TWK2A
Variation of parameters (Section 4.6)
Solutions
1.
y 00 + y = sin x
Homogeneous equation:
y 00 + y = 0 ) m2 + 1 = 0
) m1 = i;
m2 =
i
So yc = c1 cos(x) + c2 sin(x)
For particular solution:
cos(x) sin(x)
sin(x) cos(x)
W =
W1 =
0
sin(x)
sin(x) cos(x)
W2 =
= cos2 (x) + sin2 (x) = 1
sin2 (x)
=
cos(x)
0
sin(x) sin(x)
= cos(x) sin(x)
So
u01 =
sin2 (x) ) u1 =
x
2
u02 = cos(x) sin(x) ) u2 =
1
1
sin(2x) = sin(x) cos(x)
4
2
1
x
2
1
cos2 (x)
2
Hence
y p = u1 y 1 + u 2 y 2 =
=
1
sin(x) cos2 (x)
2
1
x cos(x)
2
1
x cos(x)
2
So
y(x) = yc + yp = c1 cos(x) + c2 sin(x)
(valid for
1
cos2 (x) sin(x)
2
1 < x < 1).
1
x cos(x)
2
2.
x
x
y
e2
e 2
x
+
y = cosh
=
2
2
2
00
Homogeneous equation:
y 00
y = 0 ) m2 1 = 0
) m= 1
) yc = c1 ex + c2 e
x
Particular solution:
W =
ex e x
ex
e
W1 =
0
cosh
W2 =
=
x
e
x
2
x
e
ex
0
ex cosh
x
2
2
=
x
=
x
e
3x
2
x
2
e +e
2
x
2
=
e
x
2
e
2
2
x
2
e
e
+
2
2
So
x
3
e2
e2
W1
=
=
+
W
4
4
x
3x
e2
e 2
) u1 =
2
6
3x
x
W2
e2
e2
u02 =
=
W
4
4
3x
x
e2
e2
) u2 =
6
2
u01
Thus
y(x) = yc + yp = yc + u1 y1 + u2 y2
x
x
(valid for all x).
= c1 e + c2 e
x
= c1 ex + c2 e
x
e2
2
x
2e 2
3
e
x
2
6
x
2e 2
3
x
e2
6
3x
2
e
x
2
2
3.
x
4y 00
y = xe 2
y 0 (0) = 0
y(0) = 1;
Homogeneous case:
4y 00
y = 0 ) 4m2
1=0
1
2
) m=
So
x
yc = c1 e 2 + c2 e
x
2
:
Particular solution:
x
x
W =
W1 =
e2
e2
1 x2
1
e
e
2
2
0
e
x
xe 2
4
W2 =
e
x
2
1
e
2
x
2
1
e
2
0
x
2
x
xe 2
4
=
1
2
=
x
4
x
2
x
2
=
1
=
2
1
xex
4
Hence,
y(x) = yc + yp
= y c + u 1 y 1 + u2 y 2
x
x (
x
x2 x
xex + ex )
= c1 e 2 + c2 e 2 + e 2 + e 2
8
4
x
x
x
x
x
x2 e 2
xe 2
e2
+
= c1 e 2 + c2 e 2 +
8
4
4
Now,
y(0) = c1 + c2 +
y 0 (0) =
c1
2
c2
2
1
=1
4
1
=0
8
So
3
4
1
=
4
1
1
=
; c2 =
2
4
c1 + c2 =
c1
c2
) c1
Thus,
x
x
x
x
e2
e2
x2 e 2
+
+
y(x) =
2
4
8
x
x
x
3e 2
e2
x2 e 2
+
+
=
4
4
8
4.
1
4
x2 y 00 + xy 0 + x2
Standard form:
W =
=
= x
1
2
x
cos x
1 23
x cos x
2
1
x
2
1
1
4
x2
1
y 00 + y 0 +
x
x2
2
x
x
1
2
sin x
cos x sin x + x
1
cos2 x + sin2 x =
1
2
x
xe 2
e2
+
4
4
x
xe 2
4
x
y = x2
1
2
y=x
sin x
3
2 sin x + x
1
x
2
1
x
2
cos2 x
1
2
cos x
2
cos x sin x
1
sin x
1
x
1
W1 =
W2 =
0 x 2 sin x
1
1 23
x2
x sin x + x
2
x
1
2
cos x
3
2 cos x
1
x
2
x
1
2
1
2
cos x
0
1
sin x x 2
=
=x
x
1
cos x
Hence,
W1
x 1 sin x
=
= sin x ) u1 = cos(x)
W
x 1
W2
x 1 cos x
=
=
= cos x ) u2 = sin x
W
x 1
u01 =
u02
x
1
sin2 x
So
y(x) = yc + yp = yc + u1 y1 + u2 y2
1
1
1
= c1 x 2 cos x + c2 x 2 sin x + x 2 cos2 x + x
1
1
1
= c1 x 2 cos x + c2 x 2 sin x + x 2
5.
y 00
2y 0 + y = 4x2
3+
1
2
sin2 x
ex
x
Homogeneous case:
y 00 2y 0 + y = 0
) m2 2m + 1 = 0
) m1 = 1; m2 = 1 (repeated real roots)
So
yc = c1 ex + c2 xex :
Now, we have
Ly = y 00
2y 0 + y = f (x) + g(x)
x
where f (x) = 4x2 3 and g(x) = ex :
Note that, if yp1 is a particular solution of Ly = f (x) and yp2 is a particular
solution of Ly = g(x), then yp1 + yp2 is a particular solution of Ly = f (x) +
g(x).
We use method of undetermined coe¢ cients to solve
y 00
2y 0 + y = 4x2
3
and variation of parameters to solve
y 00
to get particular solutions.
For
y 00
2y 0 + y =
ex
x
2y 0 + y = 4x2
3
assume the form for the solution is Ax2 + Bx + c:
So y 00 = 2A; y 0 = 2Ax + B and y = Ax2 + Bx + C which gives
(2A) 2(2Ax + B) + (Ax2 + Bx + C) = 4x2 3
) Ax2 + (B 4A)x + (2A + C 2B) = 4x2 3
) A = 4; B = 16; C = 21
Hence, yp1 = 4x2 + 16x + 21:
For
ex
y
2y + y =
x
ex xex
W =
= xe2x + e2x
ex xex + ex
00
W1 =
0
0
ex
x
x
u02
e2x
e
ex
W1
=
= 1 ) u1 = x
W
W2
e2x
1
=
= 2x =
) u2 = ln jxj
W
e x
x
W2 =
u01
xex
=
xex + ex
e2x
0
x
=
e
x
x
xe2x = e2x
So
yp2 = u1 y1 + u2 y2
=
xex + xex ln jxj
General solution:
y(x) = yc + yp1 + yp2
= c1 ax + c2 xex + 4x2 + 16x + 21
xex + xex ln jxj:
6.
y 00 + y = sec x tan x
The complementary function is found by solving the auxiliary equation
m2 + 1 = 0:
(1)
The roots are
i, and hence
(2)
yc = c1 cos x + c2 sin x:
We now assume a particular solution of the form
(3)
yp = u1 (x) cos x + u2 (x) sin x:
Then follows that
yp0 = u01 cos x + u02 sin x
u1 sin x + u2 cos x:
If we make the assumption that
u01 cos x + u02 sin x = 0;
(4)
the …rst derivative simpli…es to
yp0 =
u1 sin x + u2 cos x:
From further di¤erentiation we have
yp00 =
u01 sin x + u02 cos x
u1 cos x
u2 sin x:
(5)
We substitute (3) and (5) in (1):
yp00 + yp0 =
u01 sin x + u02 cos x = sec x tan x
(6)
We now solve (4) and (6) for u1 and u2 . From (4) we have
u01 =
u02 tan x:
(7)
Substitute (7) in (6):
cos x :
u02 sin x tan x + u02 cos x = sec x tan x
(sin2 x + cos2 x)u02 = tan x
u02 = tan x =
sin x
:
cos x
(8)
Integrate w.r.t. x:
u2 =
ln j cos xj:
(9)
Substitute (8) in (7):
u01 =
tan2 x = 1
sec2 x:
Integrate w.r.t. x:
u1 = x
(10)
tan x
Substituting (9) and (10) in (3), we have the particular solution
yp = (x tan x) cos x sin x ln j cos xj = x cos x sin x sin x ln j cos xj: (11)
The general solution follows from (2) and (11):
sin x sin x ln j cos xj
sin x ln j cos xj:
y = c1 cos x + c2 sin x + x cos x
= c1 cos x + c3 sin x + x cos x
7.
9x
:
(12)
e3x
The complementary function is found by solving the auxiliary equation
y 00
9y =
m2
which has roots
9 = 0:
3, and so
yc = c1 e3x + c2 e
3x
(13)
We assume a particular solution of the form
yp = u1 (x)e3x + u2 (x)e
3x
(14)
:
Hence,
yp0 = u01 e3x + u02 e
3x
+ 3u1 e3x
3u2 e
3x
:
Furthermore, we assume
u01 e3x + u02 e
3x
= 0;
so that
yp0 = 3u1 e3x
3u2 e
3x
(15)
and
yp00 = 3u01 e3x
3u02 e
3x
+ 9u1 e3x + 9u2 e
3x
(16)
:
Substitute (14) and (16) into (12):
yp00
9yp0 = 3u01 e3x
)
3u02 e
u01 e3x
u02 e
3x
9x
e3x
= 3xe 3x :
3x
=
(17)
We solve (15) and (17) for the ui . From (15 we have
u02 =
u01 e6x :
(18)
Substitute (18) into (17):
u01 e3x + u01 e3x = 3xe
3x
3
) u01 = xe
2
Integrate w.r.t. x using integration by parts:
Z
3
u1 =
xe 6x dx
2
Z
3
x 6x
1
=
e
e
2
6
( 6)
3
x 6x
1 6x
=
e
e
:
2
6
36
6x
6x
(19)
dx
Hence,
u1 =
1
(6x + 1)e
24
6x
(20)
:
Substitute (19) into (18):
3
x:
2
u02 =
Integrate w.r.t. x:
3 2
x:
4
We substitute (20) and (21 into (14) to …nd a particular solution:
(21)
u2 =
yp =
1
(6x + 1)e
24
6x 3x
e
3 2
xe
4
3x
=
1
(18x2 + 6x + 1)e
24
3x
:
(22)
The general solution follows from (13) and (22):
y = yc + yp
= c1 e3x + c2 e
3x
= c1 e3x + c3 e
3x
1
(18x2 + 6x + 1)e
24
1
(3x + 1)xe 3x :
4
3x
8.
2y 00 + y 0
y =x+1
(23)
y(0) = 1; y 0 (0) = 0:
(24)
with the initial conditions
The complementary function follows from the auxiliary equation
2m2 + m
1 = 0 ) (2m
The roots are m1 = 21 ; m2 =
1)(m + 1) = 0
1 and we have
1
yc = c1 e 2 x + c2 e x :
(25)
We assume a particular solution of the form
1
yp = u1 (x)e 2 x + u2 (x)e x :
(26)
Then follows
1
yp0 = u01 e 2 x + u02 e
x
1
1
+ u1 (x)e 2 x
2
u2 e x :
If we make the further assumption that
1
u01 e 2 x + u02 e
we have
and
1
1
yp0 = u1 e 2 x
2
1
1
yp00 = u01 e 2 x
2
u02 e
x
x
= 0;
u2 e
x
1
1
+ u1 e 2 x + u 2 e x :
4
(27)
(28)
(29)
Substitute (26), (28) and (29) into (23):
1
yp = u01 e 2 x
2yp00 + yp0
x
2u02 e
(30)
= x + 1:
We now solve (27) and (30) for the ui . From (27) we have
u02 =
3
u01 e 2 x :
(31)
Substitute (31) into (30):
1
1
1
u01 e 2 x + 2u01 e 2 x = x + 1 ) u01 = (x + 1)e
3
1
x
2
(32)
Integrate by parts:
u1
1
=
3
=
=
and so
Z
Z
1
e
dx +
xe
3
1
2 1x 1
e 2 +
2xe 2 x
3
3
1
2 1x 1 h
e 2 +
2xe 2 x
3
3
1
x
2
u1 =
1
x
2
2
x+2 e
3
dx
Z
( 2)e
i
1
4e 2 x :
1
x
2
:
1
x
2
dx
(33)
Substitute (32) into (31):
u02 =
1
(x + 1)ex :
3
Integrate w.r.t. x:
u2 =
=
=
=
Z
Z
1
1
x
xe dx
ex dx
3
3
Z
1
1 x
x
xe
ex dx
e
3
3
1
1 x
[xex ex ]
e
3
3
1 x
xe
3
(34)
We substitute (33) and (34) into (26) to obtain a particular solution:
2
x+2 e
3
2
1
x 2
x
3
3
(x + 2)
yp =
=
=
1
x
2
1
e2x
1 x
xe e
3
x
(35)
The general solution follows from (25) and (35):
1
y = c1 e 2 x + c2 e
x
(x + 2)
(36)
We now introduce the initial conditions. Firstly, we di¤erentiate (14) to
obtain
1 1
y 0 = c1 e 2 x c2 e x 1:
(37)
2
Now introduce (24):
y(0) = c1 + c2 2 = 1
1
c1 c2 1 = 0
y 0 (0) =
2
We readily solve these equations to obtain
1
8
c1 = ; c2 = :
3
3
The solution (36) now becomes
y=
1
1
8e 2 x + e
3
x
(x + 2):
9.
y 000 + 4y 0 = sec 2x
(38)
The complementary function follows from the auxiliary equation
m3 + 4m = 0 ) m(m2 + 4) = 0
which has roots 0; 2i and so
yc = c1 + c2 cos 2x + c3 sin 2x:
(39)
We assume a particular solution of the form
yp = u1 (x) + u2 (x) cos 2x + u3 (x) sin 2x:
(40)
Hence,
yp0 = u01 + u02 cos 2x + u03 sin 2x
2u2 sin 2x + 2u3 cos 2x:
Furthermore, we assume that
u01 + u02 cos 2x + u03 sin 2x = 0;
(41)
yp0 =
(42)
so that
2u2 sin 2x + 2u3 cos 2x
and
yp00 =
2u02 sin 2x + 2u03 cos 2x
4u2 cos 2x
4u3 sin 2x:
Similar to (41) we assume
2u02 sin 2x + 2u03 cos 2x = 0
(43)
so that
yp00 =
4u2 cos 2x
(44)
4u3 sin 2x
and
yp000 =
4u02 cos 2x
4u03 sin 2x + 8u2 sin 2x
8u3 cos 2x:
(45)
We substitute (42) and (45) into (38):
yp000 + 4yp0 =
4u02 cos 2x
4u03 sin 2x:
(46)
In (41), (43) and (46) we have three simultaneous equations which may be
solved for the three ui . From (43) we have
u03 = u02 tan 2x:
Substitute (47) into (46):
cos 2x :
4u02 cos 2x 4u02 tan 2x sin 2x = sec 2x
4u02 cos2 2x + sin2 2x = 1
(47)
and so
u02 =
1
:
4
(48)
u2 =
1
x:
4
(49)
Integrate w.r.t. x:
Substitute (48) into (47):
1
tan 2x =
4
u03 =
1 sin 2x
:
4 cos 2x
(50)
Integrate w.r.t. x:
1
ln j cos 2xj:
8
Substitute (48) and (50) into (41):
u3 =
u01
1
cos 2x
4
(51)
1
tan 2x sin 2x = 0:
4
Hence,
u01 =
1
1 sin2 2x
1 cos2 2x + sin2 2x
1
cos 2x +
=
= sec 2x:
4
4 cos 2x
4
cos 2x
4
Integrate w.r.t. x:
1
ln j sec 2x + tan 2xj:
8
We substitute (49), (51) and (52) into (40):
u1 =
yp =
1
ln j sec 2x + tan 2xj
8
1
1
x cos 2x + sin 2x ln j cos 2xj:
4
8
(52)
(53)
The general solution follows from (39) and (53):
1
1
1
y = c1 +c2 cos 2x+c3 sin 2x+ ln j sec 2x+tan 2xj x cos 2x+ sin 2x ln j cos 2xj:
8
4
8
This seems to be a rather long-winded approach. Let us use Wronskians
instead to set up the relevant equations for the u’s. Say
y1 = 1; y2 = cos 2x; y3 = sin 2x:
Using
A B C
D E F
G H I
= AEI
AF H
DBI + DCH + GBF
GCE:
we …nd
W =
W1 =
y1 y2 y3
y10 y20 y30
y100 y200 y300
0 y2 y3
0 y20 y30
f y200 y300
=
0
0
sec 2x
=
1
0
0
cos 2x
2 sin 2x
4 cos 2x
cos 2x
2 sin 2x
4 cos 2x
sin 2x
2 cos 2x
4 sin 2x
sin 2x
2 cos 2x
4 sin 2x
=8
= 2 sec 2x cos2 2x + 2 sec 2x sin2 2x
= 2 sec 2x
W2 =
y1 0 y3
y10 0 y30
y100 f y300
W3 =
=
y1 y2 0
y10 y20 0
y100 y200 f
1
0
0
0
0 sec 2x
=
1
0
0
sin 2x
2 cos 2x
4 sin 2x
=
cos 2x
0
2 sin 2x
0
4 cos 2x sec 2x
2 cos 2x sec 2x =
=
2 sin 2x sec 2x
Hence,
2 sec 2x
sec 2x
W1
=
=
W
8
4
W2
2
1
=
=
=
W
8
4
W3
2 sin 2x sec 2x
=
=
=
W
8
u01 =
u02
u03
which seems to be a much easier approach.
10.
y 00 + y = tan x:
Auxiliary equation:
m2 + 1 = 0
)m =
i
sin 2x
=
4 cos 2x
tan 2x
4
2
Complementary function:
yc = c1 cos x + c2 sin x:
Assume a particular solution of the form
yp = u1 (x) cos x + u2 (x) sin x:
Hence,
cos x
sin x
= cos2 x
W =
( sin2 x) = 1
sin x cos x
0
sin x
W1 =
=
sin x tan x
tan x cos x
cos x
0
W2 =
= cos x tan x = sin x
sin x tan x
and so
W1
W
W2
=
W
u01 =
=
u02
= sin x:
sin x tan x
Integration yields
u1 =
Z
sin x tan x dx
Z
cos x tan x
( cos x) sec2 x dx
Z
= cos x tan x
sec x dx
=
= sin x
ln j sec x + tan xj:
and
u2 =
Z
sin x dx =
cos x:
(54)
Substitute this expression into (54):
yp = (sin x ln j sec x + tan xj) cos x
=
cos x ln j sec x + tan xj:
cos x sin x
The general solution is thus
y = yc + yp
= (c1 ln j sec x + tan xj) cos x + c2 sin x:
11.
y 00
4y 0 + 4y = 12x2
6x e2x
(55)
with initial-values
y(0) = 1;
y 0 (0) = 0:
(56)
Auxiliary equation:
m2
4m + 4 = 0 ) (m 2)2 = 0
) m = 2 (twice).
Complementary function:
yc = c1 e2x + c2 xe2x :
(57)
Assume a particular solution of the form
yp = u1 (x)e2x + u2 (x)xe2x :
(58)
Hence,
W =
e2x
xe2x
2x
2e
(1 + 2x)e2x
0
xe2x
(1 + 2x)e2x
W1 =
(12x2
W2 =
e2x
0
2x
2
2e
(12x
6x)e2x
6x)e2x
= (1 + 2x)e4x
=
= (12x2
and so
W1
= 12x3 + 6x2
W
W2
=
= 12x2 6x:
W
u01 =
u02
2xe4x = e4x
(12x3
6x)e4x
6x2 )e4x
Integration yields
u1 =
u2 =
Z
Z
( 12x3 + 6x2 ) dx =
(12x2
3x4 + 2x3 :
6x) dx = 4x3
3x2 :
Substitute into (57):
yp = ( 3x4 + 2x3 )e2x + (4x3
= (x4 x3 )e2x :
3x2 )xe2x
The general solution is thus
y = yc + yp = (c1 + c2 x
x3 + x4 )e2x :
(59)
Hence,
y0 =
=
c2 3x2 + 4x3 + 2c1 + 2c2 x 2x3 + 2x4 e2x
2c1 + c2 + 2c2 x 3x2 + 2x3 + 2x4 e2x :
Substitute y (0) = 1 into (59) ) c1 = 1:
Substitute y 0 (0) = 0 into (60) ) 2c1 + c2 = 0; c2 =
Solution to the initial-value problem:
y = (x4
12.
y
y 00
00
x3
2x + 1)e2x :
e2x
:
4y =
x
4y = 0
) m2 4 = 0
) m= 2
) yc (x) = c1 e2x + c2 e 2x
) y1 = e2x and y2 = e 2x
) y10 = 2e2x and y20 = 2e
2x
2c1 =
2:
(60)
0
Hence, W1 =
e2x
x
2x
e
2e2x
W2 =
2x
e
2e2x
W =
e 2x
2e 2x
=
=
e4x
x
e 2x
2e 2x
=
0
e2x
x
1
x
4
W1
1
ln jxj
=
) u1 =
W
4x
4 Z
W2
e4x
e4x
=
=
) u2 =
dx
W
4x
4x
and so u01 =
u02
(The above integral is nonelementary and may be left in integral form.
See Z&C 6th ed., sec 4.6, example 3.)
y p = u 1 y 1 + u2 y 2
e2x ln jxj
e
=
4
2x
) y(x) = c1 e
Note:
Z
ax
Z
+ c2 e
2x
2x
Z
e4x
dx
4x
e2x ln jxj
+
4
e
2x
Z
e4x
dx
4x
eax
dx does have a series solution:
x
2
3
e
ax
(ax)
(ax)
dx = ln x +
+
+
+
x
1 1!
2 2!
3 3!
13.
y 00 + 2y 0 + y = e
x
ln x :
y 00 + 2y 0 + y = 0
) m2 + 2m + 1 = 0
) m1 = 1; m2 = 1
#
) yc (x) = c1 e x + c2 xe x
) y1 = e x and y2 = xe
) y10 = e x and y20 = e
Hence, W1 =
0
xe x
e x ln x e x xe
W2 =
e x
0
e x e x ln x
W =
e x
xe x
x
x
e
e
xe
x
x
x
=e
x
x
xe
=
2x
xe
2x
ln x
ln x
=e
2x
W1
x2 ln x x2
= x ln x ) u1 =
+
W
2
4
W2
=
= ln x ) u2 = x ln x x
W
and so u01 =
u02
y p = u1 y 1 + u 2 y 2
x2 ln x x2
=
+
2
4
e
x
+ (x ln x
x) xe
x
3x2 e x
+ x2 e
2
4
ln x 3
= c1 e x + c2 xe x + x2 e x
:
2
4
) y(x) = c1 e
x
+ c2 xe
x
x2 e
x
ln x
x
ln x
14. First we solve for the complimentary solution in
yc00 + yc = 0
which has the auxiliary equation m2 + 1 = 0 and solutions m1 = i m2 =
so that
yc = c1 cos x + c2 sin x:
i
For the particular solution we assume yp = u1 (x) cos x+u2 (x) sin x and using
the technique of variation of parameters we obtain (y1 = cos x, y2 = sin x,
f (x) = cos2 x)
cos x sin x
sin x cos x
W =
0
sin x
2
cos x cos x
W1 =
cos x
0
sin x cos2 x
W2 =
u01 =
u1 =
W1
=
WZ
= cos2 x + sin2 x = 1:
sin x cos2 x:
=
= cos3 x = cos x
cos x sin2 x:
sin x cos2 x:
sin x cos2 x dx =
1
cos3 x:
3
u02 = cos x cos x sin2 x:
Z
u2 =
(cos x cos x sin2 x)dx = sin x
y p = u1 y 1 + u2 y 2 =
1 3
sin x:
3
1
cos4 x + sin2 x
3
1 4
sin x:
3
Thus the general solution is
y(x) = c1 cos x + c2 sin x +
1
cos4 x + sin2 x
3
1 4
sin x:
3
15. The auxiliary equation is m2 + 3m + 2 = (m + 2)(m + 1) = 0 with solutions
m1 = 1 and m2 = 2 and so yc = c1 e x + c2 e 2x . For the particular
solution yp = u1 e x + u2 e 2x we …nd (y1 = e x , y2 = e 2x , f (x) = sin ex )
W =
e x
e x
e 2x
2e 2x
=
e
3x
W1 =
0
sin ex
e 2x
2e 2x
=
e
2x
W2 =
e x
0
x
e
sin ex
=e
x
:
sin ex :
sin ex :
u01 =
u1 =
u02 =
u2 =
=
W1
= ex sin ex :
W
Z
ex sin ex dx =
W1
=
WZ
cos ex :
e2x sin ex :
e2x sin ex dx
(u = ex ; du = ex dx)
Z
Z
u sin u du = u cos u
cos u du = u cos u
= ex cos ex
sin u
sin ex :
Thus the general solution is
y(x) = c1 e
x
+ c2 e
2x
e
x
cos ex + e
x
cos ex
2x
2x
e
2x
sin ex :
or
y(x) = c1 e
x
+ c2 e
e
sin ex :
16. In standard form
1
1
1
y 00 + y 0 + 2 y = 2 sec(ln x):
x
x
x
Thus we have, with y1 = cos(ln x), y2 = sin(ln x), f (x) =
W =
W1 =
W2 =
cos(ln x) sin(ln x)
sin(ln x)
x
cos(ln x)
x
=
0
sin(ln x)
x)
sec(ln x) cos(ln
x
1
x2
cos(ln x)
sin(ln x)
x
u02
u2
0
sec(ln x)
sec(ln x),
1
1
(cos2 (ln x) + sin2 (ln x)) = :
x
x
=
=
1
tan(ln x):
x2
1
:
x2
W1
1
=
tan(ln x):
WZ
x
1
=
tan(ln x)dx = ln j cos(ln x)j:
x
W2
1
=
= :
x
ZW
1
=
dx = ln x:
x
u01 =
u1
1
x2
1
x2
So the general solution is
y(x) = c1 y1 + c2 y2 + u1 y1 + u2 y2
= c1 cos(ln x) + c2 sin(ln x) + cos(ln x) ln j cos(ln x)j + ln(x) sin(ln x):
17. In standard form we have
y 00
2y 0 + 10y = 5 sin x +
ex
tan 3x:
3
The auxiliary equation is m2 2m + 10 = 0 with solutions m1 = 1 + 3i and
m2 = 1 3i and consequently the complimentary solution is
yc (x) = c1 ex cos 3x + c2 ex sin 3x:
We use yp = yp1 + yp2 where
00
yp1
0
2yp1
+ 10yp1 = 5 sin x
ex
tan 3x
3
and the superposition principle for non-homogeneous linear di¤erential equations. We use undetermined coe¢ cients to …nd yp1 = A cos x + B sin x in
00
yp2
00
yp1
0
2yp2
+ 10yp2 =
0
2yp1
+ 10yp1 = ( A cos x B sin x) 2( A sin x + B cos x) + 10(A cos x
= (9A 2B) cos x + (9B + 2A) sin x = 5 sin x
B sin x)
2
and comparing coe¢ cients for cos x and sin x yields B = 29 A and A = 17
9
and B = 17 so that
2
9
yp1 =
cos x +
sin x:
17
17
For yp2 we use variation of parameters where y1 ex cos 3x, y2 ex sin 3x
x
and f (x) e3 tan 3x to obtain
W =
W1 =
ex cos 3x
ex sin 3x
ex (cos 3x 3 sin 3x) ex (sin 3x + 3 cos 3x)
ex
3
W2 =
0
ex sin 3x
x
tan 3x e (sin 3x + 3 cos 3x)
ex cos 3x
ex (cos 3x 3 sin 3x)
ex
3
0
tan 3x
=
=
= 3e2x :
e2x sin2 3x
=
3 cos 3x
e2x
sin 3x:
3
e2x
(sec 3x
3
cos 3x):
u01 =
u1 =
u02 =
u2 =
yp2 =
=
W1
1
=
(sec 3x cos 3x):
W Z
9
1
1
(sec 3x cos 3x)dx =
(ln j sec 3x + tan 3xj sin 3x):
9
27
W2
1
= sin 3x:
WZ
9
1
1
sin 3x dx =
cos 3x:
9
27
ex
ex
u1 y 1 + u 2 y 2 =
(ln j sec 3x + tan 3xj cos 3x sin 3x cos 3x)
cos 3x sin 3x
27
27
ex
ln j sec 3x + tan 3xj cos 3x:
27
Thus we …nd the general solution
y(x) = yc + yp = yc + yp1 + yp2
= c1 ex cos 3x + c2 ex sin 3x +
2
9
cos x +
sin x
17
17
ex
ln j sec 3x + tan 3xj cos 3x:
27
18. In standard form on (0; 1)
1
y 00 + y 0
x
4
1
y = 4:
2
x
x
Since y1 = x2 is a solution to the associated homogeneous equation we can
generate a second linearly independent solution
y2 = x
2
= x
2
Z
Z
e
R
1
dx
x
x4
dx
x 5 dx =
1
x
4
2
it is simpler, however, to discard the constant and use y2 = x 2 . Thus we
have
yc = c1 x2 + c2 x 2 :
Using variation of parameters we have yp = u1 y1 + u2 y2 and identifying
f (x)
x
4
we …nd
W =
x2
2x
x 2
2x 3
W1 =
0
x 4
W2 =
x2 0
2x x 4
x 2
2x 3
W1
1
= 5:
4x
ZW
1
1
=
dx =
:
5
4x
16x4
W2
1
=
=
:
WZ
4x
1
1
=
ln x:
=
4x
4
4x 1 :
=
=
x 6:
= x 2:
u01 =
u1
u2
u2
yp = u1 x2 + u2 x
2
=
1
16x2
1
ln x =
4x2
1
4x2
so that the general solution is
y(x) = c1 x2 + c2 x
2
1
4x2
1
4
ln x :
1
4
ln x :