14.
hnposition of the initial conditions y (2) = 10, /(2) = 15 on the general solution
y (x) = (.', x: +(\X-J yield,; the t,vo equatioll.'i 4t; +c/8 = JO, 4c1 - 3c/ I6 = IS with
solution c, = 3, c, = - 16. Hence the desired pa,ticular solution is J(x) =
IS.
3xl -
16/J.
Imposition of the initial conditions y ( I)= 7, y'( I) = 2 on the general solution
J'(x) = t;X + C:-f lnx yield,; the two equations c, = 7, t; + c1 = 2 with solution
c, = 7, c, = -5. Hence the de.~ red pa1ticular solution i, y(x) = 1x - 5x lnx.
16.
Imposition of the initial conditions y ( I)= 2, y'(I) = 3 on the general solution
y(x) = c, cos(ln x) +c, sin(ln x) yield, the two equatio1» c, = 2, c, = 3. Hence the
desired pa,ticular solution is y(x) = 2 co.,(lnx) + 3 sin(lnx).
17.
If y=cl x then y' +y' = -cl x'+c' !x' = c(c- l)!x';, 0 w1lesseither c= O
or e= !.
18.
If y=cx' then yy' = cr' ·6cr = 6c'x' ;, 6x' w1less
19.
If y =J+ J; tl,en >:r'+(y')' = (J+ J ;)(-x"'"l 4)+(x· '"l 2)' = -x-'"1 4
20.
Li11early dependent, becau.se
c' = I.
;I,
0.
f{x) = 1r = n(co.,'x+sin'x) = 1rg(x)
21.
Linearlyindependent,because x3= + r l~ ifx > O, whereas x l =-.r jxl if x < O.
2-Z.
Linearly independent, because I + x = c(I +lxl) would require. that c = I with x = 0,
but c = 0 with x = · I. Thll'i there is no such 0011.,;tant c.
23.
Li11early independent, because /{x) = +g(x) if x > 0, wl1erea, f(x) = - g(x) if x < 0.
24.
Li11early dependent, becau.se g(x) = 2 /(x).
25.
/(x) = e' sin x and g(x) = e' co., x are li11early independent, becau.se f{x) = k g(x)
wouldi,npty that sinx = kcosx, whereas sinx and cos x are linearly independent.
26.
To see tll3t /(x) and g(x) are linearly independent, assume tll3t f{x) = c g(x), and tl,en
substitute both x = 0 and x = nf2.
27.
Let L[y] = y' + py + qy. T l,en L[y,] = 0 and L[y.J = J; so
L[y, +y.J = L[y,J + [>~] = 0 +f =
234
Chapter 5
f
28.
If .r(x) = I +c1 cos x+c1 sin x then y'(x) = ·c1 sin x+c1cos x, so the initial condition.~
J(O) = y'(O) = - I y ield c, = - 2, c, = - 1. Hence y = I - 2co., x - sinx.
29.
There i~ no conb·adiction becau.~ if the.given differential equation is divided ~Y J.1 to
get tl1e fonn in F.quation (8) in tl1e text, then tl1e resulting functions p(_x) = -4/x and
q(x) = 6/Jr are not continuous at X = 0.
30.
(a)
>': = x and >': = lx 1 are linearly independent becau.~ x = c lx 1 would
require. that c = I with x = I , but c = • I with x = · I .
3
3
3
3
(b)
The fact tl1at f"{y,,y,) = 0 everi.where doe.snot contradict Theorem 3, hecau.,e
when the given equation is written in the required fonn
:!' - (3/x))f + (3/.r)y
= 0,
the coefficient func tioll.~ p(x) = • 3/x and q(x) = 3h.2 are not continuou.~ at x = 0.
31.
fl'{y.,y2) = - 2x vanisl,es at x = 0, whereas if y, and y, were (linearly independent)
solutioll.~ of an equation :Y + J~Y + qy = 0 with p and q both continuou.~ on an open
interval I containing x = 0, then T heorem 3 would itnply that W ;t O on /.
32.
(a)
W = Y 1..V{ • y 1)1, so
, , +Y1Y1" · Yi".~ · Y1Y1
, ')
/ tfv
. ·-- / t(Y1Y1
= y,(Ay{) - y,(Ayt)
= y,( • By,'
• C:!'l) • y,( • Bvi' • Qt,)
= - B(Y,.l'l' • y ,'.Y,)
and thu., AW' = - BW:
(b)
Ju.st separate tl,e variable.,
(c)
Becau.~ the exponential factor is never zero.
In ProblenL~ 33-42 we give the c haracte,istic equation, it~ root~ and theco1Tesponding general
solution.
33.
r'- 3r +2 = O;
r= I, 2;
34.
r' +2r-lS
= 0;
35.
r' +Sr= O;
r
y(x) = eii! +c,l'
r= 3,-S;
= 0, -5;
J(x)
J(x)
= c.e·"
+c,l'
= c, + c,e·"
Section 5.1
235
2r: + 3r
37.
2r'-r-2=0;
39.
4r'+4r+ I = O;
40.
9r' - 12r+ 4 = 0,
r = -2/ 3, -2/ 3;
41.
6r'-7r-20 = O;
r= - 4 / 3,5/ 2;
y(x) = c, e·'·" +c,e'"'
42.
35r'-r-12 = O;
r= - 4 / 7,3/ 5;
y(x) = c,e·'"' +c,e""
=
0~
r
=
0, - 3/ 2;
.)(x)
= c1 + Cie ·lt!l
36.
r= l,- 1/ 2;
)(x)= c, e·,n +c,e'
r= - 1/ 2,-1 / 2;
y(x) = (c, +c,x)e·'n
y(x) = (c1 + c,x}l"'
In ProblenL~ 43-48 we first write and sin1plify the equation with the indicated characte,istic
root~. and then write the co1Tesponding differential equation.
43.
(r-OXr+IO)=l+IOr=O;
y'+IO/
=O
44.
(r-lO)(r+IO) = r'- 100 = O;
45.
(r+IO)(r+IO) = r'+20r+ IOO = O;
46.
(r-lO)(r-100) = r'- IIOr+ IOOO = 0,
47 .
(r-OXr-0) = r' = O;
48.
(r-l- -./2)(r-l + .fi) = r'-2r-l = O;
49.
lhe solution cmve witl, y (O) = I, / (0) =6 is y(x) = 8e·· - 7 e-'' . We find that
/ (x) = 0 when x = ln(7/4) so e-· = 4 / 7 and e-'· = 16/ 49 . It follows tl,at
y (ln(7 / 4)) = 16/7, so thehigh point on tl,ecwvei, (ln(7 / 4)),16/7) = (0.56,2 .29),
which looks consi~tent with Fig. 3.1 .6.
SO.
lhe two solution cu,ves witl1 y(O) = a and y(O) = b (as well as / (0) =I) are
y'-IOOy = O
y ' +20y' +100y = O
y'-110/ +IOOOy = O
y' = O
y' - 2/ - y = 0
y = (2a+l)e_, -(a+ l)e-'· ,
y = (2b+ J)e-· -(b+ l)e-''.
236
Chapter 5
I
7.
8.
W =
w=
X
x'
0 I 2x
0 0 2
=2
ii; nonzero everywhere.
e'
,i·
e'
2e'' 3e'' = 2 e"·
e'
4e.!•
e''
1s never zero.
91'
10.
W = x ·'e(x + 1Xx+ 4 ) is nonzero for x > 0.
II.
W =
12.
W = x ·' [2 co.s'(lnx) + 2 sin' (ln x)] = 2x·' is nonzero for x > 0.
x'l-' is nonzero if x,oO.
In each of Problems I3- 20 we first fonn the general solution
y(x) = c,y,(x) + <i.>'>(x) + cly,(x),
then calculate y'(x) and _y'(x), and finally impo.se tl,e given initial conditions to detennio,e tl,e
values of the coefficients c.. '1, CJ.
13.
hnpo.,ition of tl,e initial conditions y(O) =I, y'(O) =2, y'(O) =0 on tl,e ge11eral solution
y(x) = c.e· + c1e·· +cf!· 1• yields the three equatio11.o;
with solution c, = 4 / 3, c 1 = 0, cJ = - I / 3. Hence the desired patticular solution i,;
given by y(x) = (4 <" - e·")/3.
14.
lmpo.,ition of tl,e initial conditions y(O) =0, y'(O) =0, y'(O) =3 on tl,e ge11eral solution
..v(x) = c.e· + c1 i• +c 3e3' yield,; the three equatioll.o;
with solution c, = 3/ 2, c 1 = - 3, cJ = 3/ 2. Hel'K:e the desired patticular solution ii;
given by y(x) = (3e' - W-' + 3.?')12.
IS.
hnpo., ition oftl,e initial conditions y(O) =2, y'(O) =0, y' (O) =0 on tl,e ge11eral
solution y(x)
= c1, ( +c1xe· +c~i·
yield,; the three equatio11.,;
Section 5.2
239
with solution c, = 2, c1 = - 1. cJ = I. Heoce the desired pat1icular solution is given by
.l(x) = 2x - x· 2 + x~2ln x.
In each of Problems 2 I · 24 ,ve fi rst fo nn tl1e general solution
J(x) = y<(x) +.v.(x) = c.y,(x) + <).~X) + ;ix),
tl1en calculate y'(x). and finalty impose the given initial conditions to detennine the values of
tl1e coefficients c 1 and c1 .
21.
Imposition of the initial conditions y(O) = 2, y'(O) = - 2 on tl,e general solution
y( x) = c. cosx+c 1 sinx+3x yields tl1e twoequatio11.s c1 = 2. c1 +3 = -2 with
solution c, = 2, c 1 = - S. Hence the desired particular solution is given by
.,{x) = 2cosx - Ssinx + lt.
22.
Imposition of tl,e initial conditions y(O) = 0, y'(O) = 10 on tl,e general solution
y(x) = c.e1 ' +c1e· 1 • -3 yields tl1envoequatio11.s c1 +c: -3 = 0, 2c. -2c1 = 10 with
solution c1 = 4, c1 = - I. Hence tl1e desired parriculai· solution is given by
J(X) = 4/' - e·" - 3.
23.
Imposition of the initial conditions ,1f O) =3, y'(O) = 11 on the !,'<?Iler.ii solution
.r(x) = c.e· • +c 1i• -2 yields the twoequatiol\.s c, +c1 -2 = 3, - c. +3c1 = 11 with
solution c1 = I, c1 = 4. Hence the de.c;i.red particular solution is given by
)(X) =
e-" + 4eJ,: ... 2.
24.
Imposition of the initial conditions y(O) = 4, y '(O) = 8 on the general solution
y( x) = c.e' cosx +c1e~ cosx+ x + I yields tl1e twoequations G +1 = 4, c, +c 1 + 1= 8
with solution c, = 3, c1 = 4. Hence the desired particular solution is given by
J(x) = ef(3co., x+ 4sin x) +x+ I.
25.
L[v] = L[y, + y,] = L[v, J + L[I'> ] = J + g
26.
(a)
27 .
The equations
(h)
Yi = 2 and Y1 = 3x
c, +c,x+c,I = 0,
V
.,
=
V
..i
c.,+ 2c.,x+ O,
Section 5.2
+.,.V,-
= 2 + 3x
2c, = 0
241
SECTION 5.3
HOMOGENEOUS EQUATIONS WITH CONSTANT COEFFICIENTS
This ii; a pw·efy computational section devoted to the single most widely applicable type of
higher order differential equations - linear ones with con.stant coefficients. In ProblenLs 1- 20,
we write first the characteiistic equation and it,; list of roots, then the corresponding general
solution of the given differential e.q uation. Explanato,y conunent,; are included only when the
solution of the characteristic equation is not routine.
I.
?- 4 = (r- 2)(r+2) = O;
2.
2r'-3r = r(2r- 3) = O;
3.
? +3r- JO = (r+5)(r-2} = O; r =-5, 2;
4.
2r'-7r +3 = (2r- l)(r- 3) = 0;
5.
r1
6.
r +5r +5 = O;
+6r+9 = (r+3t = O;
r = -2,2;
r = 0,3/ 2;
y(x)= c,e"+c,e·'-'
y(x)= c,+c,e'"'
y(x)= c,e"+c,e·''
r = 1/ 2, 3;
r =-3,- 3;
y(x)= eien+c,il'
y (x)=
c,e-.h+c:¥"£>.u
r =(-5±./5)12
J(x) = , -,,n[c, e., p(x,/sl2)+ c2exp(- x,/s12)]
8.
?- 6r +l3 = O;
r = (6±../=iG)t2 = 3±2i;
9.
r +8r+25 = O;
r = (-&±.J-36)12 =-4 ±3i;
10.
5r'+3r' = r'(5r +3) = O;
11.
r' -Sr + 16r ·' = r·" (r-4)' = O;
12.
r' -3r' +3r'-r
13.
9r +12r·' +4r= r ( Jr+2 )'• = O;
r = 0, 0, 0,- 3/ 5;
'
= r(r-t)' = O;
'
r
r
.>'(x)= e·"(c, co.,3x+c2sin3x)
y(x) = c, +c,x+~+c,e·""
= 0,0, 4, 4;
= 0, I, I, I;
r =O. - 2 / 3. - 2 / 3
J{x) = ea+ 02.e·2t' 3 + C)Xe· 1»3
246
.>'(x)= <"'( c, co.,2x+c,sin 2x)
Chapter 5
14.
i'+3r:-4 = (r 1 -l}(r:+4) = O:
r =-l,l,±2i
.,{x) = c1e" + ~e·x + c1co.,; 2r + c., sin 2x
IS.
4r'-sr'+l6 = (r'-4)' = (r-2 )'(r+2)' = O;
.,{x) = c1c?'- + ci){eh + c1 e·1,, + c"xe·1.,;
16.
/+Jsi+81=(r:+9f=o;
r = 2,2. - 2,-2
r=±3i,±3i
)\X) = (Ci + ()X)co.s l\" + (c; + C,X~Sill 3x
17.
6r' + I Ir' +4 = (2r + 1){3r +4) = O; r = ±il./2,±2il../J,
Jix) = c1co.,(xl ./i )+c, si~xl ./i )+c,cos(2x! ./3 ) + c,sinl,2x/.,/3 )
18.
r' -16 = (r'-4)(r'+4) = O; r =-2, 2. ± 21
J(X) = {).e!i. + cie· 1" + C)COS 2x + C4Sin 2x
19.
r' +r'-r- 1 = r(r'- l)+(r'- 1) = (r- l)(r+ l)' = O;
J{X) = Cle" + cie~x+ l'j..\tt""'
20.
r'+2?+3?+2r+ I = (?+r+ l)' = O;
r = 1,- 1,- 1;
(- 1±../3 1)1 2, (- 1± ./3 1)12
y = e·'a(c1 +()X)co.s(.x./3!2)+e·'"(c, +c,x)sir(x./312)
21.
Imposition of the initial conditions y(O) = 7, y'(O) = 11 011 tl,e general solution
y(x} = ,·,e-' + t·:i' yields the two equations c1 + c1 = 7, c1 + X'2 = I I with solution
c, = 5, c, = 2. Hence tl1e desired patticular solution is J(x) = 5d + 2J' .
22.
h1110.sition of the initial conditions y(O) = 3, /(0) = 4
y(x) = e- ·a [ c, cos( x t.{3') + c2 sin( x I
011
the general solution
J3)J yield,; the two equation.,;
c, = 3, -c, / 3+ c2 t../3 = 4 with solution c, = 3, c 2 =sJi Hence the desired pa,ticular
solution is y(x) = .,-·" [3co.,(xt ../J)+s./3sin(xt./3)].
23.
lmpo.,ition of the initial conditions y(O) = 3, /(0) = I on the general solution
y(x) = e''(c, co.,4x+c,sin4x) yiekls tl1e twoequario1» c, = 3, 3c,+4c,= I witl1
solution c, = 3, c 1 =- 2. Hence ~1e desired particul..u· solution is
J(x) = e"(3 co., 4x - 2 sin 4x).
Section 5.3
247
14.
i'+3r:-4 = (r 1 -l}(r:+4) = O:
r =-l,l,±2i
.,{x) = c1e" + ~e·x + c1co.,; 2r + c., sin 2x
IS.
4r'-sr'+l6 = (r'-4)' = (r-2 )'(r+2)' = O;
.,{x) = c1c?'- + ci){eh + c1 e·1,, + c"xe·1.,;
16.
/+Jsi+81=(r:+9f=o;
r = 2,2. - 2,-2
r=±3i,±3i
)\X) = (Ci + ()X)co.s l\" + (c; + C,X~Sill 3x
17.
6r' + I Ir' +4 = (2r + 1){3r +4) = O; r = ±il./2,±2il../J,
Jix) = c1co.,(xl ./i )+c, si~xl ./i )+c,cos(2x! ./3 ) + c,sinl,2x/.,/3 )
18.
r' -16 = (r'-4)(r'+4) = O; r =-2, 2. ± 21
J(X) = {).e!i. + cie· 1" + C)COS 2x + C4Sin 2x
19.
r' +r'-r- 1 = r(r'- l)+(r'- 1) = (r- l)(r+ l)' = O;
J{X) = Cle" + cie~x+ l'j..\tt""'
20.
r'+2?+3?+2r+ I = (?+r+ l)' = O;
r = 1,- 1,- 1;
(- 1±../3 1)1 2, (- 1± ./3 1)12
y = e·'a(c1 +()X)co.s(.x./3!2)+e·'"(c, +c,x)sir(x./312)
21.
Imposition of the initial conditions y(O) = 7, y'(O) = 11 011 tl,e general solution
y(x} = ,·,e-' + t·:i' yields the two equations c1 + c1 = 7, c1 + X'2 = I I with solution
c, = 5, c, = 2. Hence tl1e desired patticular solution is J(x) = 5d + 2J' .
22.
h1110.sition of the initial conditions y(O) = 3, /(0) = 4
y(x) = e- ·a [ c, cos( x t.{3') + c2 sin( x I
011
the general solution
J3)J yield,; the two equation.,;
c, = 3, -c, / 3+ c2 t../3 = 4 with solution c, = 3, c 2 =sJi Hence the desired pa,ticular
solution is y(x) = .,-·" [3co.,(xt ../J)+s./3sin(xt./3)].
23.
lmpo.,ition of the initial conditions y(O) = 3, /(0) = I on the general solution
y(x) = e''(c, co.,4x+c,sin4x) yiekls tl1e twoequario1» c, = 3, 3c,+4c,= I witl1
solution c, = 3, c 1 =- 2. Hence ~1e desired particul..u· solution is
J(x) = e"(3 co., 4x - 2 sin 4x).
Section 5.3
247
24.
Imposition of the initial conditions y(O) = I, /(0) = -1. >•' (O) = 3 on the!."''"''"'
solution y(.t') = t.·1 + C1e 1' + C / w~t: yields ~le three equation..,;
with solution c, =- 7 / 2, c 1 = I/ 2, c3 = 4. Hence the desired patticular solution is
)(X) = ( • 7 + / ' + g.,·'·')12.
25.
Imposition oftl,e initial conditions y (O) = -1, /(0) =O, y '(O) = I on tl,e general
solution y(x) = '-', + <\X + C/F:~,J yields the three equations
with solution c, =- 13/ 4, <; =3/ 2, c3 =9/ 4. Hence the desire.d particular solution is
)\X) = (- 13 + fu- + 9e·"'" )f4.
26.
ln1,o.sition of the initial conditions y (O) = I, y'(O) = -1. y '(O) = 3 on tl1e general
.
· ,u.<;
0 · he l
.
soIuuon
y(x) = ,·, + t·:e -5 , +,•,xe...S, y1e
t t 1ree equatK>ns
with solution c, = 24/ 5, c: =- 9/ 5, c, =-5. Hence thedesired pa,t icular solurion ii;
)(X) = (24 - 9e·" - 25xe·'')l5.
27.
First we spot the root r = I. T hen long division of the polynomial l + 3r 1 -4
by r - I yields tl,e quadratic fac tor r' +4r+ 4 = (r + 2)' with roots
r = - 2, - 2. Hence the general solution is J{x) = c1e..( + 0ie· 2 " + (.)Xe~2 .t.
28.
Firs t we spot the root r = 2. T l,en long division of the polynomial 2l • ,:i - Sr - 2
by tl,e fac tor r - 2 yields tl,e quadratic fac tor 2? + 3r + I = (2r + I )(r + I) with roots
r = -
29.
I. - 112. Hence the general solution is y(x) = c 1eu + ci.e~"· + CJe ·v 2.
First ,vespot the root r = - 3. Then long division of the JX)lynomial r l + 27 by
r+3 yietds thequadraticfactor r 1 - 3r+9 with roots r=3( 1±i.J3)t2. Hence the
general sohrtion LS )(<) = c,e·" + e' ' " [c,cos(3x,h/ 2) + c, sin(lr ,h12)].
30.
First ,ve spotthe root r = • I. Then long division of the polynomial
l
~
....
rJ - r+,,-.
- :,r
- 6
248
Chapter 5