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Unformatted text preview: a1 equations are introduced and
These key concepts include
solutions (Theorem 2),
lutions (Theorem 4). This
treatment of nth order linear
ewhat abstract to .diﬂ‘erenﬁ
ns. Sfy the given differential equaiirm yr ﬁnd 1’1 “ﬂ . . .
rial conditions on the general _ )m the given ini ya” =0, y'(0) = 5 on the general solution
c, = 0 with solution hmequations cl+cz=0, 6‘1"
= 5(9" — «(H/2. the desired particular solution is ﬁx) ‘o‘ns y(_0) = — l, y'(0) = 15 on the general solution
‘  twoequations c, +cz = —1, 3c, — 3c2 = 15 with solution
particular solution is y(x) = 2e” — 39'3". " y(0) = 3, y'(0} = 3 on the general solution
ﬂietwo equations c, = 3, ?.c2 '= 8 with solution
, solution is y(x} = 3 cos 2x + 4 sin 2x. =19. yl'(0) = —10 on the general solution
ensc, =10, 5cI =lD with solution
Q9 l5 ﬁx) = 10 cos 5;: — 2 sin 51:. It“? ytx .
6154’ c, .. 3. Hence the desiredm Imposition of the initial conditions 3&0 10. ll. 12. 13. 05mm, of the initial current )I(x) ﬁ C'I. Imp Impositi
y”) 2 ﬁg" + clx e" yields the two equations of C"— ~ imposition of the initial conditions y(t)) = 3, 3:10 ) :: Clea: + £2843 yields that“r5 + cze" yields the two equations
: 6, 52 = — 8. Hence the desired puﬁ'culu_ osition of the initial conditions ﬂit) =4 fit! on of the initial conditions y(0) =2I 3,! ) = ‘r 2 6'2 : 1. Hence the desired particular solutto ytx) : 5,9" +ch e” yieldsthc two equations _c c, = 3‘ 92 : ﬂ 2. Hence the desired particular solution Imposition of the initial conditions y(0) =0, y'({)) =, ‘ _
cle‘ cos x + cle‘ sin 3: yields the two equations o‘ y( x) z
: 5. Hence the desired particular solutionis ' cl = 0. ('3 imposition of the initial conditions yUJ) =2, 3110);;
ytx) = cle'“ cos 2x + cze'“ 5'1an yields thetwo cg, solution c, = 2, r:2 = 3. Hence the desired? eul e'3xt2 cos 2): + 3 sin'lx).
tial conditions y(1)=3;_ ﬁg. "
yix) = up: + c,):2 yields the two equations c,
I . .. "1"
c2 = a 2. Hence the des'tredp . Imposition of the ini c, =5, will), 40,—3c1I16=15 With ﬂsoluﬁunis XI) = 3x: — 16!;3_ :2 on the gonna! solution
5‘ a: 1, c. +r:2 = 2 with solutiOn )= Tatlenx. cumsolution is 10‘ encrul solution #2. 1.10:3 ontheg
2. r:1 =3. Hence the dsmtwuquaﬁms ”I =
2M:) + 3 Sillﬂn I). “,2”: = 46.1)”: ¢ 0 unless either c:[) = n+JIx—x'mr4)+(x"”/2J’ = «"214 s 0. . x~= dmzx+sin2ﬂ = ﬁght) J=+lexl ifx>0, whereas xJ=fx1lxl if x<0, ;.l+x=c(l+x) wouldrequirethat c = l with x = O
I‘tttereisao such constant c. ‘ “Ninenrlly independent,hecause ﬁx) = kg{x)
 — SIM: and can atelinearly independent. '5 no contradiction . l
'19 There form in Equation (8) Ix: are not continuous if x; ‘ The fact that Ill/(yum) = 0 everywher d b) . . .
( the given equation ts written in the 1.equ "'i when r' — new +919), the coefﬁcient functions p(x) = 3”: and 9(3) = 3 WU’LM) 5 ‘2I vanishes at x = 0, whereas if yt cudi: 31. .
solutions of an equation y" + p); + qy = 0 with p an
interval I containing 3: = 0’ then Thwrem3wouldIgmp 32. (a) W : J’U/l' *Jh'yz, so A W' = Abﬁ'yz' + you"  m:  n'yz'l)”
: yt(Ay2")  MAM”
= yttByz'  Cyz) yz(Byt' .'
= eBer' yl'y2) and thus AW' = —BW. (b) Just separate the variables.
(9) Because the exponential factor is never? in Problems 3342 we give the characteristic equatio '
solution. 33. r2 —3r + 2 = 0;
34. 3S. [illitna‘ln .3. I it“. aqunﬁﬂﬂ with the indicated characteristic equation. ‘2’" = a; yI_2yl_y = 0 ﬂm=6 is HI) = Bed78'“. We ﬁnd that
. =4]? and e‘z‘=16l49. It follows that
intonthecurveis (lﬂl7/4))a16’7) = (055.229) subtraction and then di . __
NOW substitution in. eitherf r.
H“ 4) ... (a) The substitution 1.! = m . . d2
y =—‘:1=d
‘ =_L.dl
xidu Substitution of these expressions for y and ' _ . _ .
immediately the desired Eq, (22}; ' . _ " "‘"‘' " (b) If the roots 1'" and r1 of the chemo
distinct, then a general solution of the origins y(x) _. C,Er"+cle = 52, The substitution v = in 1: yields the convenedequal
equation r1 l =0 has roots ’1 =1' chaIactetistic
corresponding general solution is 53. The substitution v: whose chaxaeteristie equation r' + r—12=‘D e" : x,thc corresponding general solutionzip. y ____ c124r+czg1r 54 The substitution v=lnx yields the? I whose characteristic equation 41? Because a“ = x, the cone111% scction to see that, in the case, ' .. riding theorem in Section 5.1. Similarly, the computational
_ parallel those for the previous section. By the end of Section
u . that, although we do not prove the existenceuniqueness theorem Ihr everything we do with linear differential equations, listed in Problems 1—6 were discovered ' .of'rl'heorems I through 4 in this 'by inspection" — that is, by l. W 12 In each n then cal
values 0 13. 14. 3211 = x is nonzero if x5e W = {In cosztln x) + 2 sinlﬂn f Problems 13—20 we ﬁrst form the culate y'th) and fix}. and ﬁnally 1mm f the coefﬁcients a], (:2, c3. unposition ofthc initial conditions yﬂ) ﬁx) = cle’ + cze" +c,e"‘ yields the
c, + c2 +c3 =1, c, —c,—2c,=2. with solution C. =4l3, c:2 =0, c,=ll3 given by you) = (441’:  e'kﬂl Imposition of the i
yt x) = o‘e‘ + ole“ +cle c1+c1+c3=L with solution c1 = 312, c2 = 
given by y(x) = as" — 62" +3; imposition of the initial co solution y(x) = Clei'l'ﬂi 7‘ ”unlit.[mriiimpu =2: nitial conditions y(0 '_ I
3* yields the e1 +‘2.ez +‘3cé ' Somalia... ._a.., . "I"» ‘
''.' i".!li‘_l‘\_ ‘ ._“ ‘lll
.. ”Hi I i , :30— Hmu the desiléd panicular SOIulionis (Uh: 71.!"Ié‘f Mr) = Clcﬂsx+czsinx+3x .. .i
 .. 'Ilplimmk h
solution 61:2! c1=_5_ HE J . .
' Elillil‘ll‘ﬂu‘l y(x} = Ecosx  Ssinx+3x_ 2mm H.: Imposition of the initial canditio”_
you = we“ +c2e'“ 3 m ‘
solution cl .= 4, c: =_ 1. Hence .. '22. . _ c =—]l3. Hence the desired panicular 301““ is
~ I '31:!9.
I. x  sm ) .vix) : 4e” — 2413. . _ 0 =1. 1,19): 0, y'(0) : 0 on the genera] SOIut‘  ‘ I
ans ﬂ ) _ ion , _ lnipoSition of the initial conditions y mph”‘,mi.'l..__‘.r_‘” . J _. 1. .
_. g}: _2 yieldslhe' _ Err”H: , 1;: ., . _ '==—.1 __1_ Hencethe desired panicular solutionis iven . .. . . ‘  .
3:51_ 5: g . imposition of the initial conditions 3(1) , f .,._lI1=...,,W‘.I;l _ r + ix : ce'cosxlc e‘cosx+x+l yiel
J‘. I 1 with solution c,=—3,c2=4.l1encctl_1_ conditions y(l)=6, y'(l) = 14, y'(l) = 22 on the general
y{__x) — E. (3 cosx+4sinx)+x+l. 5x1 +i';,:n:3 yields 1116 three equations :6, t:.+2¢:1+3r:3 =14, 2t:2 +661: ‘2
I . . .1 = = 3::
=3. Hence [he desned particular solution is given by (a) "H 2 and Y2 The equations .J'(1)= 1. JP'U) =5. y'll) = — 11 on the general Cl + czx + 6311 =
yields the ihrec equations diam”! ..., .3". Upon substitution of
=. 0, so it follows that all them: 2v 5 0 is rewritten in standard form ,9 = .2]; and pix) = 2b:2 arenot continuous at x =
'a 3 are not satisfied. ﬁn—lhediﬁttrcntial equation gives y'(a) = — p y'(a) _ q(a}. ‘." D, then theequatinn y“ — 2y' — 5y = 0 implies that y, = Vy: +
in the differential actuation 7.
ivy? + 21:5,; + 9": ,
”[yf "' Dy: + 9'. But the terms within brackets var“ equation
W" + (2y: +' that We can solve by writing V. y'
—r = — 2—1. _ p :7:
‘’ y1 ' ', V'ix) = SlatI'm)” :5}
M H _. With C :1 and K=0 this givestheeeco ] (xi
yr“) = Fitﬁle p 37. When we substitute y = this in the given (1'
separable equation xv” + v' = 0 that we sole I
V
_7 => IHV' = I
V r V b i 43.
rentiﬂl equation and simplify, We 1
anew] solution v(x) = Ax + B_ mime aim M“
on y' = 0 With g le)=.r and hence yzl") = ”m' _ in ﬂagVB“ diﬂ'emntial equation and simplify, We get [he 4': 0 that we solve by ”ﬁling a [1111' = x+lnAv 3 v(x) = Ae‘+B. 5):“? in the giVen differential equation and simplify, we get the —l+——!— => lnv’=—x+ln(l+x)+lnA,
.4”: a yo) = Aj(1+x)e'*dx = —A(2+x)e" +3. '(2+x)e" and hence y2(x) = 2 + x. 4x)v'+xv’= 0 thatwe solve by uniting 44. When we substitute y = v x 'lll'tlhe
separable equation x (x2 1” v
ln‘?’ =
x' 20”) “it
. _ 1 l . 1
12(1) — A[ x+§ln(1+x)__ilnu"x)f : With A =—l and 3:0 we get . l 1 l
=—"—l 1 _ _ I’ .
v(xj x 2 n( +x)+2ln(1 x) z: VI
When we substitute y = vx"”eosx in the give '
eventually get the separable equation (eosx) v" 25inx v 6051
v' = Aseczx =>  = COEFFICIENTS i
I widely applicable type of 1 '5‘ 4r‘ _Br +16 2:: (r' ._ tooeﬂicients. 1n Fromm“5 1520 i yo!) g (.122: + 82x92“? corresponding general  'cludcd only when the I 16 r41+13r +81 = (r‘ +9)__
1 31(1) : (c. 4 warms 3x+("3+04£
i 11 Gr' +llr2 +4 = (2r2+l)(3r'+4).
i ytx) .— c.cos(xlﬁ)+c;stn(xif)+¢:
 '—5 2‘. Hr): cii¢,'z‘+.r:;ge'sI 13 rd “[6 = (r1—4)(r1+4) = 03 i"
z: 4: : ytI) = €19 +C1‘3 +93cva52x+ '  m + 62931 cistnﬂxlh‘
. 19. r1+r1~r—1 = r(r=i)+(r1_i) = (ii J’(x)= cie'3‘+ cm"3x ym = Cl?! + 029: + core"
20. r“+2r3+3ir1+2r+1 =(r‘+r+1f=o (on. y : Edam + Cmcoso‘ﬁmJ'e'ﬂtcﬁcixhinbo  no 21, Imposition of the initial conditions y(0) = 1, 3m] yixi= or” + 6219’
ﬁx) : cle‘ +c1e3‘ yields the two equations cwogat‘fiﬁ
c] : 5. c1 : 2. Hence the desired particular solution‘istiyt: (5'1‘L15Jn = 3:2:; y(.r)= e"(cicos 2x + Cgsin 2x)
22, Imposition of the initial conditions y(0)= 3, y'(0)' ' H36 12 =—4i3i; ytx)= 24%: cos 3x+  i
i ) ' (15'1”!) y(x) : e'" ’[c1cos(le3)+czsin(xiJ3)] yiei’gi’ thati
c :3 —c!3+c2lJ3= —4withsolution cl: =3, e 3 5' 1' =0.0.0.—3f5; y(x) = c. + ng+c3x2 + means
soiutionis y(x) = e"”[3003(le3)+5J3;5m : = U,D,4,4; J’U‘) : Cl + 62x + 6384‘ + axe“
23. Imposition ofthe initial conditions nor ‘ y(x)—  e" (c,cos4x+c sin4x) yields, solution cl =3, c =—2. Kenneth .ileé HI) = e3’(3 cos 4::  2 sin 41:). i \ J'Uf) = Cl + (:22; + c3xe‘ + c4xze" out; 4. Hence the desired particular 501mm“ is vision of the polynomial r3 + 3r2 — 4
= {r+2)I with roots
21* e‘ + ege'l" + cyce‘ . 'l‘henlongdi
_ *ﬁtotor r1+4r+4
solution is y(x] = or mial 2r3 — J"2 — 5r — 2 (2" + ”(f + 1) With mots
nl'z Thou long division of the polyno
tiodmio factor Zrz + 3r + l =
Solution is y(x) = clez' + ale" + (:39 mnlongdivision ofthepolynomial :—3 +27 by
3r+9 withroots r=3(tisJ§)/2. Hence the . [mosﬂxﬁlb + (:3 sin(3xJ§ m]. The characteristic _ .
division then yields th
complex ooniugato roots The characteristic equation r‘ + r:1 _ ii
found by trial and error, and long diViaron
i lair.=‘.tltl_w._ 33.  7 M] . .
Knowmg lhﬁt Y  8 15 one solutron,wedivideme 3,.3 _ 2,3 + 12r  8 by 3r — 2 and gettho quadraticfacto solution is 34. _ m3 '
Pix) — 018 + cacos 2x + Cgsin 9x _ 35. The fact that y = c0523: is one solution tells usthat 3+ a. characteristic polynomial 6r" + s? + 25r2 + 20': + 4. Then long division yields the quadratic factor or1 + 5? ' r = — 1/2. # l I 3. Hence the general solution is
yet) = cle‘m + (an!Jug + canon»
one solution tells 'usr f 36. The fact that y = e"sinx is
polynomial A 7 is a factor of the characteristic L‘J '“Jm'r'iil i The characteristic tools at: r Cie‘m‘ﬁ" +cze(InJij ‘ )‘(X) = 5] into the characteristic
am ,. 2+1oo_ Thus the . The general solution is 1:02"
p i (—1 iiJEVZ. Impusiﬁ' .‘ I "x l'!il\..\ ....'w510x+C sin [0x . ftions yields the equations 521+:10C = 10, 2544—1003 = 250 C=0 Hence the desired particular solution is that we solve for A211: given realvalued solution. L 28, some. differential equation is _
49. The generai solution is y = A 1'
initial conditions yields theeq ' fvﬁy’HZJf—By =
A+B+C 2P+4r8, so Ihe differential equation is '
244 B
4A+ B 4? 811—3 that we solve for A = 2, B= 50. if .t > 0 then the differential
y— = A coax + Bsinxu Butif il‘iittir‘i'i‘iiiii‘. ...
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 Spring '08
 Toland
 Equations

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