twk2a variation of parameters (section 4.6) solutions

TWK2AVariation of parameters (Section 4.6)Solutions

1.y00 + y = sinx

Homogeneous equation:

y00 + y = 0 ) m2 + 1 = 0

) m1 = i; m2 = �i

So yc = c1 cos(x) + c2 sin(x)For particular solution:

W =

�� cos(x) sin(x)� sin(x) cos(x)

�� = cos2(x) + sin2(x) = 1W1 =

�� 0 sin(x)sin(x) cos(x)

�� = � sin2(x)W2 =

�� cos(x) 0� sin(x) sin(x)

�� = cos(x) sin(x)So

u01 = � sin2(x)) u1 = �x

2� 14sin(2x) =

1

2sin(x) cos(x)� 1

2x

u02 = cos(x) sin(x)) u2 = �1

2cos2(x)

Hence

yp = u1y1 + u2y2 =1

2sin(x) cos2(x)� 1

2x cos(x)� 1

2cos2(x) sin(x)

= �12x cos(x)

Soy(x) = yc + yp = c1 cos(x) + c2 sin(x)�

1

2x cos(x)

(valid for �1 < x <1).

2.

y00 � y = cosh�x2

�=ex2

2+e�

x2

2

Homogeneous equation:

y00 � y = 0 ) m2 � 1 = 0) m = �1) yc = c1e

x + c2e�x

Particular solution:

W =

�� ex e�x

ex �e�x�� = �2

W1 =

�� 0 e�x

cosh�x2

��e�x

�� = �e�x2 �ex2 + e

�x2

�= �e

�x2

2� e

�3x2

2

W2 =

�� ex 0ex cosh

�x2

� �� = e3x2

2+ex2

2

So

u01 =W1

W=

e�x2

4+e�32

4

) u1 = �e�x2

2� e

�3x2

6

u02 =W2

W= �e

3x2

4� e

x2

4

) u2 = �e3x2

6� e

x2

2

Thus

y(x) = yc + yp = yc + u1y1 + u2y2

= c1ex + c2e

�x � ex2

2� e

�x2

6� e

x2

6� e

�x2

2

= c1ex + c2e

�x � 2ex2

3� 2e

�x2

3

(valid for all x).

3.4y00 � y = xex2 y(0) = 1; y0(0) = 0

Homogeneous case:

4y00 � y = 0 ) 4m2 � 1 = 0

) m = �12

Soyc = c1e

x2 + c2e

�x2 :

Particular solution:

W =

�� ex2 e

�x2

12ex2 �1

2e�x2

�� = �12 � 12 = �1W1 =

�� 0 e�x2

xex2

4�12e�x2

�� = �x4W2 =

�� ex2 012ex2

xex2

4

�� = xex

4

Hence,

y(x) = yc + yp

= yc + u1y1 + u2y2

= c1ex2 + c2e

�x2 +

x2

8ex2 + e

�x2(�xex + ex)

4

= c1ex2 + c2e

�x2 +

x2ex2

8� xe

x2

4+ex2

4

Now,

y(0) = c1 + c2 +1

4= 1

y0(0) =c12� c22� 18= 0

So

c1 + c2 =3

4

c1 � c2 =1

4

) c1 =1

2; c2 =

1

4

Thus,

y(x) =ex2

2+e�x2

4+x2e

x2

8� xe

x2

4+ex2

4

=3e

x2

4+e�x2

4+x2e

x2

8� xe

x2

4

4.

x2y00 + xy0 +

�x2 � 1

4

�y = x

x2

Standard form:

y00 +1

xy0 +

�x2 � 1

4

�x2

y = x�12

W =

�� x�12 cosx x

�12 sin x

�12x�32 cosx� x�1

2 sin x �12x�32 sin x+ x

�12 cosx

��= �1

2x�2 cosx sin x+ x�1 cos2 x�

��12x�2 cosx sin x� x�1 sin2 x

�= x�1

�cos2 x+ sin2 x

�=1

x

W1 =

�� 0 x�12 sin x

x�12 �1

2x�32 sin x+ x

�12 cosx

�� = �x�1 sin xW2 =

�� x�12 cosx 0

�12x�32 cosx� x�1

2 sin x x�12

�� = x�1 cosxHence,

u01 =W1

W=�x�1 sin xx�1

= � sin x) u1 = cos(x)

u02 =W2

W=x�1 cosx

x�1= cos x) u2 = sinx

So

y(x) = yc + yp = yc + u1y1 + u2y2

= c1x12 cosx+ c2x

�12 sin x+ x

�12 cos2 x+ x

�12 sin2 x

= c1x�12 cosx+ c2x

�12 sin x+ x

�12

5.y00 � 2y0 + y = 4x2 � 3 + e

x

x

Homogeneous case:

y00 � 2y0 + y = 0) m2 � 2m+ 1 = 0) m1 = 1;m2 = 1 (repeated real roots)

Soyc = c1e

x + c2xex:

Now, we have

Ly = y00 � 2y0 + y = f(x) + g(x)

where f(x) = 4x2 � 3 and g(x) = ex

x:

Note that, if yp1 is a particular solution of Ly = f(x) and yp2 is a particularsolution 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

y00 � 2y0 + y = 4x2 � 3

and variation of parameters to solve

y00 � 2y0 + y = ex

x

to get particular solutions.For

y00 � 2y0 + y = 4x2 � 3

assume the form for the solution is Ax2 +Bx+ c:So y00 = 2A; y0 = 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

y00 � 2y0 + y = ex

x

W =

�� ex xex

ex xex + ex

�� = xe2x + e2x � xe2x = e2xW1 =

�� 0 xexex

xxex + ex

�� = �e2xW2 =

�� ex 0ex ex

x

�� = e2x

x

u01 =W1

W= �1 ) u1 = �x

u02 =W2

W=e2x

e2xx=1

x) u2 = ln jxj

So

yp2 = u1y1 + u2y2

= �xex + xex ln jxj

General solution:

y(x) = yc + yp1 + yp2= c1a

x + c2xex + 4x2 + 16x+ 21� xex + xex ln jxj:

6.y00 + y = secx tan x (1)

The complementary function is found by solving the auxiliary equation

m2 + 1 = 0:

The roots are �i, and hence

yc = c1 cosx+ c2 sin x: (2)

We now assume a particular solution of the form

yp = u1(x) cos x+ u2(x) sinx: (3)

Then follows that

y0p = u01 cosx+ u

02 sin x� u1 sin x+ u2 cosx:

If we make the assumption that

u01 cosx+ u02 sin x = 0; (4)

the �rst derivative simpli�es to

y0p = �u1 sin x+ u2 cosx:

From further di¤erentiation we have

y00p = �u01 sin x+ u02 cosx� u1 cosx� u2 sin x: (5)

We substitute (3) and (5) in (1):

y00p + y0p = �u01 sinx+ u02 cosx = secx 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):

u02 sin x tan x+ u02 cosx = sec x tan x

� cosx : (sin2 x+ cos2 x)u02 = tan x

u02 = tanx =sinx

cosx: (8)

Integrate w.r.t. x:u2 = � ln j cosxj: (9)

Substitute (8) in (7):

u01 = � tan2 x = 1� sec2 x:

Integrate w.r.t. x:u1 = x� tan x (10)

Substituting (9) and (10) in (3), we have the particular solution

yp = (x�tan x) cos x�sin x ln j cosxj = x cosx�sin x�sin x ln j cosxj: (11)

The general solution follows from (2) and (11):

y = c1 cosx+ c2 sin x+ x cosx� sin x� sin x ln j cosxj= c1 cosx+ c3 sin x+ x cosx� sin x ln j cosxj:

7.y00 � 9y = 9x

e3x: (12)

The complementary function is found by solving the auxiliary equation

m2 � 9 = 0:

which has roots �3, and so

yc = c1e3x + c2e

�3x (13)

We assume a particular solution of the form

yp = u1(x)e3x + u2(x)e

�3x: (14)

Hence,y0p = u

01e3x + u02e

�3x + 3u1e3x � 3u2e�3x:

Furthermore, we assume

u01e3x + u02e

�3x = 0; (15)

so thaty0p = 3u1e

3x � 3u2e�3x

andy00p = 3u

01e3x � 3u02e�3x + 9u1e3x + 9u2e�3x: (16)

Substitute (14) and (16) into (12):

y00p � 9y0p = 3u01e3x � 3u02e�3x =

9x

e3x

) u01e3x � u02e�3x = 3xe�3x: (17)

We solve (15) and (17) for the ui. From (15 we have

u02 = �u01e6x: (18)

Substitute (18) into (17):

u01e3x + u01e

3x = 3xe�3x ) u01 =3

2xe�6x (19)

Integrate w.r.t. x using integration by parts:

u1 =3

2

Zxe�6x dx

=3

2

��x6e�6x �

Z1

(�6)e�6x dx

�=

3

2

��x6e�6x � 1

36e�6x

�:

Hence,

u1 = �1

24(6x+ 1)e�6x: (20)


u02 = �3

2x:

Integrate w.r.t. x:

u2 = �3

4x2: (21)

We substitute (20) and (21 into (14) to �nd a particular solution:

yp = �1

24(6x+ 1)e�6xe3x � 3

4x2e�3x = � 1

24(18x2 + 6x+ 1)e�3x: (22)


y = yc + yp

= c1e3x + c2e

�3x � 1

24(18x2 + 6x+ 1)e�3x

= c1e3x + c3e

�3x � 14(3x+ 1)xe�3x:

8.2y00 + y0 � y = x+ 1 (23)

with the initial conditions

y(0) = 1; y0(0) = 0: (24)

The complementary function follows from the auxiliary equation

2m2 +m� 1 = 0) (2m� 1)(m+ 1) = 0

The roots are m1 =12; m2 = �1 and we have

yc = c1e12x + c2e

�x: (25)


yp = u1(x)e12x + u2(x)e

�x: (26)

Then follows

y0p = u01e

12x + u02e

�x +1

2u1(x)e

12x � u2e�x:

If we make the further assumption that

u01e12x + u02e

�x = 0; (27)

we havey0p =

1

2u1e

12x � u2e�x (28)

andy00p =

1

2u01e

12x � u02e�x +

1

4u1e

12x + u2e

�x: (29)

Substitute (26), (28) and (29) into (23):

2y00p + y0p � yp = u01e

12x � 2u02e�x = x+ 1: (30)

We now solve (27) and (30) for the ui. From (27) we have

u02 = �u01e32x: (31)


u01e12x + 2u01e

12x = x+ 1) u01 =

1

3(x+ 1)e�

12x (32)

Integrate by parts:

u1 =1

3

Ze�

12x dx+

1

3

Zxe�

12x dx

= �23e�

12x +

1

3

��2xe� 1

2x �

Z(�2)e� 1

2x dx

�= �2

3e�

12x +

1

3

h�2xe� 1

2x � 4e� 1

2xi:

and so

u1 = ��2

3x+ 2

�e�

12x: (33)


u02 = �1

3(x+ 1)ex:

Integrate w.r.t. x:

u2 = �13

Zxex dx� 1

3

Zex dx

= �13

�xex �

Zex dx

�� 13ex

= �13[xex � ex]� 1

3ex

= �13xex (34)

We substitute (33) and (34) into (26) to obtain a particular solution:

yp = ��2

3x+ 2

�e�

12xe

12x � 1

3xexe�x

= �23x� 2� 1

3x

= �(x+ 2) (35)


y = c1e12x + c2e

�x � (x+ 2) (36)

We now introduce the initial conditions. Firstly, we di¤erentiate (14) toobtain

y0 =1

2c1e

12x � c2e�x � 1: (37)

Now introduce (24):

y(0) = c1 + c2 � 2 = 1

y0(0) =1

2c1 � c2 � 1 = 0

We readily solve these equations to obtain

c1 =8

3; c2 =

1

3:

The solution (36) now becomes

y =1

3

�8e

12x + e�x

�� (x+ 2):

9.y000 + 4y0 = 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)


yp = u1(x) + u2(x) cos 2x+ u3(x) sin 2x: (40)

Hence,

y0p = u01 + u

02 cos 2x+ u

03 sin 2x� 2u2 sin 2x+ 2u3 cos 2x:

Furthermore, we assume that

u01 + u02 cos 2x+ u

03 sin 2x = 0; (41)

so thaty0p = �2u2 sin 2x+ 2u3 cos 2x (42)

andy00p = �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 thaty00p = �4u2 cos 2x� 4u3 sin 2x (44)

andy000p = �4u02 cos 2x� 4u03 sin 2x+ 8u2 sin 2x� 8u3 cos 2x: (45)

We substitute (42) and (45) into (38):

y000p + 4y0p = �4u02 cos 2x� 4u03 sin 2x: (46)

In (41), (43) and (46) we have three simultaneous equations which may besolved for the three ui. From (43) we have

u03 = u02 tan 2x: (47)


�4u02 cos 2x� 4u02 tan 2x sin 2x = sec 2x

� cos 2x : �4u02�cos2 2x+ sin2 2x

�= 1

and sou02 = �

1

4: (48)

Integrate w.r.t. x:

u2 = �1

4x: (49)


u03 = �1

4tan 2x = �1

4

sin 2x

cos 2x: (50)

Integrate w.r.t. x:

u3 =1

8ln j cos 2xj: (51)

Substitute (48) and (50) into (41):

u01 �1

4cos 2x� 1

4tan 2x sin 2x = 0:

Hence,

u01 =1

4cos 2x+

1

4

sin2 2x

cos 2x=1

4

cos2 2x+ sin2 2x

cos 2x=1

4sec 2x:

Integrate w.r.t. x:

u1 =1

8ln j sec 2x+ tan 2xj: (52)

We substitute (49), (51) and (52) into (40):

yp =1

8ln j sec 2x+ tan 2xj � 1

4x cos 2x+

1

8sin 2x ln j cos 2xj: (53)


y = c1+c2 cos 2x+c3 sin 2x+1

8ln j sec 2x+tan 2xj�1

4x cos 2x+

1

8sin 2x ln j cos 2xj:

This seems to be a rather long-winded approach. Let us use Wronskiansinstead to set up the relevant equations for the u�s. Say

y1 = 1; y2 = cos 2x; y3 = sin 2x:

Using ��A B CD E FG H I

�� = AEI � AFH �DBI +DCH +GBF �GCE:we �nd

W =

��y1 y2 y3y01 y02 y03y001 y002 y003

�� =��1 cos 2x sin 2x0 �2 sin 2x 2 cos 2x0 �4 cos 2x �4 sin 2x

�� = 8

W1 =

��0 y2 y30 y02 y03f y002 y003

�� =��

0 cos 2x sin 2x0 �2 sin 2x 2 cos 2x

sec 2x �4 cos 2x �4 sin 2x

�� = 2 sec 2x cos2 2x+ 2 sec 2x sin2 2x

= 2 sec 2x

W2 =

��y1 0 y3y01 0 y03y001 f y003

�� =��1 0 sin 2x0 0 2 cos 2x0 sec 2x �4 sin 2x

�� = �2 cos 2x sec 2x = �2W3 =

��y1 y2 0y01 y02 0y001 y002 f

�� =��1 cos 2x 00 �2 sin 2x 00 �4 cos 2x sec 2x

�� = �2 sin 2x sec 2xHence,

u01 =W1

W=2 sec 2x

8=sec 2x

4

u02 =W2

W=�28= �1

4

u03 =W3

W=�2 sin 2x sec 2x

8= � sin 2x

4 cos 2x= �tan 2x

4

which seems to be a much easier approach.

10.y00 + y = tanx:

Auxiliary equation:

m2 + 1 = 0

) m = �i

Complementary function:

yc = c1 cosx+ c2 sin x:

Assume a particular solution of the form

yp = u1(x) cos x+ u2(x) sinx: (54)

Hence,

W =

��cosx sin x

� sin x cosx

�� = cos2 x� (� sin2 x) = 1

W1 =

��0 sinx

tan x cosx

�� = � sin x tan x

W2 =

��cosx 0

� sin x tan x

�� = cos x tan x = sinx

and so

u01 =W1

W= � sin x tan x

u02 =W2

W= sinx:

Integration yields

u1 = �Zsin x tan x dx

= �� cosx tan x�

Z(� cosx) sec2 x dx

�= cosx tan x�

Zsec x dx

= sinx� ln j sec x+ tanxj:

and

u2 =

Zsin x dx = � cosx:

Substitute this expression into (54):

yp = (sinx� ln j sec x+ tan xj) cos x� cosx sin x= � cosx ln j sec x+ tan xj:

The general solution is thus

y = yc + yp

= (c1 � ln j sec x+ tanxj) cos x+ c2 sin x:

11.y00 � 4y0 + 4y =

�12x2 � 6x

�e2x (55)

with initial-valuesy(0) = 1; y0(0) = 0: (56)

Auxiliary equation:

m2 � 4m+ 4 = 0) (m� 2)2 = 0) m = 2 (twice).

Complementary function:

yc = c1e2x + c2xe

2x: (57)

Assume a particular solution of the form

yp = u1(x)e2x + u2(x)xe

2x: (58)

Hence,

W =

�� e2x xe2x

2e2x (1 + 2x)e2x

�� = (1 + 2x)e4x � 2xe4x = e4x

W1 =

�� 0 xe2x

(12x2 � 6x)e2x (1 + 2x)e2x

�� = �(12x3 � 6x2)e4x

W2 =

�� e2x 02e2x (12x2 � 6x)e2x

�� = (12x2 � 6x)e4x

and so

u01 =W1

W= �12x3 + 6x2

u02 =W2

W= 12x2 � 6x:

Integration yields

u1 =

Z(�12x3 + 6x2) dx = �3x4 + 2x3:

u2 =

Z(12x2 � 6x) dx = 4x3 � 3x2:

Substitute into (57):

yp = (�3x4 + 2x3)e2x + (4x3 � 3x2)xe2x

= (x4 � x3)e2x:

The general solution is thus

y = yc + yp = (c1 + c2x� x3 + x4)e2x: (59)

Hence,

y0 =�c2 � 3x2 + 4x3 + 2c1 + 2c2x� 2x3 + 2x4

�e2x

=�2c1 + c2 + 2c2x� 3x2 + 2x3 + 2x4

�e2x: (60)

Substitute y (0) = 1 into (59) ) c1 = 1:Substitute y0 (0) = 0 into (60) ) 2c1 + c2 = 0; c2 = �2c1 = �2:Solution to the initial-value problem:

y = (x4 � x3 � 2x+ 1)e2x:

12.

y00 � 4y = e2x

x:

y00 � 4y = 0

) m2 � 4 = 0) m = �2

) yc(x) = c1e2x + c2e

�2x

) y1 = e2x and y2 = e

�2x

) y01 = 2e2x and y02 = �2e�2x

Hence, W1 =

�� 0 e�2x

e2x

x�2e�2x

�� = �1xW2 =

�� e2x 0

2e2x e2x

x

�� = e4x

x

W =

�� e2x e�2x

2e2x �2e�2x�� = �4

and so u01 =W1

W=1

4x) u1 =

ln jxj4

u02 =W2

W= �e

4x

4x) u2 = �

Ze4x

4xdx

(The above integral is nonelementary and may be left in integral form.

See Z&C 6th ed., sec 4.6, example 3.)

yp = u1y1 + u2y2

=e2x ln jxj

4� e�2x

Ze4x

4xdx

) y(x) = c1e2x + c2e

�2x +e2x ln jxj

4� e�2x

Ze4x

4xdx

�Note:

Zeax

xdx does have a series solution:Z

eax

xdx = ln x+

ax

1 � 1! +(ax)2

2 � 2! +(ax)3

3 � 3! + � � �#

13.y00 + 2y0 + y = e�x lnx :

y00 + 2y0 + y = 0

) m2 + 2m+ 1 = 0

) m1 = �1;m2 = �1

) yc(x) = c1e�x + c2xe

�x

) y1 = e�x and y2 = xe

�x

) y01 = �e�x and y02 = e�x � xe�x

Hence, W1 =

�� 0 xe�x

e�x lnx e�x � xe�x�� = �xe�2x lnx

W2 =

�� e�x 0�e�x e�x lnx

�� = e�2x lnxW =

�� e�x xe�x

�e�x e�x � xe�x�� = e�2x

and so u01 =W1

W= �x lnx) u1 = �

x2 lnx

2+x2

4

u02 =W2

W= ln x) u2 = x lnx� x

yp = u1y1 + u2y2

=

��x

2 lnx

2+x2

4

�e�x + (x lnx� x)xe�x

) y(x) = c1e�x + c2xe

�x � x2e�x lnx

2� 3x

2e�x

4+ x2e�x lnx

= c1e�x + c2xe

�x + x2e�x�lnx

2� 34

�:

14. First we solve for the complimentary solution in

y00c + yc = 0

which has the auxiliary equation m2 + 1 = 0 and solutions m1 = i m2 = �iso that

yc = c1 cosx+ c2 sinx:

For the particular solution we assume yp = u1(x) cos x+u2(x) sinx and usingthe technique of variation of parameters we obtain (y1 = cos x, y2 = sinx,

f(x) = cos2 x)

W =

�� cosx sin x� sinx cosx

�� = cos2 x+ sin2 x = 1:W1 =

�� 0 sinxcos2 x cosx

�� = � sin x cos2 x:W2 =

�� cosx 0� sin x cos2 x

�� = cos3 x = cos x� cosx sin2 x:u01 =

W1

W= � sin x cos2 x:

u1 = �Zsinx cos2 x dx =

1

3cos3 x:

u02 = cosx� cosx sin2 x:

u2 =

Z(cosx� cosx sin2 x)dx = sinx� 1

3sin3 x:

yp = u1y1 + u2y2 =1

3cos4 x+ sin2 x� 1

3sin4 x:

Thus the general solution is

y(x) = c1 cosx+ c2 sin x+1

3cos4 x+ sin2 x� 1

3sin4 x:

15. The auxiliary equation is m2+3m+2 = (m+2)(m+1) = 0 with solutionsm1 = �1 and m2 = �2 and so yc = c1e

�x + c2e�2x. For the particular

solution yp = u1e�x + u2e�2x we �nd (y1 = e�x, y2 = e�2x, f(x) = sin ex)

W =

�� e�x e�2x

�e�x �2e�2x�� = �e�3x:

W1 =

�� 0 e�2x

sin ex �2e�2x�� = �e�2x sin ex:

W2 =

�� e�x 0�e�x sin ex

�� = e�x sin ex:

u01 =W1

W= ex sin ex:

u1 =

Zex sin exdx = � cos ex:

u02 =W1

W= �e2x sin ex:

u2 = �Ze2x sin exdx (u = ex; du = exdx)

= �Zu sinu du = u cosu�

Zcosu du = u cosu� sinu

= ex cos ex � sin ex:

Thus the general solution is

y(x) = c1e�x + c2e

�2x � e�x cos ex + e�x cos ex � e�2x sin ex:

ory(x) = c1e

�x + c2e�2x � e�2x sin ex:

16. In standard formy00 +

1

xy0 +

1

x2y =

1

x2sec(ln x):

Thus we have, with y1 = cos(lnx), y2 = sin(ln x), f(x) = 1x2sec(ln x),

W =

�� cos(lnx) sin(lnx)

� sin(lnx)x

cos(lnx)x

�� = 1

x(cos2(lnx) + sin2(lnx)) =

1

x:

W1 =

�� 0 sin(lnx)1x2sec(ln x) cos(lnx)

x

�� = � 1x2 tan(lnx):W2 =

�� cos(lnx) 0

� sin(lnx)x

1x2sec(ln x)

�� = 1

x2:

u01 =W1

W= �1

xtan(lnx):

u1 = �Z1

xtan(lnx)dx = ln j cos(lnx)j:

u02 =W2

W=1

x:

u2 =

Z1

xdx = ln x:

So the general solution is

y(x) = c1y1 + c2y2 + u1y1 + u2y2

= c1 cos(lnx) + c2 sin(lnx) + cos(lnx) ln j cos(lnx)j+ ln(x) sin(ln x):

17. In standard form we have

y00 � 2y0 + 10y = 5 sinx+ ex

3tan 3x:

The auxiliary equation is m2� 2m+ 10 = 0 with solutions m1 = 1+ 3i andm2 = 1� 3i and consequently the complimentary solution is

yc(x) = c1ex cos 3x+ c2e

x sin 3x:

We use yp = yp1 + yp2 where

y00p1 � 2y0p1 + 10yp1 = 5 sinx

y00p2 � 2y0p2 + 10yp2 =ex

3tan 3x

and the superposition principle for non-homogeneous linear di¤erential equa-tions. We use undetermined coe¢ cients to �nd yp1 = A cosx+B sin x in

y00p1 � 2y0p1 + 10yp1 = (�A cosx�B sin x)� 2(�A sinx+B cosx) + 10(A cosx�B sin x)= (9A� 2B) cos x+ (9B + 2A) sinx = 5 sinx

and comparing coe¢ cients for cosx and sin x yields B = 92A and A = 2

17

and B = 917so that

yp1 =2

17cosx+

9

17sin x:

For yp2 we use variation of parameters where y1 � ex cos 3x, y2 � ex sin 3xand f(x) � ex

3tan 3x to obtain

W =

�� ex cos 3x ex sin 3xex (cos 3x� 3 sin 3x) ex (sin 3x+ 3 cos 3x)

�� = 3e2x:W1 =

�� 0 ex sin 3xex

3tan 3x ex (sin 3x+ 3 cos 3x)

�� = �e2x3 sin2 3xcos 3x= �e

2x

3(sec 3x� cos 3x):

W2 =

�� ex cos 3x 0ex (cos 3x� 3 sin 3x) ex

3tan 3x

�� = e2x

3sin 3x:

u01 =W1

W= �1

9(sec 3x� cos 3x):

u1 = �19

Z(sec 3x� cos 3x)dx = � 1

27(ln j sec 3x+ tan 3xj � sin 3x):

u02 =W2

W=1

9sin 3x:

u2 =1

9

Zsin 3x dx = � 1

27cos 3x:

yp2 = u1y1 + u2y2 = �ex

27(ln j sec 3x+ tan 3xj cos 3x� sin 3x cos 3x)� ex

27cos 3x sin 3x

= � ex

27ln j sec 3x+ tan 3xj cos 3x:

Thus we �nd the general solution

y(x) = yc + yp = yc + yp1 + yp2

= c1ex cos 3x+ c2e

x sin 3x+2

17cosx+

9

17sin x� ex

27ln j sec 3x+ tan 3xj cos 3x:

18. In standard form on (0;1)

y00 +1

xy0 � 4

x2y =

1

x4:

Since y1 = x2 is a solution to the associated homogeneous equation we cangenerate a second linearly independent solution

y2 = x2Ze�

R1xdx

x4dx

= x2Zx�5dx = �1

4x�2

it is simpler, however, to discard the constant and use y2 = x�2. Thus wehave

yc = c1x2 + c2x

�2:

Using variation of parameters we have yp = u1y1 + u2y2 and identifying

f(x) � x�4 we �nd

W =

�� x2 x�2

2x �2x�3�� = �4x�1:

W1 =

�� 0 x�2

x�4 �2x�3�� = �x�6:

W2 =

�� x2 02x x�4

�� = x�2:u01 =

W1

W=

1

4x5:

u1 =

Z1

4x5dx = � 1

16x4:

u2 =W2

W= � 1

4x:

u2 = �Z

1

4x= �1

4lnx:

yp = u1x2 + u2x

�2 = � 1

16x2� 1

4x2lnx = � 1

4x2

�1

4� lnx

�:

so that the general solution is

y(x) = c1x2 + c2x

�2 � 1

4x2

�1

4� lnx

�:

twk2a variation of parameters (section 4.6) solutions

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