181202(MA 2161) – MATHEMATICS - II UNIT 1 notes

CLICK HERE TO DOWNLOAD

Unit. 1 Ordinary Differential Equations
1
MISRIMAL NAVAJEE MUNOTH JAIN ENGINEERING
COLLEGE, CHENNAI - 97
DEPARTMENT OFMATHEMATICS
MATHEMATICS (MA2161)
FOR
SECOND SEMESTER ENGINEERING STUDENTS
ANNA UNIVERSITY SYLLABUS
This text contains some of the most important short-answer (Part A) and longanswer
questions (Part B) and their answers. Each unit contains 30 university
questions. Thus, a total of 150 questions and their solutions are given. A student
who studies these model problems will be able to get pass mark (hopefully!!).
Prepared by the faculty of Department of Mathematics
JANUARY, 2009
www.engg-maths.com
Unit. 1 Ordinary Differential Equations
2
UNIT I ORDINARY DIFFERENTIAL EQUATIONS
Part – A
Problem 1 Solve the equation D2  D 1 y  0
Solution:
The A.E is 2 1 0 1 1 4 1 3
2 2
m m m i
  
      .
1 3 and 1 ; 3
2 2 2
m i  

  
G.S : y  e x  Acos x  Bsin  x
1
2 . : cos 3 sin 3
2 2
G S y e x A x B x
 
   
 
where A, B are arbitrary constants.
Problem 2 Find the particular integral of D2  a2  y  bcos ax  c sin ax.
Solution:
Given D2  a2  y  bcos ax  c sin ax.
P.I 2 2 2 2
b 1 cos ax c. 1 sin ax.
D a D a
 
 
sin cos
2 2
bx ax cx ax
a a
 
 sin cos .
2
x b ax c ax
a
 
Problem 3 Find the particular integral of  2 D 1 y  ex cos x .
Solution:
P.I
 2
1 cos
1
e x x
D
 

 2 cos
1 1
e x x
D


 
2
e x 1 cos x
D
 
e x 1 sin x
D
 
 ex cos x.
Unit. 1 Ordinary Differential Equations
3
Problem 4 Find the particular integral of D2  4 y  x4 .
Solution:
P.I 4
2
1
4
x
D


4
2
1
4 1
4
x
D

 
  
 
2 1
1 1 4
4 4
D x
  
   
 
2 4
1 1 4
4 4 16
D D x
 
    
 
2
1 4 4.3 4.3.2.1
4 4 16
x x
 
    
 
1 4 3 2 3 .
4 2
  x  x  
 
Problem 5 Solve D2  6D  9 y  e2 xx3 .
Solution:
The A.E is m2  6m 9  0
 2  m 3  0
m  3,3
C.F:  A Bxe3x
 
2 3
2
1
3
e x x
D
 

 
2
3
2 2 3
e x x
D


 
 
 
2
3 2 2 3
2 1
1
x
e x e x D x
D

    

P.I  e2x 1 2D  3D2  4D3  x3
 e2x x3  23x2  33.2x  43.2.1
 e2x x3  6x2 18x  24
G.S. y   A Bxe3x  x3  6x2 18x  24e2x .
Unit. 1 Ordinary Differential Equations
4
Problem 6 Solve D2  2D 1 y  x
Solution:
The A.E is 2 2 1 0 2 4 4
2
m m m
  
     y
m  1 2
C.F: Ae 1 2x Be 1 2 x Ae xe 2x Be xe 2x          
P.I  2 
1
2 1
x
D D

 
 2 
1
1 2
x
D D

  
  1 2D D2 1 x
      
P.I   1 2D  D2  x  x  2
G.S: y  ex Ae 2x  Be 2x x  2.
Problem 7 Find the particular integral D2  4D  5 y  e2x cos x
Solution:
P.I 2
2
1 cos
4 5
e x x
D D
 
 
 
 2 
2
1 cos
2 1
e x x
D
 
 
 
2
2
1 cos
2 2 1
e x x
D
 
  
2
2
1 cos
1
e x x
D
 

P.I
2
sin .
2
xe x x


Problem 8 Solve for x from the equations x  y  t and x  y 1.
Solution:
x  y  t (1) x  y 1 x 1  y
x  y 1(2) x  x 1 1
Thus x  x  2 (or) D2 1 x  2
m2 1  0m   i
C.F. Acos t  Bsin t
Unit. 1 Ordinary Differential Equations
5
P.I      2  1  
2
1 2 1 2 2
1
D
D
    

G.S: x  Acos t  Bsin t  2.
Problem 9 Solve D3  3D2  6D  8 y  x .
Solution:
The A.E is m3  3m2  6m  8  0
m1m 2m 4  0
m 1,  2, 4
2 4
1 2 3 C.F is C ex C e x C e x
P.I 3 2
1
3 6 8
x
D D D

  
3 2
1
8 1 3 6
8
x
D D D

   
  
 
1 3 3 2 6 1 1
8 8
D D D x
    
   
 
1 3 3 2 6 1 ..
8 8
D D D x
     
      
   
1 6 1 3 .
8 8 8 4
 x    x      
Complete solution is y  C.F  P.I
2 4
1 2 3
1 3
8 4
y  C ex C e x C e x  x    
.
Problem 10 Solve the equation D2  4D 13 y  e2x
Solution:
Given D2  4D 13 y  e2x
The A.E is m2  4m13  0
4 16 52
2
m
 

4 36
2
 

4 6 2 3
2
i i 
  
C.F y  e2x  Acos3x  Bsin 3x
Unit. 1 Ordinary Differential Equations
6
P.I. 2
2
1
4 13
e x
D D

 
1 2 1 2
4 8 13 9
 e x  e x
 
G.S: y  C.F  P.I
y  
2
2 cos3 sin 3 .
9
x
 e x A x  B x  e
Problem 11 Solve the equation D5  D y 12ex
Solution:
Given D5  D y 12ex
The A.E is m5 m  0
mm4 1  0
m4 1  0
m  0or m4 1  0
m2 1m2 1  0
m  0, m  1, m  i
C.F 0  
1 2 3 4 5  C e x C ex C ex  C cos x C sin x
P.I 5
1 12ex
D D


1 12 Replacing D by 1
1 1
 ex

  4 12 Replacing D by 1
5 1
x ex
D


12 12 3
5 1 4
 x ex  x ex  xex

G.S. y  C.F  P.I
  1 2 3 4 5  C C ex C ex  C cos x C sin x  3xex .
Problem 12 Solve the equation D2  5D  6 y  e7 x sinh 3x
Solution:
The A.E is m5  5m 6  0
m 2m 3  0
m  2,3
C.F. is 2 3
1 2
C e x C e x
P.I 7
2
1 sinh 3
5 6
e x x
D D
 
 
Unit. 1 Ordinary Differential Equations
7
3 3
7
2
1
5 6 2
x x
e x e e
D D

   
      
4 10
2 2
1 1 1
2 5 6 5 6
e x e x
D D D D
           
1 4 10
2 16 20 6 10 50 6
 e x e x 
         
1 4 10
2 2 34
e x e x 
   
 
G.S.
4 10
2 3
1 2 .
4 68
x x
y C e x C e x e e
 
     
Problem 13 Solve the equation D3  3D2  4D  2 y  ex
Solution:
Given m3  3m2  4m  2  0
m1m2  2m 2  0
m 1 (or) m 1 i
Complementary function  Aex  ex Bcos x Csin x
P.I 3 2
1
3 4 2
ex
D D D

  
     
  3 2
1 Replacing D by 1
1 3 1 4 1 2
 ex
  
1 1
1 3 4 2 0
 ex  ex
  
3 2 6 4
x ex
D D

 
1 Replacing D by 1
3 6 4
 ex
 
 xex
G.S: y  C.F. P.I
 Aex  ex Bcos x Csin x  xex .
Problem 14 Solve the equation
2
2
2 d y 4 dy 4y e x
dx dx
   
Solution:
Given D2  4D  4 y  e2x
The A.E is m2  4m 4  0
m2  2m 2  0
Unit. 1 Ordinary Differential Equations
8
m  2,2
C.F.:  Ax  Be2x
P.I 2
2
1
4 4
e x
D D
 
 
   
2  
2
1 Replacing D by - 2
2 4 2 4
 e x
   
1 2  Dr is 0
4 8 4
 e x 
 
2 Replacing D by -2
2 4
x e x
D
 

 
2
2 2 4
 x e x
 
 
2
2 Dr is 0
2
 x e x 
G.S is  
2
2 2 .
2
y  Ax  B e x  x e x
Problem 15 Solve the equation D2  2D 1 y  ex  3
Solution:
Given D2  2D 1 y  ex  3
The A.E is m2  2m1  0
m1m1  0
m  1,1
C.F.:  Ax  Bex
P.I = 1 2 P.I  P.I
1 2
. 1
2 1
P I e x
D D
 
 
   
  2
1 Replacing D by -1
1 2 1 1
 ex
   
1
1 2 1
 ex
 
 Dr is 0
2 2
x e x
D
 


  Replacing D by -1
2 1 2
 x ex
 
0
2 2
1 3
2 1
PI e x
D D

 
Unit. 1 Ordinary Differential Equations
9
   
0  
2
1 3 Replacing D by 0
0 2 0 1
 e x
 
G.S is  
2
3.
2
y  Ax  B ex  x ex 
Part-B
Problem 1 Solve    D2  2D 8 y  4cosh x sinh 3x  e2 x  ex 2 1.
Solution:
The A.E. is m2  2m8  0
m 4m 2  0
m  2,4
C.F.: Ae2x  Be4x
R.H.S   4cosh xsin 3x  e2x  ex 2 1
  3 3 2 4 2 1
2 2
x x x x
e e e e e x ex
      
       
  
 e4x  e2x  e2x  e4x  e4x  2e3x  e2x 1
 e2 x  e4 x  2e3x 1e0 x
P.I.         1 2 1  4 2 3 0 
4 2 4 2
e x e x e x e x
D D D D
      
   
           
4 3
1 2 2 1
2 4 2 8 2 1 5 4 2
x x
e x e e
D

 
   
      
2 4 2 3 1
6 16 5 8
xe x e x e x
   
G.S is
2 4 3
2 4 2 1 .
6 16 5 8
x x x
y Ae x Be x xe e e
 
      
Problem 2 Solve y  y  sin2 x  cos x cos 2x cos3x
Solution:
The A.E is m2 1  0m2  1m   i
C.F.: Acos x  Bsin x
R.H.S
cos 2cos 2 cos3  cos   cos5 cos
2 2
x x x x x x   
1 2cos cos5 2cos2
4
  x x  x
Unit. 1 Ordinary Differential Equations
10
1 cos 6 cos 4 1 cos 2 
4
 x  x   x
2cos Acos B  cos  A B  cos  A B A  3x, B  2x
P.I.  
2
2
1 sin cos cos 2 cos3
1
x x x x
D
    
0 0
2
1 cos 2 cos 6 cos 4 cos 2
1 2 2 4 4 4 4
e x x x x x e x
D
 
          
 2      
1 cos 2 cos6 cos 4 cos 2 1
2 4 1 4 36 1 4 16 1 4 4 1 4
  x  x  x  x 
       
3 cos 2 cos 6 cos 4 cos 6 3 .
4 6 140 60 140 4
  x  x  x  x 
G.S. is cos sin cos 2 cos 4 cos 6 3 .
12 60 140 4
y  A x  B x  x  x  x 
Problem 3 Solve
2
2 d x 10x cos8y.
dy
 
Solution:
Here y is independent and x is dependent variable
Let D d
dy
 .
The A.E is m2 10  0
m2  10
m   10i
C.F.: Acos 10y  Bsin 10y
P.I  2 
1 cos8 cos8 cos8
10 64 10 54
y y y
D

  
  
G.S. is cos 10 sin 10 cos8
54
x  A y  B y  y
Problem 4 Solve
2
2 d y 6 dy 9y sin x cos 2x
dx dx
   .
Solution:
The A.E is m2  6m 9  0
 2 m 3  0
m  3,  3.
C.F.:  A Bxe3x
R.H.S 2sin cos 2 1 sin 3 sin  
2 2
 x x   x  x   
Unit. 1 Ordinary Differential Equations
11
1 sin 3 sin 
2
 x  x
2sin Acos B  sin  A B  sin  A B A  x, B  2x
P.I.
 2  2
1 1 sin 3 1 1 sin
2 3 2 3
x x
D D
 
 
2 2
1 1 sin 3 1 1 sin
2 6 9 2 6 9
x x
D D D D
 
   
P.I. 1 1 sin 3 1 1 sin
2 9 6 9 2 1 6 9
x x
D D
   
     
1 1 sin 3 1 1 sin
12 2 8 6
x x
D D
   

 
 
  
cos3 4 3 sin
12 3 4 4 3 4 3
x D x
D D
 
 
 
  2
1cos3 1 1 4sin 3cos
36 4 16 9
x x x
D

  

1cos3 1 4sin 3cos
36 4 16 9
x x x  
  

cos3 sin 3cos
36 25 100
 x x x
  
Problem 5 Solve D2  4 y  x4  cos2 x
Solution:
The A.E. is m2  4  0
m  2i
C.F.: Acos 2x  Bsin 2x
P.I 4
2 2
1 1 1 cos 2
4 4 2
x x
D D
          
 
4 0
2 2 2
1 1 1 1 1 1 cos 2
4 2 4 2 4 1
4
x e x x
D D D
  
   
  
 
 
 
  
2 1
4 1 1 sin 2 1
4 4 2 4 2 2 2
D x x x
  
     
 
2 4
1 1 4 1 sin 2
4 4 16 8 8
D D x x x
 
      
 
4 12 2 4.3.2.1 1 sin 2
4 16 64 8 8
 x  x    x x
Unit. 1 Ordinary Differential Equations
12
G.S. is
4 3 2 4 sin 2 cos 2 sin 2
8 4 4 8
y  A x  B x   x  x  x x
Problem 6 Solve    D2  2D 1 y  x  ex 2  cos 2x cosh x .
Solution:
The A.E is m2  2m1  0
2 4 4 1 2
2
m   
   
C.F.:  1 2x  1 2 x Ae Be     
P.I.      
  2
2 2
1 2 1 cos 2
2 1 2 1 2
x x
x x e e
x xe e x
D D D D
 
   
   
   
2 2
2 2
1 1
2 1 1 2
x x
D D D D
 
     
     1 2D  D2  2D  D2 2 .... x2  
  1 2D  D2  4D2  x2  x2  
 x2  4x  52
2 2
2
1 4 10
2 1
x x x
D D
   
 
2  2  
2 2
2 1 1 2 1 1
x
xex e x
D D D D
 
  
         
2
2 1
2 1 2 2 1
ex x
D D D

    
  2
2
4 2
ex x
D D

 
2
2 1
2
1 2
2
ex x
D D

  
    
  
2 1
1 2
2
ex D D x
   
     
  
2
1 2 ...
2
ex D D x
   
       
   
 ex 1 2D x
      2
2 2 2
2 1
xex ex x x ex
D D
   
 
Unit. 1 Ordinary Differential Equations
13
 
2
2 2
2
1 1
2 1 4 4 1 7
x
e x e x e
D D
 
   
2 2
1 cos 2 1 cos 2
2 1 2 2 1 2
ex x e x x
D D D D


   
 2    2  
1 cos 2 1 cos 2
2 1 2 1 1 2 1 2 1 1
ex e x x x
D D D D

 
       
2
1 cos 2 1 cos 2
2 2 1 2 2 1 2 4 2
ex e x x x
D D D

 
      
 
  
2 1 cos 2
cos 2
2 2 2 1 2 1 12
ex D x e x x
D D
 
 
 
    2
1 2.2sin 2 cos 2 cos 2
2 2 4 1 12
ex e x x x x
D

   

 
 
4sin 2 cos 2 cos 2
4 16 1 12
ex  x  x ex x
 
 
cos 2 4sin 2  cos 2
17 12
ex x  x ex x
  
The General Solution is
     
 
2
1 2 1 2 10 4 2 2
7
cos 2 4sin 2 cos 2
17 12
x
x x x
x x
y Ae Be x x e x e
e x x e x
   

       
  
Problem 7 Solve D2  4 y  x2 cos 2x
Solution:
The A.E is m2  4  0
m2  4
m  2i
C.F.: Acos 2x  Bsin 2x
P.I  2 
2
1 cos 2
4
x x
D


 
2
2 2 2
2 2
. 1 . .
4 2 4
ix
R P of x ei x R P of e x
D D i
 
  
P.I. 2 2
2
. 1
4 4 4
R P of e ix x
D iD

  
 
2 2 2 2
2
. 1 . 1
4 4
R P of e ix x R P of e ix x
D iD D D i
 
 
Unit. 1 Ordinary Differential Equations
14
. 2 1 1 2
4 1
4
R P of e ix x
D i D
i

    
 
2 1
. 1 1 2
4 4
e ix D R P of x
i D i
      
 
2 2
. 1 1 2
4 4 16
e ix D D R P of x
i D i
 
    
 
2 3 2
.
4 3 4 8
ie ix x x x R P of
i
   
     
  
2 3 2
.
4 3 4 8
ie ix x ix x R P of
   
     
  
2 3 2
.
4 3 4 8
e ix x i x ix R P of
   
     
  
cos 2 sin 2  3 2
.
4 3 4 8
x x x i x ix R P of
   
    
 
1 2 cos 2 3 sin 2 sin 2
4 4 3 8
 x x x x x x 
    
 
P.I.
1 2 cos 2 3 sin 2 sin 2
4 4 3 8
 x x x x x x 
    
 
G.S.:
2 cos 2 3 sin 2 sin 2 cos 2 sin 2 .
16 12 32
y  A x  B x  x x  x x  x x
Problem 8 Solve D2  a2  y  sec a x .
Solution:
The A.E. is m2  a2  0
m2  a2
m   ai
C.F.: Acos ax  Bsin a x
P.I      1 sec a x 1
D ai D ai
 
 
Using partial fractions
1 2
2 2
1 C C
D a D ai D ai
        
    1 2 1  C D  ai C D  ai
1 2
1 , 1
2 2
C C
ia ia
  
Unit. 1 Ordinary Differential Equations
15
P.I.    
1 1 sec 1 1 sec
2 2
ax a x
ia D ai ia D ai
  
 
 
1 1 sec sec
2 2
aix
ax e e aix a x dx
ia D ai ia
   
  
sec sec
2 2
aix aix
e eaix a x dx e e aix a x dx
ia ia

     
P.I.
cos i sin  cos i sin 
2 cos 2 cos
e aix ax ax eaix ax ax dx dx
ia ax ia ax
  
    
P.I. 1 tan  1 tan 
2 2
e aix eaix i ax dx i ax dx
ia ia

      
log sec logsec
2 2
e aix i eaix i x ax x ax
ia a ia a
             
  2
2 2 log sec
2 2 2 2
x eaix e aix i eaix e aix ax
a i ia
       
     
   
   2
x sin ax 1 log sec ax cos ax
a a
 
  2
1 ax sin ax cos ax log cos ax
a
 
G.S. is y  C.F  P.I.
Problem 9 Solve D2  4D  3 y  ex sin x  xe3x .
Solution:
The A.E is m2  4m 3  0
m1m 3  0
m  1,3
C.F.: Aex  Be3x
P.I      
1 sin 1 3
3 1 1 3
e x x xe x
D D D D
  
   
         
3
sin
1 3 1 1 3 1 3 3
e x e x x x
D D D D

 
       
    
1 sin 3 1
2 4 6
e x x e x x
D D D D
  
  
  
3
2
2 cos 1
2 2 10 24
e x D x e x x
D D D D
 
  
   
   
3
2 2
1 sin 2cos 1
4 24 1 10
24 24
x
e x x x e x
D D D
     
  
Unit. 1 Ordinary Differential Equations
16
P.I.    
3 2 1
2
1 sin 2cos 1 5
4 24 12 24
x
e x x x e D D x
D

  
         
 
 
sin 2cos 3 5 1
1 4 24 12
e x x x e x D x
            
 
3 5 sin 2cos
5 24 12
e x e x x x x
         
 
G.S. is  
3
3 sin 2cos 5 .
5 24 12
x x
y Ae x Be x e x x e x

         
 
Problem 10 Solve the Legendre’s linear equation
3x  22 D2  33x  2D 36 y  3x2  4x 1
Solution:
Let 3x  22  D2  33x  2D 36 y  3x2  4x 1
Let 3x  2  et or t  log 3x  2
3
3 2
dt
dx x
  

3x  ez  2
1 2
3 3
x  ez 
Let 3x  2D  3D
3x  22 D2  9DD 1
9  1 33  36 3 1 2 4 1 2 1
3 3 3 3
 D D   D   y   ez     ez       
9 2 9 9 36 3 1 2 4 4 4 8 1
9 9 9 3 3
 D  D  D   y   e z   ez   ez      
9 2 36 1 2 4 4 4 8 1
3 3 3 3 3
 D   y  e z   ez  ez    
1 2 1
3 3
 e z 
A.E is 9m2 36  0
9m2  36
m2  4
m   2
C.F  Ae2 z  Be2 z
 2   2 A 3x 2 B 3x 2     
Unit. 1 Ordinary Differential Equations
17
2
1 2
. 1
9 36 3
e z P I
D

 
1 . 1 2
3 36 36
 e z

1 1 2
3 18
z e z
D


1 2
54 2
e z  z
1 2
108
 ze z
1 log 3 23 22
108
 x  x 
0
2 12
. 1
9 36 3
e z P I
D


1. 1 0
3 36
 e z

1
108
 
1 2 y  C.F  P.I  PI
3 22 3 2 2 1 3 22 log3 2 1
108 108
A x B x x x         
3 22 3 2 2 1 3 22 log 3 2 1 .
108
A x B x x x             
Problem 11 Solve D2 5D 4 e x sin 2x x2 1whereD d
dx
       .
Solution:
The A.E m2  5m 4  0
m  4or m  1
C.F  Ae4x  Bex
 2 
2
. 1 sin 2 2 1
5 4
P I e x x x
D D
   
 
   
3  2 
2 2
1 sin 2 1 1
1 5 1 4 4 1 5
4
e x x x
D D D D
  
         
 
  2
3 2
2
1 sin 2 1 1 5 5 1
3 4 4 16
e x x D D x
D D
 
         
2sin 2 3 2  1 2 5 13
26 4 2 8
e x x cos x x x
          
 
G.S : y  C.F. P.I
Unit. 1 Ordinary Differential Equations
18
4 2sin 2 3cos 2  1 2 5 13 .
26 4 2 8
x
y Ae x Be x e x x x x

          
 
Problem 12 Solve  
2
2
2 x d y 4x dy 6y sin log x .
dx dx
  
Solution:
Given equation is  
2
2
2 x d y 4x dy 6y sin log x .
dx dx
  
x2D2  4xD  6 y  sin log x1
Put x  ez or  z  log x
xD  D2
x2D2  DD 13 Where D denotes d
dz
Sub (2) & (3) in (1) we get
DD 1  4D  2 y  sin z
i.eD2  D  4D  2 y  sin z
D2  3D  2 y  sin z 4
The A.E is m2  3m 2  0
m1m 2  0
m  1,2
C.F.: Aez  Be2z
P.I.= 2
1 sin
3 2
z
D D    
1 sin
1 3 2
z
D

   
1 sin
3 1
z
D

 
2
3 1 sin
9 1
D z
D
 

 
 
 
2 3 1 sin
Replace D 1
9 1 1
D z
by
 
      
3 sin  sin
10
D z  z


3cos sin
10
z  z


The solution of (4) is
2 3cos sin
10
y Aez Be z z  z
  

Unit. 1 Ordinary Differential Equations
19
Sub z  log x or x  ez , we get
    log 2log 3cos log sin log
10
x x x x
y Ae Be 
  
    1 2 3cos log sin log
10
x x
y Ax Bx 
  
   
2
3cos log sin log
10
A B x x y
x x

  
This gives the solution of the given differential equation.
Problem 13 Solve the simultaneous ordinary differential equation
D  4 x  3y  t , 2x  D  5 y  e2t
Solution:
Given D  4 x  3y  t 1
2x  D 5 y  e2t 2
21 D 42
6y  D  4D  5 y  2t  D  4e2t
6  D2  9D  20 y  2t  2e2t  4ezt  
D2  9D 14 y  6e2t  2t
The A.E. is m2  9m14  0
m 7m 2  0
m  2,  7
C.F.: Ae2t  Be7t
P.I.    
2
2 2
6 2
9 14 9 14
e t t
D D D D
 
   
 
2
2
6 2 1
4 18 14 14 1 9
14 14
e t t
D D
 
   
 
6 2 1 9 2 1 1
36 7 14 14
e t D D t
  
     
 
 
2 1 9 2 1 9 1
6 7 14 6 7 14
e t D e t      t    t  
   
G.S. is
2
2 7 9
6 7 98
t
y  Ae t  Be t  e  t 
To Calculate x
2
2 2 7 7 2 1
6 7
t
Dy   Ae t  Be t  e 
Unit. 1 Ordinary Differential Equations
20
2
5 5 2 5 7 5 5 45
6 7 98
t
y  Ae t  Be t  e  t 
 
2
5 3 2 2 7 7 5 1 45
6 7 7 98
t
D  y  Ae t  Be t  e  t  
22x  D  5 y  e2t
2
3 2 2 7 7 5 31 2
6 7 98
t
  Ae t  Be t  e  t   e t
3 2 7 7 2 5 31
2 72 14 196
x  A e t Be t e t t
    
The General solution is
2
3 2 7 5 31
2 12 14 196
t
x  A e t Be t e t
    
2
2 7 9 .
6 7 98
t
y  Ae t  Be t  e  t 
Problem 14 Solve:
2
2 d y y tan x
dx
  by method of variation of parameters
Solution:
A.E is m2 1  0
m  i
C.F = 1 2 c cos x  c sin x
1 2 P.I  PI  PI
1 f  cos x ; 2 f  sin x
1 f   sin x ; 2 f   cos x
2 1 1 2 f  f  f f 1
Now, 2
1 2 1 2
P f X dx
f f f f
     
=  sin x tan xdx
=
sin 2 ( 1 co s 2 )
co s co s
x d x x d x
x x
 
   
= sec xdx  cos xdx
= logsec x  tan x  sin x
1
1 2 1 2
Q f X dx
f f f f
    
=  cos x tan xdx
= cos x
 y = C.F + Pf1 + Qf2
Unit. 1 Ordinary Differential Equations
21
= c1 cos x  c2 sin x +[log sec x  tan x  sin x]cos x cos x sin x
= 1 2 c cos x  c sin x log(sec x  tan x) cos cos x .
Problem 15 Solve by the method of variation of parameters
2
2 d y 4y sec 2x
dx
 
Solution:
The A.E is m2  4  0
m  2i
C.F = 1 2 c cos 2x  c sin 2x
P.I = Pf1 + Qf2
f1 = cos 2x; f2 = sin 2x
1 f   2sin 2x ; 2 f   2cos 2x
2 1 1 2 f  f  f  f  2
Now, 2
1 2 1 2
P f X dx
f f f f
     
= sin 2 sec 2
2
 x xdx
= 1 tan 2 1 log cos 2 
2 4
  xdx  x
1
1 2 1 2
Q f X dx
f f f f
    
= 1 cos 2 sec 2 1
2 2
 x xdx  x
 y = C.F + Pf1 + Qf2
= 1 2 c cos 2x  c sin 2x 1 log cos 2 cos 2
4
 x x 1 sin 2
2
 x x .
Previous
Next Post »