2 }~ Cgt . Setting (:1 = 1 and £2 = O, we obtain yﬁt) = t '. 26. Set ya“) 2 to(t). Substitution into the ODE results in U” — 'U’ = [1. This ODE
is linear in the variable in = 11’. It follows that 'u’('(.) m clef, and 110‘.) x (:1 e" + (32.
Thus 31136) : clie‘ «1— cat. Setting c1 : 1 and (:3 2 {1, we obtain 3130‘.) 2 te‘. 28. Set yaks) = e‘“*u(:a). Substitution into the ODE results in :1: — 2
T} H + {Ewle’zﬂ This ODE is linear in the variable to = 1;” . An integrating factor is mwﬂ p. x e m—1 (in: m (3’1. . I
Rewrite the equation as [ ' ] = D, from which it follows that *u’ (re) 2 c(:L‘ w 1)e“"i 37”]
Hence Mm) 2 else”: + (12 and yam) = cla + ege‘". Setting (:1 = 1 and c3 2 0, we
obtain yg(.7:) : an 29. Set yg(:r) 2 311m} ’UffL‘), in which y; (:u) = 3:1/‘le2V/E. It can be veriﬁed that yl
is a solution of the ODE, that is7 :L'gyl” — (:u w 0.1875)y1 2 0. Substitution of the
given form of in results in the differential equation leg/4v" + Hit?“ l— res/‘1)u' = 0 .
This ODE is linear in the variable in = U’. An integrating factor is a H : ef[2mﬁi’f'+ﬁ3d1n in I
Rewrite the equation as a”; 10’] = 0, from which it follows that o"(:i:) = Integrating7 11(3)) 2 clem'ﬁ + C2 and as a result1
yﬁm) m owl/"64¢"? + egmlf‘lez‘ﬂ.
Setting (:1 = 1 and Cg = 0, we obtain yam) 2531/4194“?
32. Direct substitution veriﬁes that y. = em‘hﬂ/2 's a solution of the ODE. Now
set yam) x 1111(35) Substitution of ya into the ODE results in
11"» ﬁrst" x O. This ODE is linear in the variable w = u’. An integrating factor is p = 6532/2. Rewrite the equation as [ B‘ISm'J/q‘vf] = 07 from which it follows that gems/2 . 'U I (3:) Cl ...
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This note was uploaded on 01/10/2012 for the course MATH 2233 taught by Professor Binegar during the Fall '08 term at Oklahoma State.
 Fall '08
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