JSS MAHAVIDYAPEETHA

JSS ACADEMY OF TECHNICAL EDUCATION, NOIDA

DEPARTMENT OF MATHEMATICS

UNIT-1
Ordinary Differential Equations of Higher Order

C121.1 Remember the concept of differentiation to evaluate LDE of nth order with
constant coefficient and LDE with variable coefficient of 2nd order.
 TOPICS COVERED

1.1 Definition

1.2 Order and Degree of an ODE

1.3 Formation of Ordinary Differential Equations

1.4 Linear Differential Equation with constant coefficient

1.4.1 Rules for finding C.F.

Questions Based on C.F

1.4.2 Rules for finding P.I.

Questions Based on P.I

1.5 Linear Differential Equation With Variable coefficient

1.5.1 Cauchy Euler Linear Differential Equation

Questions Based on Cauchy Euler Differential Equation

1.6 Simultaneous differential equation

Question Based on Simultaneous Differential Equations

1.7 Second Order Linear Differential Equation with Variable Coefficient

1.7.1 Method of Variation of Parameters for Finding Particular Integral

Question Based on Method of Variation of Parameters

1.7.2 Changing the Independent Variable

Question Based on Change of Independent Variable

1.8 Applications of Differential equations

Question Based on Application of Ordinary Differential Equation

ANSWER KEY
1.1 Definition

Introduction to Ordinary Differential Equations (ODE)

A differential equation is an equation which involves differentials or differential coefficients.
Differential equation is an equation involving one or more functions with its derivatives. The
derivatives of the function define the rate of change of a function. Differential equations are used
to model physical phenomena that involves rate of change.
d2y dy dy
Examples: (i) 2
 2  y  sin x (ii)  x2
dx dx dx

1.2 Order and Degree of a ODE

Order is the highest derivative present in the differential equation and degree is the exponent of
the highest derivative.

 d 2 y   dy 
2

Examples: (i)  2   2   y  2 Order = 2 and Degree = 3

 dx   dx 

1.3 Formation of Ordinary Differential Equations

The following are the steps involved in forming an ODE:
Given the general solution of a ODE in the form f x, y, c1 , c2 ,.........cn   0
f x, y, c1 , c2 ,.........cn   0 (1)
Step 1: Find the number of arbitrary constants ‘n’ present in equation (1).
Step 2: Differentiate (1) w.r.t. independent variable x, present in (1).
Step 3: Keep differentiating ‘n’ times, so that (n+1) equations are obtained.
Step 4: Using the (n+1) equations are obtained, eliminate the constants c1 , c2 ,.........cn .

1.4 Linear Differential Equation with constant coefficient

dny d n 1 y d n2 y
a0  a  a           an y  Q (1)
dx n 1 dx n  2
1 3
dx n
Here a0 , a1 , a3..........an are constant. Q is a function of x .

Equation (1) can be represented by

f ( D) y  Q (2)

(A) In equation (2) If Q  0 then f ( D) y  0 . The solution will be

y  Complement ary Function  C.F.
 (B) In equation (2) If Q  0 then f ( D) y  Q . The solution will be
y  Complement ary Function  Particular Integratio n  C.F.  P.I.

1.4.1 Rules for finding C.F.

Case A f ( D) y  0

Step 1. Make Auxiliary : Put y  emx in f ( D) y  0 so that it becomes

f (m)  0 (3)
Step2. Solve (3) for m , so following cases arise

(i) If values of m are real and distinct e.g. (m  m1, m2 ,m3 ........mn ) then solution will
be y  c1em1 x  c2em 2 x      
(ii) If values of m are real and repeated e.g. (m  m1  m2 m3  ........mn ) then solution
will be y  c1  c 2 x  c 3 x 2  ..........cn xn emx
(iii) If values of m are Imaginary and distinct e.g. (m  (1  i1, 2  i 2 ............))
then solution will be
y  e1 x c1 cos 1 x  c2 sin 1 x   e 2 x c3 cos  2 x  c4 sin  2 x   ..........
(iv) If values of m are Imaginary and repeated e.g. (m  (1  i1  2  i2 ............))
then solution will be
y  e1x (c1  c 2 x  c3 x 2  .......) cos 1 x  (d1  d 2 x  d 3 x 2  ......) sin 1 x 
(v) If values of m are surds and distinct e.g. (m  (1  1 ,  2   2 ............)) then
solution will be
y  e1 x c1 cosh 1 x  c2 sinh 1 x   e 2 x c3 cosh  2 x  c4 sinh  2 x   ..........
OR
(  2  2 ) x
y  c1e(1  1 ) x  c2 e 
(vi) If values of m are surds and repeated e.g. (m  (1  1 ,  2   2 ............)) then
solution will be
y  e1x (c1  c 2 x  c3 x 2  .......) cosh 1 x  (d1  d 2 x  d 3 x 2  ......) sinh 1 x 

OR

y  c1  c2 x  c3 x 2  ..........cn x n e  ( 1  1 ) x
Questions Based on C.F
Solve the following Differential equations
Ques1. ( D 4  10D3  35D 2  50D  24) y  0 SOLVED

Ques2. ( D  2D  D) y  0
3 2
SOLVED

Ques3. ( D 4  2D 2  1) y  0 SOLVED

Ques4. ( D3  2D  4) y  0 SOLVED

Solve the differential equation where R C  4 L and R, C , L are constants

2
Ques5.
d 2i R di i
2
  0
dt L dt LC

1.4.2 Rules for finding P.I.

Case B f ( D) y  Q Q0

y  C.F  P.I

**Rules for finding C.F we have discussed earlier.

Rules for finding P.I
1
P.I  Q
f ( D)
Resolve f (D ) into Linear factors
General Method
1
For P.I  Q
(D  a)(D  b)
For D  ay  Q

The general Rule will be P.I = e ax  Qe  ax dx …………..4(i)

For D  by  Q
The general Rule will be P.I = e bx  Qe bx dx …………..4(ii)
Steps to Follow

Step 1. Take f (D ) into denominator for P.I.

1
i.e. P.I. = Q
f ( D)
Step 2. Now factorize f (D ) into linear factor.
Step 3. Apply Partial fraction to separate Linear factor
Step 4. Now apply Linear factor on Q as in 4(i) and 4(ii).

Short cuts for Different Q

Case 1. When Q  eax

1 ax 1 ax
e  e f (a)  0
f ( D) f (a)

If f (a )  0 for Q  eax then multiply numerator by x and differentiate denominator with

respect to D .Repeat until the denominator is not zero.

Example
1
Find P.I. of eax
( D  a) 2
x
Solution P.I .  eax  f (a)  0
2( D  a)
x 2 ax
Again P.I .  e
2
Remarks:
This case can also be applied on following functions
 If Q  cons tan t so we can write it Q  e 0 x
 If Q  a x so we can write as Q  e x log a .
e x  e x
 If Q  cosh x so we can write as Q 
2 .
e x  e x
 If Q  sinh x so we can write as Q 
2 .

Case 2. When Q  sin ax or cos ax

 Case 3. When Q  x n (n is a positive integer)

Case 4. When Q  eaxV where V is a function of x.

1 ax 1
e V  eax V
f ( D) f ( D  a)

Note 1 . If V is sin ax or cos ax then proceed as in case 2

2. If V is x n then proceed as in case 3

When Q  x nV where V is a function of x.

Note 1 . If V is e ax then solve it as in case 4 by writing e ax x n

2. If V is sin ax then write it Imaginary part of e iax then solve it as in case 4 by
writing IM.P of e iax x n .
3. If V is cos ax then write it Real part of e iax then solve it as in case 4 by writing
R.P of e iax x n .
Questions Based on P.I
Solve the following Differential equations

Ques6. ( D 2  D  2) y  e x SOLVED

Ques7. 
( D3  1) y  e x  1 2

Ques8. ( D2  4) y  sinh( 2 x  1)  4 x  5 SOLVED

Ques9. ( D  D  2) y  sin x
2

Ques10. ( D  4 D  1) y  cos x cos 4 x  sin

2 2
x
Ques11. ( D 2  2D  1) y  cos 2x SOLVED

Ques12. ( D 2  9) y  sin 2 x cos x

Ques13. ( D  1) y  5 x  2
2
SOLVED

Ques14. ( D  6 D  9) y  1  x  x
2 2
SOLVED

Ques15. ( D 2  2) y  x 2e x SOLVED

Ques16.
( D 2  2D  1) y  e x sin x
Ques17.
( D 2  4) y  x sinh x

Ques18. ( D 2  1) y  x 2 sin 2 x SOLVED

Ques19. ( D 2  1) y  x sin x  (1  x 2 )e x SOLVED

Ques20.
( D 4  6D3  11D 2  6D) y  20e 2 x sin x

Ques21.
( D 2  2D  2) y  e  x sec3 x
( D2  3D  2) y  ee
x
Ques22. SOLVED

Ques23. ( D  4) y  tan 2 x
2

Ques24. ( D 2  4D  4) y  8x 2e2 x sin 2 x

Ques25. ( D  2D  4) y  e cos x  sin x cos 3x
2 x
1.5 Linear Differential Equation With Variable coefficient
Differential Equations Reducible to Linear Form with Constant Coefficients

Some special type of homogenous and non homogeneous linear differential equations with
variable coefficients after suitable substitutions can be reduced to linear differential equations
with constant coefficients.

1.5.1 Cauchy Euler Linear Differential Equation

n 1 n2
dny n 1 d y n2 d y
a0 x n n
 a1 x n 1
 a3 x n2
          an y  Q
dx dx dx
is called Cauchy’s linear equation and it can be reduced to linear differential equations with
constant coefficients by following substitutions:

Legendre’s Linear Differential Equation (Content Beyond Syllabus)

The differential equation of the form:
dny d n 1 y d n2 y
a0 (ax  b) n n  a1 (ax  b) n 1 n 1  a3 (ax  b) n  2 n  2           an y  Q
dx dx dx
is called Legendre’s linear equation and it can be reduced to linear differential equations with
constant coefficients by following substitutions:
Questions Based on Cauchy Euler Differential Equation

Solve the following Differential equations

d3y 2
2 d y dy
Ques26. x 3
 3x
3
2
 x  8 y  13 log x x0 SOLVED
dx dx dx

d2y dy
Ques27. x 2 2
 2 x  4 y  x 2  2 log x x0
dx dx

d2y
Ques28. 3x  2  33x  2  36 y  3x 2  4 x  1
2 dy
2 SOLVED
dx dx

d2y
Ques29. x  1 2  x  1 dy  y  2 sin(log( x  1)), x  1
2

dx dx
2

Ques30. x  12 d y2  x  1 dy  (2 x  3)(2 x  4)

dx dx

d2y dy
Ques31. x 2
 x  y  x 3e x
2

dx dx

1.6 Simultaneous differential equation

Linear differential equations having two or more dependent variables with single independent
variable are called simultaneous differential equations

d
f1 ( D) x  f 2 ( D) y  F (t ) g1 ( D) x  g 2 ( D) y  G (t ) D
dt
Given system can be solved as follows:
1. Eliminate y from the given system of equations resulting a differential equation
exclusively in x .
2. Solve the differential equation in x by usual methods to obtain x as a function of t
3. Substitute value of x and its derivatives in one of the simultaneous equations to get an
equation in y .
4. Solve for y by usual methods to obtain its value as a function of t
Question Based on Simultaneous Differential Equations

Solve the following Simultaneous Differential equations

dx dy
Ques32.  7 x  y  0,  2x  5 y  0
dt dt
dx dy
Ques33.  y  et ,  x  e t SOLVED
dt dt

d 2x d2y
Ques34.  3x  4 y  0, x y 0 SOLVED
dt 2 dt 2
d 2x dx d2y dy
Ques35.  4  4 x  y,  4  4 y  25 x  16et
dt 2 dt dt 2
dt
d 2x d2y
Ques36. 2
 4x  5 y  t 2 ,  5x  4 y  t  1
dt dt 2

1.7 Second Order Linear Differential Equation with Variable Coefficient

1.7.1 Method of Variation of Parameters for Finding Particular Integral
Method of Variation of Parameters enables us to find the solution of 2 nd and higher order
differential equations with constant coefficients as well as variable coefficients.
Working rule Consider a 2nd order linear differential equation:

d2y dy
2
 P  Qy  R( x) (1)
dx dx

1. Find complimentary function given as: C.F.  c1 y1 c2 y2 where y1 and y2 are two linearly
independent solutions of (1).
y y2
2. Calculate 1' , W is called Wronskian of y1 and y2.
y1 y2'
y 2 R( x ) y R( x)
3. Compute A   dx and B   1 dx
W W
4. Find Solution by P.I  Ay1  By2
5. Complete solution is given by: C.F. + P.I

Note: Method is commonly used to solve 2nd order differential but it can be extended to solve
differential equations of higher orders.
Question Based on Method of Variation of Parameters

Ques37. Solve the differential equation by using the method of variation of parameter
d2y dy
2
 3  2 y  sin( e x )
dx dx

Ques38. Solve the differential equation by using the method of variation of parameter
d2y dy
2
 2  2 y  e x tan x
dx dx
Ques39. Solve the differential equation by using the method of variation of parameter
d2y dy e3 x
 6  9y  2 SOLVED
dx 2 dx x

Ques40. Solve the differential equation by using the method of variation of parameter
d2y 2
y
dx 2
1  ex
Ques41. Solve the differential equation by using the method of variation of parameter
d2y
 y  sec x tan x SOLVED
dx 2
1.7.2 Changing the Independent Variable
Let the linear differential equation of second order be 𝑦"+𝑃𝑦′+𝑄𝑦=𝑅 (1) where P, Q and R are
functions of x.
Procedure to solve equation (1)
Step 1: If the independent variable x is changed to z (z being a function of x) then (1)
d2y dy
becomes 2
 P1  Q1 y  R1
dz dz
d2y dy
P
Step 2: Here P1  dz
2
dz Q  Q , R  R
2 1 2 1 2
 dz  ,  dz   dz 
     
 dx   dx   dx 

 dz 
Step 3: To compute   , Q1 is equated to a constant.
 dx 
 dz 
Step 4: To find z integrate   and substitute in R1 .
 dx 
Step 5: Solve equation (2) by finding 𝑦=𝐶.𝐹.+𝑃.𝐼.
Step 6: Replace the value of z by a function of x.
Normal Form or Changing the dependent Variable(Content Beyond Syllabus)

(ii) By changing the dependent variable y to v

Procedure to solve equation (1)
d2y
Step 1: Compare with equation (1) and find P, Q, R (Make coefficient of  1 ).
dx 2
Step 2: Let complete solution is 𝑦=𝑢𝑣.
1
1 dP P 2 R   Pdx
Step 3: Calculate I  Q   and S  , where u  e 2
2 dx 4 u
2
d v
Step 4: Here v is obtained from  Iv  S .............(1)
dx 2
(i) If 𝐼= constant then solve (1) as LDE with constant coefficients
cons tan t
(ii) If I  then solve (1) as Cauchy Euler.
x2
Step 5: After Solving (1) put 𝑢 and 𝑣 in 𝑦=𝑢𝑣.

Question Based on Change of Independent Variable

Ques42. Solve the differential equation by changing the independent variable

d2y dy
2
 cot x  4 y cos ec 2 x  0
dx dx
Ques43. Solve the differential equation by changing the independent variable
d2y

x 2  4x2  1
dx
 dy
dx
 4 x3 y  2 x3 SOLVED

Ques44. Solve the differential equation by changing the independent variable

d2y dy
cos x 2  sin x  2 y cos3 x  2 cos5 x SOLVED
dx dx
Ques45. Solve the differential equation by changing the independent variable
d 2 y dy
x 2   4 x3 y  8 x3 sin x 2
dx dx
Ques46. Solve the differential equation by reducing it into normal form
d2y dy
2
 2 tan x  5 y  sec xex
dx dx

1.8 Applications of Differential equations

A) Applications of Second Order Differential Equations To Mechanical and Electrical
Oscillatory Circuits
a) Free Oscillations
d 2x k
Mechanical Systems 2
2x  0 2 
dt m
 Solution of the Systems is
x  Acos(t  B)

d 2q 1
Electrical Circuits 2
  2q  0 2 
dt LC
Solution of the Systems is
q  Acos(t  B)

b) Damped Free Oscillations

d 2x dx k
Mechanical Systems 2
 2p 2x  0 2 
dt dt m
Solution of the Systems is
x  e  pt  c1e p  t  c2 e  p  t  p 
2 2 2 2

 
x  e pt c1  c2t  p 

x  e  pt c1 cos w2  p 2 t  c1 sin w2  p 2 t  p 
2
d q dq 1
Electrical Circuits 2
 2p  2q  0 2 
dt dt LC
q  e  pt  c1e p   c2 e  p  t  p 
2 2 2 2
t

 
q  e c1  c2t 
 pt
p 

q  e  pt c1 cos w 2  p 2 t  c1 sin w 2  p 2 t  p 

c) Forced Oscillations
d 2x k
Mechanical Systems 2
  2 x  E cos nt 2 
dt m
Solution of the Systems is
E
x  Acos(t  B)  2 cos nt n
  n2
d 2q 1
Electrical Circuits 2
  2 q  E cos nt 2 
dt LC
Solution of the Systems is
E
q  Acos(t  B)  2 cos nt n
  n2

d) Damped Forced Oscillations

d 2x dx k
Mechanical Systems 2
 2 p   2 x  E cos nt 2 
dt dt m
 Solution of the Systems is
  2 pn 
E cos nt  tan 1  2  
  n 
2
x  e  pt  c1e p 2  2 t
 c2 e  p 2  2 t  

 
 
 2  n2  4 p2n2
2

d 2q dq 1
Electrical Circuits 2
 2p   2q  E cos nt 2 
dt dt LC
  2 pn  
E cos nt  tan 1  2 
2 
    n 
q  e pt  c1e p 2  2 t
 c2e  p 2  2 t 

   2  n2 2  4 p 2n2
B) Applications of Second Order Differential Equations in SHM
B) d 2x
The equation of SHM is 2
  2 x
dt
Question Based on Application of Ordinary Differential Equation

Ques47. A inductance of 2H and a resistance of 20ohm are connected in series with an

e.m.f E volts . If the current is zero when t=0, find the current st the end of 0.01s,
if E-100V.
Ques48. The equation of emf in terms of current i for an electric circuit having resistance
i
R and a condenser of a capacity C in series is E  Ri   dt. Find the current i
c
at any time t when E  E0 sin t SOLVED

Ques49. A body executes damped forced vibrations given by the equation

d 2x dx
2
 2k  b 2 x  e  kt sin t Solve the equations for both the cases
dt dt
  b  k 2 and when  2  b 2  k 2
2 2
SOLVED

Ques50. A mass M suspended from the end of a helical spring is subjected to a periodic
force f  F cos t in the direction of its length. The force f is measured
positive vertically downwards and at zero time M is at rest. If the spring stiffness
is k , prove that the displacement M at a time t from the commencement of
motion is given by x 
F
cos t  cos pt  where p 2  k and
M(p  w )
2 2
M
damping effects are neglecting.
Ques51. A particle moving in a straight line with SHM has velocities  1 and  2 when its
distances from the centre are x1 and x2 . Show that the period of motion is
x12  x22
2
 22   12
ANSWER KEY
Ans1. y  c1e x  c2 e 2 x c 3 e3 x c 3 e 4 x
Ans2. y  c1  c2 c 3 x e  x
y  (c1  c2 x)e x  (c3  c4 x)e  x
Ans3.

Ans4. y  c1e 2 x  c2 cos x c 3 sin x e x

R

i  c1 c 2 t e
t
2L
Ans5.

2 x xex
Ans6. y  c1e  c2e  x

 
x
3 3   2  1
Ans7. y  e  c1 cos2
x  c2 sin x   e x  c3  x   e 2 x  1
 2 2   3  7
x 2 x 4x 5
Ans8. y  c1e  c2 e  cosh( 2 x  1) 
2x

4 (log 4)  4 4
2

y  c1e  2 x  c2 e x  cos x  3 sin x 

1
Ans9.
10
Ans10. y  c1e ( 2 3)x
 c2 e ( 2  3)x

1
5 sin 5x  6 cos 5 x   1 3 sin 3x  2 cos 3x   1  1 8 sin 2 x  3 cos 2 x 
488 104 73

y  c1  c2 x e x    4 sin 2 x  3 cos 2 x 

1 1
Ans11.
2 50

x 1
Ans12. y  c1 cos 3x  c2 sin 3x  cos 3x  sin x
12 16

Ans13. y  c1e x  c2e  x  5x  2

 1  7 7x 
Ans14. y  e 3 x c1  c2 x      x2 
93 3 

Ans15. 
y  c1 cos 2 x  c2 sin 2 x 
e3 x  50 12 x
 
11  121 11
 
 x2 


Ans16. y  c1  c2 x e x  e x sin x

Ans17.  x
2
y  c1e 2 x  c2 e  2 x  sinh x  cosh x
3 9

Ans18.  x
 2
y  c1e 2 x  c2 e 2 x  sinh x  cosh x
3 9

Ans19. y  c1 cos x  c2 sin x  

8x
9
cos 2 x 
1
27

26  9 x 2 sin 2 x 

Ans20. 
y  c1  c2e  x  c3e 2 x  c4 e 3 x  2e 2 x (sin x  2 cos x) 
 sin x tan x   x
Ans21. y   c1 cos x  c2 sin x  e
 2 

e x  x 3 x 2 3x 
Ans22. 
y  c1e  c2ex x
  x sin x  cos x      
1
2 23 2 2

Ans23.  
y  c1e x  c2e2 x  e2 x ee
x

Ans24. 
y  c1  c2 x e2 x  e2 x  2 x 2 sin 2 x  4 x cos 2 x  3 sin 2 x 

Ans25. 
y  e x c1 cos 3x  c2 sin 3x   e x cos x 1
2
 cos 2 x 
8
1
104
2 cos 4 x  3 sin 4 x 

Ans26. y
c1
x 2
 5

 x c2 cos( 3 log x)  c3 sin( 3 log x)  8 cos(log x)  sin(log x) 
1
 c1 x2 1  3
Ans27. y  c2 x    log x  
x 6 2 4

Ans28. y
c1
3x  22
 c2 3x  2 
2 1
108

3x  22 log( 3x  2)  1 
Ans29. y  c1 coslog x  1  c2 sin log x  1  log x  1coslog x  1

y  c1  c2 log( x  1)  ( x  1) 2  6( x  1)  log( x  1)

2
Ans30.

1 x2 1  3
Ans31. y  c1x  c2    x  3  e x
x 6 2 x

Ans32. x  e6t c1 cos t  c2 sin t  y  e6t c1  c2 cos t  c1  c2 sin t 

x  c1 cos t  c2 sin t   et  e  t y  c1 sin t  c2 cos t   et  e  t

1 1 1 1
Ans33. 2 2 2 2
Ans34. y  c1  c 2 t e  c3  c 4 t e
t t

x   2c1  2c 2  2c 2 t e  2c3  2c 4  2c 4 t e t
t


Ans35. x  c1e3t  c2e 3t  c3 cos t  c4 sin t  8et , y  c1e3t  25c2e 3t  3c3  4c4 cos t  4c3  3c4 sin t  8et

8 1 44 
Ans36. x  c1et  c2e t  9c3 cos t  9c4 sin t  , y  c1et  c2e t  c3 cos t  c4 sin t   5t 2  4t  
9 9 9 

Ans37. y  c1e x  c2e2 x  e2 x sin e x

Ans38. y  e x c1 cos x  c2 sin x   e x cos x log sec x  tan x 

Ans39. y  e x c1  c2 x   e3 x  e3 x log x

Ans40.   
y  c1e x  c1e x  e x log 1  e x  e x log 1  e x  1 
Ans41. y  c1 cos x  c1 sin x  x cos x  sin x  sin x log sec x
  x  x
Ans42. y  c1 cos 2 log tan   c2 sin  2 log tan 
 2  2

Ans43.  
y  c1  c2 x 2 e  x 
2 1
2

Ans44. y  c1e 2 sin x

 c2e 2 sin x
 sin 2 x

y  c1e x  c2e x  sin x2

2 2

Ans45.

 ex 
Ans46. y  sec x c1 cos 6 x  c2 sin 6 x  
 7

Ans47. i=0.475A

E0C
t

i cos t  RC sin t   ae RC
Ans48. 1  C R
2 2 2

 b  k  t  B b ek  
 kt

Ans49.
kt
Case1 : x  Ae cos
2 2
2 2 2
sin t

e kt t
x  Ae  kt
cos t  B  cos t
Case2: 2