TWK2A Variation of parameters (Section 4.6) Solutions 1. y 00 + y = sin x Homogeneous equation: y 00 + y =0 ) m 2 +1=0 ) m 1 = i; m 2 = i So y c = c 1 cos(x)+ c 2 sin(x) For particular solution: W = cos(x) sin(x) sin(x) cos(x) = cos 2 (x) + sin 2 (x)=1 W 1 = 0 sin(x) sin(x) cos(x) = sin 2 (x) W 2 = cos(x) 0 sin(x) sin(x) = cos(x) sin(x) So u 0 1 = sin 2 (x) ) u 1 = x 2 1 4 sin(2x)= 1 2 sin(x) cos(x) 1 2 x u 0 2 = cos(x) sin(x) ) u 2 = 1 2 cos 2 (x) Hence y p = u 1 y 1 + u 2 y 2 = 1 2 sin(x) cos 2 (x) 1 2 x cos(x) 1 2 cos 2 (x) sin(x) = 1 2 x cos(x) So y(x)= y c + y p = c 1 cos(x)+ c 2 sin(x) 1 2 x cos(x) (valid for 1 <x< 1).
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TWK2A Variation of parameters (Section 4.6) Solutions
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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:
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)
Substitute (19) into (18):
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)
The general solution follows from (13) and (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)
We assume a particular solution of the form
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)
Substitute (31) into (30):
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)
Substitute (32) into (31):
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)
The general solution follows from (25) and (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)
We assume a particular solution of the form
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)
Substitute (47) into (46):
�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)
Substitute (48) into (47):
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)
The general solution follows from (39) and (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),