Lecture notes on Ordinary Differential Equations
Khanday M. A.
Unit-II
Solution in series
For solving linear differential equations by this method, we start with the assumption that
it is possible to find a convergent series arranged according to the powers of the
independent variables which approximately satisfies the equation i.e., expresses the value
of the variable.
Let the linear equation
P0
dy n
dy n 1

P
 ...  Pn y  0
1
dx n
dx n 1
have a solution of the form
y  A0 x m0  A1 x m1  A2 x m2  ....  Ar x mr
where the series is either finite or a convergent series for some value or values of x. Now
we have to find the initial term, the relation between the powers of x and the relation
between the coefficients.
Law of exponents: Substituting y = xm,
So that
dy
 mx m 1
dx
d2y
 m(m  1) x m2 , etc., in the given equation.
2
dx
Let we have,
f1 (m) x m  f 2 (m) x m  ...
/
//
Now the successive differences of the powers of x in the series in general should be
m// - m/ = s (say).
Then the series can be written as
y  A0 x  A1 x ms  A2 x
m
m2 s
 ... Ar x mrs  ...

or
y   Ar s m  rs
r 0
So that

dy
  Ar (m  rs ) x m  rs1
dx r 0
d2y 
  Ar (m  rs )m  rs  1x m rs2
2
dx
r 0
etc.
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
1
Lecture notes on Ordinary Differential Equations
Khanday M. A.
Substituting these values in the given differential equation, we obtain an identity with its
right hand member equal to zero. Equating the zero the coefficients of various powers of
x we get a number of equations which determine the value or values of m and the
relations between the various A’s which result in giving a law of means of which Ar+1 can
be determined in terms of Ar whatever may be. Thus the A’s entire are determined in
terms of A0 which remains undetermined.
Remark. For determining the initial power of x for an equation of nth order, put that
coefficient equal to zero which is of nth order in m. Thus we shall get an equation known
as Indicial equations. From this we shall get the value of m.
Case I. Roots of the indicial equation unequal and differing by a quantity not an integer.
d 2 y dy

 6 xy  0
dx 2 dx
Exercise:
Solve (2 x  x 3 )
Solution.
Substituting xm for y in left hand side of the given equation, we have
(2 x  x 3 )m(m  1) x m2  mx m1  6 x m1
(2m 2  2m  m) x m1  (m 2  m  6) x m1
or
Clearly, the difference of successive powers is 2.

Let the solutions be
y

A x
r 0
So that
m 2r
r

dy
  Ar (m  2r ) x m  2 r 1
dx r 0
d2y 
  Ar (m  2r )( m  2r  1) x m 2 r 2
2
dx
r 0
and
Substituting these values in the given equation, we have

 Ar[(2 x  x
3
)( m  2r )( m  2r  1) x m  2 r  2  (m  2r ) x m  2 r 1  6 x m  2 r 1 ]  0
r 0

 Ar[2(m  2r )(m  2r  1)  (m  2r )x
m  2 r 1
 (m  2r )( m  2r  1)  6x m  2 r 1 ]  0
r 0

 Ar[(m  2r ){2(m  2r  1)  1}]x
m  2 r 1
 (m  2r )( m  2r  1)  6x m  2 r 1 ]  0
r 0
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
2
Lecture notes on Ordinary Differential Equations

or
 Ar[{( m  2r )
2
Khanday M. A.
 (m  2r )  6}x m  2 r 1  {2(m  2r ) 2  3(m  2r )}x m  2 r 1 ]  0
r 0

 Ar[(m  2r  3)(m  2r  2) x
m  2 r 1
 {( m  2r )( 2m  4r  3)}x m  2 r 1 ]  0
r 0
which being an identity, we can equate to zero the coefficients of various powers of x.

Equating to zero the coefficient of lowest powers of x. i.e., of xm-1, we have
A0 m(2m  3)  0
Now A0  0 , as it is the coefficient of the first term with which we start to write the
series.

m  0,
m
or
3
2
Again equating to zero, the coefficient of the general term i.e., of xm+2p+1, we have
Ap (m  2 p  3)( m  2 p  2)  (m  2 p  2)( 2m  4 p  1) Ap 1  0

Ap 1  
(m  2 p  3)( m  2 p  2)
Ap
(m  2 p  2)( 2m  4 p  1)
Ap 1  
m  2p 3
Ap
2m  4 p  1
Putting p = 0, 1, 2,… we have
A1  
m3
A0
2m  1
A2  
(m  1)
(m  1)( m  3)
A1  (1) 2
A0
2m  5
(2m  1)( 2m  5)
A3  
(m  1)
(m  1)( m  3)( m  1)
A2  (1) 3
A0
2m  9
(2m  1)( 2m  5)( 2m  9)
etc.

y   Ar x m  2 r  A0 x m  A1 x m  2  A2 x m  4  A3 x m  6  ...
r 0
 A0 x m [1 

m3 2
(m  3)( m  1) 4
(m  3)( m  1)( m  1)
x 
x 
x 6  ...]
2m  1
(2m  1)( 2m  5)
(2m  1)( 2m  5)( 2m  9)
when m  0, taking A0  a , we have
3
1


y  a 1  3x 2  x 4  x 6  ...  au
5
15


(say)
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
3
Lecture notes on Ordinary Differential Equations
Khanday M. A.
This is one solution of the given equation.
3
, taking A0  b , we have
2
Again when m 
3
1.3 4
1.35 6
y  bx 3 / 2 [1  x 2 
x 
x  ...]  bv (say)
8
8.16
8.16.24
Thus y  au  bv is the complete solution of the given equation
Exercise 2. Solve the Besell’s equation of order n, (taking 2n as non integral) completely.
x2
d2y
dy
 x.  ( x 2  n 2 ) y  0
2
dx
dx
Solution. Substituting xm for y in the left hand side of the given equation, we have
x 2 .m(m  1) x m2  x.mx m1  ( x 2  n 2 ).x m
{m(m  1)  m  n 2 }x m  x m 2
or
Clearly difference of successive powers is 2.

Let the solution be
y 

A
r 0
So that
r
x m 2 r

dy
  Ar (m  2r ) x m  2 r 1
dx r 0
d2y 
  Ar (m  2r )(m  2r  1) x m 2 r 2
2
dx
r 0
Substituting these values in the equation, we have

 A [(m  2r )(m  2r  1) x
r 0
r

 A [{( m  2r )
r 0
r
2
m2r
 ( m  2r ) x m  2 r  ( x 2  n 2 ) x m  2 r ]  0
 n 2 }]x m  2 r  x m  2 r  2  0
which being an identity, we can equate to zero the coefficient of various powers of x.

Equating to zero the coefficient of lowest power of x i.e., of x m , we have
A0 m  n (m  n)  0
Now

A0  0
m  n, or m  n
Difference of roots of indicial equation is 2n (not an integer).
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
4
Lecture notes on Ordinary Differential Equations
Khanday M. A.
Again equating to zero the coefficient of the general term i.e., x m  2 p , we have
Ap (m  2p  n)(m  2p - n)  A p-1  0.

Ap  
Ap 1
(m  2 p  n)( m  2 p  n)
Putting P = 1, 2, 3… we have
A1  
1
A0
(m  2  n)( m  2  n)
A2  
1
1
A1  (1) 2
A0 etc.
(m  4  n)( m  4  n)
(m  2  n)( m  2  n)( m  4  n)( m  4  n)


y   Ar x m  2 r  A0 x m  A1 x m  2  A2 x m  4  ..
r 0


1
 A0 x m 1 
x 2  
 (m  2  n)( m  2  n) 


1

x 4  ...
 (m  2  n)( m  2  n)( m  4  n)( m  4  n)

Therefore, when taking m  n, and A0  a, we have


1
1
 ax n 1 
x2 
x 4  ...
2.4.(2n  2)( 2n  4)
 2(2n  2)


x2
x4
= ax n  1  2
 4
 ...
2 .1!(n  1) 2 .2!(n  1)( n  2)


  au (say)

This is one solution of the give equation. Again when m  n , taking A0 = a,
we have


1
1
y  bx n 1 
x2 
x 4 ...
2.4.(2n  2)( 2n  4)
 2.(2n  2)


x2
x4
 bx  n  1  2
 4
 ...
2 .1!(n  1) 2 .2!(n  1)( n  2)


  bv (say)

This is another solution of the given equation.
Hence y  au  bv is the complete solution of the given equation.
Note.
If
a
1
, then au
2 (n  1)
n
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
5
Lecture notes on Ordinary Differential Equations
Khanday M. A.

1
x2
x4
n
x
1


 ... 

n
2
4
2 (n  1) 
2 .1!(n  1) 2 .2!(n  1)( n  2)

i.e.,
is called the Bessel’s function of order n and is denoted by Jn(x)
Solve the Legendre’s Equation in the descending powers of x,
Exercise.
(1  x 2 )m(m  1) x m2  2 x.mx m1  n(n  1) x m
Solution:-
(m 2  m  n 2  n) x m  m(m  1) x m2
or
Clearly the common difference of powers is 2.

Let the solution be

y   Ar x m  2 r
r 0

dy
  Ar (m  2r ) x m  2 r 1
dx r 0
So that
d2y 
  Ar (m  2r )( m  2r  1) x m2 r 2
2
dx
r 0
and
Substituting these values in the given equation, we have

 A [(1  x
r 0
2
r
)( m  2r )( m  2r  1) x m  2 r  2 x.(m  2r ) x m  2 r 1  n(n  1) x m  2 r ]  0

 A [{(m  2r )(m  2r  1)  2(m  2r )  n(n  1)}x
r 0
r

or
 A [{n
r 0
r
2
 A [(n  m  2r )(n  m  2r  1) x
r 0
 (m  2r )( m  2r  1) x m  2 r 2 ]  0
 (m  2r ) 2  (n  m  2r ) x m  2 r  (m  2r )( m  2r  1) x m  2 r  2 ]  0

or
m2r
r
m2 r
 (m  2r )( m  2r  1) x m  2 r  2 ]  0
which being an identity, we can equate to zero the coefficients of various powers of x.

Equating to zero, the coefficient of lowest power of x i.e., of xm, we have
A0 (n  m)(n  m  1)  0 .
Now A0 ≠ 0, it follows that m  n or m  (n  1)
Again equating to zero, the coefficient of general term i.e., x m  2 p , we have
A p (n - m  2p)(n  m  2p  1)  (m - 2p  2)(m  2p  1)A p 1  0.
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
6
Lecture notes on Ordinary Differential Equations
Ap  
or
Khanday M. A.
(m  2 p  2)( m  2 p  1)
Ap 1
(n  m  2 p)( n  m  2 p  1)
Putting P = 1, 2… we have
A1  
m(m  1)
A0
(n  m  2)( n  m  1)
(m  2)( m  3)
A1
(n  m  4)( n  m  3)
m(m  1)( m  2)( m  3)
 (1) 2
A0
(n  m  2)( n  m  4)( n  m  1)( n  m  3)
A2  


y   Ar x m  2 r
r 0
= A0 x m  A1 x m2  A2 x m4  ...


m(m  1)
 A0  x m 
x m  2 
(n  m  2)( n  m  1)




m(m  1)( m  2)( m  3)
 
x m  4  ....
 (n  m  2)( n  m  4)( n  m  1)( n  m  3)

when m = n, taking A0 = a, we have
y  a[ x n 
where u  [ x n 
n(n  1) n 2 n(n  1)( n  2)( n  3) n 4
x 
x  ....]  au
2(2n  1)
2.4.(2n  2)( 2n  3)
n(n  1) n 2 n(n  1)( n  2)( n  3) n 4
x 
x  ....]
2(2n  1)
2.4.(2n  2)( 2n  3)
Again for m  (n  1) , and A0  b , we have

(n  1)( n  2) n 3 (n  1)( n  2)( n  3)( n  4) n5 
y = bv ; where v   x n 1 
x

x ... 
2.(2n  3)
2.4.(2n  3)( 2n  5)


Hence y  au  bv is the complete solution of the given equation.
Note. If we replace ‘a’ by
1.3.5...(2n  1)
n!
and ‘b’ by
the ‘au’ and ‘bv’ are
n!
1.3...(2n  1)
defined by Pn(x) and Qn(x) respectively and are called Legendre’s Polynomials.
Exercise. Integrate in series
2x 2
d2y
dy
 x  (1  x 2 ) y  x 2
2
dx
dx
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
…(1)
7
Lecture notes on Ordinary Differential Equations
Khanday M. A.
Solution. First we shall find the complementary function of the given equation i.e., the
solution of the equation
2x 2
d2y
dy
 x  (1  x 2 ) y  0
2
dx
dx
…(2)
Putting y = xm in the left hand side of equation (2)
So that
or
2 x 2 .m(m  1) x m2  x.mx m1  (1  x 2 ) x m
 x m 2  (2m 2  3m  1) x m
Clearly the difference of powers is 2.


Let the solution of (2) be y   Ar x m  2 r
r 0
So that

dy
  Ar (m  2r ) x m  2 r 1
dx r 0
d2y 
  Ar (m  2r )( m  2r  1) x m 2 r 2
2
dx
r 0
Substituting in (2), we have

 A [(2(m  2r )(m  2r  1) x
r 0

or
 A [ x
r 0
m2r
r
m2r 2
r
 (m  2r ) x m  2 r  (1  x 2 ) x m  2 r ]  0
 (2m  4r  1)( m  2r  1) x m  2 r ]  0
…(3)
which being an identity, we can equate to zero the coefficients of various powers of x.
 Equating to zero the coefficient of the lowest powers of x. i.e., of xm, we have
A0 (2m  1)(m  1)  0
Now A0 ≠ 0, as it is the coefficient of the first term

m = 1, or
1
2
Now equating to zero the coefficient of the general term i.e., of x m2p2 , we have
- A p  (2m  4 p  3)( m  2 p  1) A p  1  0

Ap 1 
1
Ap
(m  2 p  1)( 2m  4 p  3)
Putting P = 0, 1, 2… in (4) we have
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
8
Lecture notes on Ordinary Differential Equations
A1 
Khanday M. A.
1
A0
(m  1)( 2m  3)
A2 
1
1
A1 
A0
(m  3)( 2m  7)
(m  1)( m  3)( 2m  3)( 2m  7)
A3 
1
1
A2 
A0
(m  5)( 2m  11)
(m  1)( m  3)( m  5)( 2m  3)( 2m  7)( 2m  11)

.

y   Ar x m  2 r
r 0
A0 x  A1 x
m
m 2
 A2 x m  4  A3 x m  6  .....
…(4)
é
1
1
= A0 x m ê 1+
x2 +
x4
(m
+1)(2m
+
3)
(m
+1)(m
+
3)(2m
+
3)(2m
+
7)
ë
ù
1
+
x 6 +... ú
(m +1)(m + 3)(m + 5)(2m + 3)(2m + 7)(2m +11)
û
Putting m = 1, and taking A0 = a, we have

x2
x4
x6
y  ax  1 


 ...
2.5 2.4.5.9 2.4.6.5.9.13




…(5)
This is one solution of (2)
Again putting m 
1
and taking A0  b , We have,
2


x2
x4
x6
y  bx1 / 2  1 


 ... 
2.3 2.4.3.7 2.4.6.3.7.11


which is the another solution of (2)

Complete solution of (2) is
é
ù
x2
x4
x6
y = ax ê 1+
+
+
+... ú
2.5 2.4.5.9 2.4.6.5.9.13
ë
û


x2
x4
x6
 bx  1 


 ... 
2.3 2.4.3.7 2.4.6.3.7.11


which is the complementary function of the given equation (1).
1/ 2
Now to find the P.I substitute
y  c0 x m , then from (1)
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
9
Lecture notes on Ordinary Differential Equations
Khanday M. A.
2c0 m(m  1) x m  c0 mx m  c0 (1  x 2 ) x m  x 2
i.e., we should have
c0 [2m 2  3m  1]x m  x 2
i.e., m  2, c0 .3  1

c0 
1
3
Now Putting m  2, A0  c0 
P.I 

1
3
in (4), we have
1 2
x2
x4
x  1

 ...
3 
3.7 3.5.7.11



x2
x4
x6


 ...
1.3 1.3.3.7 1.3.5.3.7.11
Hence the complete primitive of the given equation (1) is


x2
x4
y  ax  1 

 ... 
2.5 2.4.5.9




x2
x4
 bx1 / 2  1 

 ... 
2.3 2.4.3.7


x2
x4
x6



 ...
1.3 1.3.3.7 1.3.5.3.7.11
Exercise.
Solution.
Solve 4 x
d2y
dy
2  y 0
2
dx
dx
Substituting y  x m in the left hand side of the given equation, we have
4 xm(m  1) x m2  2mx m1  x m
(4m 2  2m) x m1  x m
or
Clearly, the difference of powers is 1.

Let
y   Ar x m  r be the solution of the given equation, So that
r 0

dy
  Ar (m  r ) x m  r 1
dx r 0
and
d2y 
  Ar (m  r )( m  r  1) x m r 2
2
dx
r 0
Putting these in the given equation, we have
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
10
Lecture notes on Ordinary Differential Equations

 A [4(m  r )(m  r  1) x
r 0
m  r 1
r

 A [2(m  r )(2m  2r  1) x
r 0
Khanday M. A.
 2(m  r ) x m  r 1  x m  r ]  0
m  r 1
r
 x m r ]  0
which being an identity, we can equate to zero the coefficient of various powers of x.

Equating to zero, the coefficient of the lowest powers of x. i.e., of xm-1, we have
2 A0 m(2m  1)  0
Since 2A0 ≠ 0, it follows that m = 0 or
1
2
The equating to zero the coefficient of the general term i.e., of xm+p-1, we have
2 A p (m  p)( 2m  2 p  1)  A p 1
Ap  
or
1
Ap 1
2(m  p)( 2m  2 p  1)
Putting P = 0, 1, 2… in (4) we have
A1  
1
A0
2(m  1)( 2m  1)
A2  
1
1
A1  (1) 2
A0
2(m  2)( 2m  3)
4(m  1)( m  2)( 2m  1)( 2m  3)
A3  
1
1
A2  (1) 3 3
A0
2(m  3)( 2m  5)
2 (m  1)( m  2)( m  3)( 2m  1)( 2m  5)
etc.
Thus

y   Ar x m 2r  A0 x m [1 
r 0
1
1
x 2
x2 
2(m  1)( 2m  1)
2 .(m  1)( m  2)(2m  1)(2m  3)
-
Now for m  0
and
1
x 3 +….]
2 (m +1)(m + 2)(m + 3)(2m +1)(2m + 3)(2m + 5)
3
A0  a,
1
1
y = a[1- x + 3 x 2 -......] = au
2
2 .3
Also for m 
1
2
and
A0  b ,
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
11
Lecture notes on Ordinary Differential Equations
y  b[1 
Khanday M. A.
1
1 2
x
x  ...]  bv
6
120
which is another solution of the equation.
HOME ASSIGNMENT
1.
9 x1  x 
d2y
dy
 12  4 y  0
2
dx
dx
2.
2 x1  x 
d2y
dy
 1  x   3 y  0
2
dx
dx
3.
2 x 2 x  1
d2y
dy
 3x3x  1  6 x  1 y  0
2
dx
dx
ROOTS OF THE INDICIAL EQUATION EQUAL:
Exercise: Solve x
d 2 y dy

 xy  0
dx 2 dx
…(1)
Solution: Substitute y  x m in the left hand side of the given equation, we have
x.m(m  1).x m2  mx m1  x.x m
x m1  m 2 .x m1
or
Clearly the difference of successive powers is 2.

Let us assume that the solution of the given equation (i) is

y   Ar x m  2 r
r 0
dy 
  Ar (m  2r ) x m  2 r 1
dx r 0
So that
and
d2y 
  Ar (m  2r )( m  2r  1) x m 2 r 2
2
dx
r 0
Substituting in (i), we have

 A [(m  2r )(m  2r  1) x
r
r 0

or
 A [x
r 0

or
 A [x
r 0
m  2 r 1
r
r
m  2 r 1
m  2 r 1
 (m  2r ) x m  2 r 1  x m  2 r 1 ]  0
 (m  2r )( m  2r  1  1) x m  2 r 1 ]  0
 (m  2r ) 2 x m  2 r 1 ]  0 ---------- (2)
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
12
Lecture notes on Ordinary Differential Equations
Khanday M. A.
which is an identity, we can equate to zero the coefficients of various powers of x.

Equating to zero the coefficient of lowest powers of x i.e., of x m 1 , we have
A0 m 2  0 .
Now A0  0, as it is the coefficient of the first term with which we start to write the
series.
.  m  0 (two co-incident values)
Equation to zero the coefficient of the next higher power of x i,e., of x m  2 p 1 , we have
Ap  (m  2 p  2) 2 Ap1
Ap+1 = i.e.,
and
1
( m + 2 p + 2)
we have A1 = -
Now putting, p  1,2,...
1
( m + 2)
2
2
…(3)
Ap
A0
in above equation, we have
1
A1
(m  4) 2
1
 (1) 2
A
2
m  2 m  42 0
A2  
A3  
1
A2
(m  6) 2
 (1) 3

1
m  2 m  42 m  62
2
A0

y   Ar x m  2 r
r 0
= A0 x m  A1 x m 2  A2 x m  4  A3 x m  6  .....
= A0 x m [1 
x2
x4

 .....]
(m  2) 2 m  22 m  42
…(4)
Putting m  0
y  A0 [1 
x2
x4
x6


.....]
2 2 2 2 4 2 2 2.4 2.6 2
…(5)
This is a solution of the given series.
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
13
Lecture notes on Ordinary Differential Equations
Khanday M. A.
Now from (4), we have

dy
(m  2) x m 1
m  4x m 3 ....
 A0 mx m 1 


dx
(m  2) 2
m  22 m  42 

and

d2y
(m  2)m  1x m m  4m  3x m 2 
m2



A
m
m

1
x


....
0
dx 2
(m  2) 2
m  22 m  42 

Substitute these values in the left hand side of the given equation, we get

(m  2)m  1x m m  4 m  3x m  2 
x. A0 mm  1x m  2 

.... 
2
2
2
(
m

2
)




m

2
m

4



m  4x m3 ....
(m  2) x m 1
A0 mx m 1 

( m  2) 2
m  22 m  42 



x m 2
x m 4
 x. A0  x m 

 .....
2
2
2
(m  2)
m  2 m  4


i.e.,
A0 m 2 x m1
(All other terms being cancelled)
As this involves the square of m, its partial differential coefficient w.r.t. m, i.e.,
2 A0 mx m 1  A0 m 2 x m 1 log x, will also vanish when m = 0,
i.e.,

  d2
d
 x y
x 2 
m  dx
dx

 2 A0 mx m 1  A0 m 2 x m 1 log x.
Differential operators being commutative this may be written as
 d2
 y
d
 x
 2 A0 mx m1  A0 m 2 x m1 log x.
x 2 
dx

m
dx


Hence
y
is a second solution of the given differential equation if m is equal to zero
m
after differentiating.
Differentiating (4) , Partially w.r.t., m, we have
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
14
Lecture notes on Ordinary Differential Equations
Khanday M. A.


y
x2
x4
 A0 x m . log x.1 

.... 
2
2
2
m
m  2 m  4 
 (m  2)
 2x 2


 4
2
2
 A0 x m 


x  ....
3
3
2
2
3
m  2 m  4 
 m  2  m  2 m  4

 2x 2

2x 4
1 
 1
 y log x  A0 x 



...



3
2
2
 m  2 m  2 m  4  m  2 m  4 

m
Putting m  0 and A0  a and b respectively in the two series, we have
 x2

x4
y  a 1  2  2 2 ...  au ( say )
2 .4 
 2
 x2

1
 y 
 bu log x  b 2  2 2 1  1 / 2x 4  ...  bv


2 .4
 m  m0
2

and
Therefore the complete solution is y  au  bv
Exercise: Solve x
d2y
dy
 1  x   2 y  0
2
dx
dx
…(1)
Solution: Substitute y  x m in the left hand side of the given equation, we have
x.m(m  1).x m2  1  x mx m1  2.x m
m  2x m  m 2 .x m1
or
Clearly the difference of successive powers is 1

Let us assume that the solution of the given equation (1) is

y   Ar x m  r
r 0

dy
  Ar (m  r ) x m  r 1
dx r 0
So that
d2y 
  Ar (m  r )( m  r  1) x m r 2
dx 2 r 0
and
Substituting these values in the given equation we have

 A [(m  r )(m  r  1) x
r
r 0

or
 A [(m  r  2) x
r 0
r
mr
m  r 1
 1  x (m  r ) x m  r 1  2 x m  r ]  0
 (m  r ) 2 x m  r 1 ]  0
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
…(2)
15
Lecture notes on Ordinary Differential Equations
Khanday M. A.
Now this being an identity, we can equate to zero the coefficient of various powers of x ,
Therefore, equating to zero the coefficients of lowest powers of x i.e., x m 1 , we have
A0 m 2  0
Now A0  0 as it is the coefficient of the first term with which we start to write series,
Therefore m  0, (two co  incident values)
Equating to zero the coefficient of the next higher power term i.e., of x m , we have
Ap (m  p  2)  (m  p  1) 2 Ap1  0
Ap 1  
Therefore
(m  p  2)
m  p  12
…(3)
Ap
Now putting p = 0, 1, 2 . . . in above, we have
m  2 A
m  12 0
m  3 A

A1  
m  2m  3 A
(m  2)
m  12 m  22 0
m  4 A  (1) 3 m  2m  3m  4
A3  
2
(m  3) 2
m  12 m  22 m  32
A2

2
1
 (1) 2
A0

y   Ar x m  r
r 0
= A0 x m  A1 x m1  A2 x m 2  A3 x m3  .....

m  2 x  m  2m  3 x 2  .....
= A0 x m 1 

2
m  12 m  22
 (m  1)

…(4)


3
4
y  A0 1  2 x  x 2  x 3 .....
2!
3!


…(5)
Putting m  0
This gives one solution of the given series.
Now if we substitute the series (4) in the left hand side of the differential equation
without putting m  0 , we shall get the single term A0 m 2 x m 1
As this involves the square of m, its partial differential coefficient w. r. t. m, i.e.,
2 A0 mx m 1  A0 m 2 x m 1 log x will also vanish when m  0 .
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
16
Lecture notes on Ordinary Differential Equations
i.e.,
Khanday M. A.

  d2
d
m 1
2 m 1
 x 2  (1  x)  2 y  2 A0 mx  A0 m x log x.
m  dx
dx

Differential operators being commutative, this may written as
 d2
 y
d
 2 A0 mx m1  A0 m 2 x m1 log x.
 x 2  (1  x)  2
dx
 dx
 m
Hence
y
is a second solution of the given differential equation if m is put to zero after
m
differentiation.
Differentiating (4), partially w. r. t., m, we have
 m  2
m  2m  3 x2  .... 
y
 A0 x m . log x.1 
x

2
m
m  12 m  22
 (m  1)

 2m  2

m  3   ....
x
1
2m  3
 A0 x m 
x
 x2 


 
3
2
2
3
m  1  m  1 m  2 m  1 m  2 m  12 m  22  
 m  1
  m  2 1  x
x2
 2m  3 m  3 
 y log x  A0 x 2
 


1 
  ...
2
2
  m  1 2  m  1 m  1 m  2  m  1 m  2  
m
Putting m  0 , and A0  a and b respectively in two series, we have
3
4


y  a 1  2 x  x 2  x 3  ....  au
2!
3!


and
 

1
3 1
1
 y 
 bu log x  b 2 2   x     2   x 2  ...  bv


2
2!  3
2
 m  m0
 

y  au  bv .
Therefore the complete solution is
Exercise: Solve


d2y
dy
xx
 1  5 x   4 y  0
2
dx
dx
2
…(1)
Solution: Substitute y  x m in the left hand side of the given equation, we have
 x  x .m(m  1).x  1  5xmx
m  4m  4x  m .x
m2
2
or
2
m
2
m 1
 4.x m
m 1
Clearly the difference of successive powers is 1

Let us assume that the solution of the given equation (1) is

y   Ar x m  r
r 0
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
17
Lecture notes on Ordinary Differential Equations
Khanday M. A.
So that

dy
  Ar (m  r ) x m  r 1
dx r 0
and
d2y 
  Ar (m  r )( m  r  1) x m r 2
dx 2 r 0
Substituting these values in the given equation we have
 A [x  x ((m  r )(m  r  1) x

2
r 0

 A [(m  r  2)
or
m r 2
r
2
r
r 0
 1  5 x (m  r ) x m  r 1  4 x m  r ]  0
x m  r  (m  r ) 2 x m  r 1 ]  0
…(2)
Now (2) being an identity, we equate to zero the coefficient of various powers of x.
Therefore equating to zero the coefficient of lowest power of x i.e., of x m 1 , we have
m 2 A0  0
Now A0  0 , as it is the coefficient of the first term of the first term with which we start
the series,
Therefore m  0, (two co  incident values)
Equating to zero the coefficient of the next higher power term i.e., of x m , we have
 A0 m  2  A1 m  1  0
2
A1 
m  22
m  12
2
A0
Again putting to zero the coefficient of the general term i.e., of x m p , we have
 Ap m  p  2  Ap 1 m  p  1  0
2
therefore
Ap 1 
2
m  p  22
m  p  12
…(3)
Ap
Now putting p  1,2,3,.... in (3), we have
A2  
A3 
m  32
(m  2) 2
m  42
(m  3) 2
A1 
A2 
m  32
m  12
m  42
m  12
A0
A0
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
18
Lecture notes on Ordinary Differential Equations

Khanday M. A.

y   Ar x m  r = A0 x m  A1 x m1  A2 x m 2  A3 x m3  .....
r 0
2
2
  m  2 2

 m  3 2  m  4 3
= A0 x 1  
 x
 x 
 x  .....
 m 1 
 m 1 
  m  1 

…(4)
y  A0 [1  2 2 x  3 2 x 2  4 2 x 3  .....]
…(5)
m
Putting m  0
which is a solution of the given series.
This gives one solution of the given series instead of two.
Now differentiating (4), partial w.r.t. m, we have
y
 A0 x m . log
m
2
2
  m  2 2

 m 3 2  m  4 3
x.1  
 x
 x 
 x  ....
 m 1 
 m 1 
  m  1 

  m  2  1

m2 
m3  2
 m  3  1
 A0 x m 2

x

2
.

x

....







2
2
 m  1  m  1 m  1 
  m  1  m  1 m  1 

  m  2  1
m  3  x 2  ...
m2 
 m  3  1
 y log x  A0 x m 2

x  2.




2 
2 
 m  1  m  1 m  1 
  m  1  m  1 m  1 

Putting m  0 and A0  a and b respectively in the two series, we have


y  a 1  2 2 x  32 x 2  4 2 x 3  ...  au (say )
 y 
 bu log x  b 2.2(1  2) x  2.3(1  3) x 2  ...



m

 m 0

and



 bu log x  2b 1.2 x  2.3x 2  3.4 x 3  ....  bv
Therefore the complete solution is y  au  bv
HOME ASSIGNEMENTS
1.
x  x  d
2
2
dx
y
2
 1  5 x 
dy
 4y  0
dx
d2y
dy
 1  x   2 y  0
2
dx
dx
2.
x
3.
x
4.
x2
d 2 y dy

y0
dx 2 dx
d2y
dy
 x  x2 y  0
2
dx
dx
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
19
Lecture notes on Ordinary Differential Equations
Khanday M. A.
ROOTS OF THE INDICIAL EQUATION DIFFERING BY A QUANTITY MAKING THE
COEFFIENT OF
“y” INFNITY.
Solve Bessel’s equation of order 2.


d2y
dy
x
 x  x2  4 y  0
2
dx
dx
2
…(1)
Solution: Substitute y  x m in the left hand side of the given equation, we have


x 2 .m(m  1).x m2  x.mx m1  x 2  4 .x m
or
m
2

 4 x m  x m 2
Clearly the difference of successive powers is 2.

Let us assume that the solution of the given equation (1) is

y   Ar x m  2 r
r 0
So that

dy
  Ar (m  2r ) x m  2 r 1
dx r 0
and
d2y 
  Ar (m  2r )(m  2r  1) x m 2 r 2
2
dx
r 0
Substituting these values in the given equation we have

 A [x
r
r 0
2

 A [(m  2r )

or
r
r 0
2

 4 x m 2r  x m 2r 2 ]  0
¥
or

(( m  2r )( m  2r  1) x m  2 r  2  x(m  2r ) x m  2 r 1  x 2  4 2 x m  2 r ]  0
å A [( m + 2r - 2) ( m + 2r + 2) x
r
m+2r
+ x m+2r+2 ] = 0
r=0
…(2)
Now this being an identity, we can equate to zero the coefficient of various powers of x ,
Therefore, equating to zero the coefficients of lowest powers of x i.e., x m , we have
A0 m  2m  2  0
Now A0  0 , as it is the coefficient of the first term of the first term with which we start
the series,
Therefore m  2 or 2
Thus the difference of the roots of the indicial equation is an integer.
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
20
Lecture notes on Ordinary Differential Equations
Khanday M. A.
Again to zero the coefficient of the general term, i.e., of x m  2 p , we have
Ap m  2 p  2m  2 p  2  A p 1  0
Ap  
or
1
m  2 p  2m  2 p  2
Ap 1
…(3)
Now putting p  1,2,3,.... in (3), we have

A1  
1
A0
m.m  4
A2  
1
1
2
A1   1
A0
(m  2)m  6
mm  2m  4m  6

y   Ar x m  2 r = A0 x m  A1 x m 2  A2 x m  4  A3 x m  6  .....
r 0


x2
x4
x6
= A0 x m 1 


 .....
2
 mm  4 mm  2m  4m  6 mm  2m  4 m  6m  8

Putting m  2 , the coefficient of terms of ‘y’ becomes infinity. To over come such a
difficulty we replace the constant
A0 by m  2k where k  0
Therefore the solution becomes


m  2x 2 
x4
x6
y  kxm m  2 

 ..... …(4)
2
mm  4 mm  4m  6 mm  4 m  6m  8


Now putting m  2 , one solution is
y  kx2 [
x4
x6
x8


 .....]  ku ( say )
2 2 .4 2 3.4.6 2 3.4 2.6.8
Now substitute (4) in the given differential equation (1) we have


d2y
dy
x
 x  x 2  4 y  kxm (m  2) 2 (m  2)
2
dx
dx
2
there is a factor (m  2) 2 , i.e., m has repeated roots. Therefore it follows that
y as well as
y
, satisfies the differential equation when m  2 . Also putting m  2
m
in (4), we get a solution. Thus we have found three solutions of the given differential
equation which is of order two only.
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
21
Lecture notes on Ordinary Differential Equations
Khanday M. A.
Now


y
m  2x 2 
x4
x6
 kx m m  2 

 ....
2
m
mm  4 mm  4m  6 mm  4 m  6 m  8


 
m  2  m  2  x 2  ......
1
 kx m 1  
 2

2 
  mm  4 m m  4 mm  4  

  1

 y 

Thus 
 ku log x  kx  2 1  
 0  0 x 2  ...


 m  m  2
   2.2

 x2

x4
 ku log x  kx  2 1  2  2 2  ...  kv ( say )
2 .4
 2

and

 y m2  kx2 4 
 x4

4 2
x4 
x6
x8
x 
...  2 2.4 2 kx2  2  3
 3 2
...
2.6
2.6.8 
 2 .4 2 .4.6 2 .4 .6.8 

 kw ( say )
Clearly it is easy to check that w  64u , so we have only two linearly independent
solutions.
Hence putting k  a and b in the two series the complete primitive is yau  bv
.
Home Assignments
1.
2.
Solve x 2
x1  x 
x  x
2

d2y
dy
 3x  y  0
2
dx
dx
3. x1  x 
4.

d2y
dy
 x  x 2  1 y  0 (Bessel’s equation of order unity.)
2
dx
dx
d2y
dy
 1  3x   y  0
2
dx
dx
 x3
d
2
dx
y
dy
 2y  0
dx
 3x 2
2
ROOTS OF THE INDICIAL EQUATION DIFFERING BY AN INTEGER MAKING THE COEFFICIENT
OF “y” INDETERMINATE

Solve 2  x 2
d
2
y
dx
2
x
dy
 1  x  y  0
dx
…(1)
Solution: Let us assume that the solution of the given equation (1) is

y   Ar x m  r
r 0
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
22
Lecture notes on Ordinary Differential Equations
So that

dy
  Ar (m  r ) x m  r 1
dx r 0
and
d2y 
  Ar (m  r )( m  r  1) x m r 2
dx 2 r 0
Khanday M. A.
Substituting these values in the given equation we have
 A [2  x ((m  r )(m  r  1) x

2
m r 2
r
r 0

or
 A [(m  r )m  r  1  m  r   1x
r 0
r
¥
or
å A [x
r
r=0
m+r+1
{
 x(m  r ) x m  r 1  1  x x m  r ]  0
mr
 x m  r 1  2m  r m  r  1x m  r  2 ]  0
}
+ ( m + r ) +1 x m+r + 2 ( m + r ) ( m + r -1) x m+r-2 ] = 0
2
…(2)
Now this being an identity, we can equate to zero the coefficient of various powers of x ,
Therefore, equating to zero the coefficients of lowest powers of x i.e., x m  2 , we have
A0 mm  1  0
Now A0  0 , as it is the coefficient of the first term of the first term with which we start
the series,
Therefore m  0 or 1
Thus the difference of the roots of the indicial equation is an integer.
Equating to zero the coefficient of the next higher power of x m 1 , we have
2 A1 m  1m  0
the coefficient of A1 vanishes when m  0.
Therefore A1 is indeterminate.
Equating to zero the coefficient of x m , we have


A0 m 2  1  2 A2 m  2m  1  0
Therefore
A2  
m
…(3)

1
A0
2(m  1)m  2
2
Again equating to zero the coefficient of the general term i.e., of x m p , we have


Ap 1  m  p   1 Ap  2m  p  2m  p  1Ap  2  0
2
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
23
Lecture notes on Ordinary Differential Equations
Khanday M. A.
{
}
or
2 ( m + p +1) ( m + p + 2) Ap+2 = -Ap-1 - ( m + p) +1 Ap
Putting
p  1,2,3,.. in (4) , we have,

2
…(4)

2m  2m  3A3   A0  m  1  1 A1
2


m  1  1 A
1
A3  
A0 
1
2(m  2)m  3
2m  2m  3
Therefore

2

2m  3m  4A4   A1  m  2  1 A2   A1 
And
Therefore


mr
r
= A0 x + A1 x
m
m



 4m  5 m 2  1
A0
2m  1m  2
2

1
m 2  1 m 2  4m  5
A4  
A1  2
A0 etc.
2(m  3)m  4
2 m  1m  2m  3m  4
 y A x
r 0
2
m+1
= A0 x m  A1 x m1  A2 x m  2  A3 x m3  .....
(m
+1)
2
é
ù
m +1) +1
A0
(
A0 x - ê
+
A1 ú x m+3
2 ( m +1) ( m + 2)
êë 2 ( m + 2) ( m + 3) 2 ( m + 2) ( m + 3) úû
é
( m2 +1) ( m2 + 4m + 5) A ùú x m+4.....
A1
+ ê+ 2
0
êë 2 ( m + 2) ( m + 3) 2 . ( m +1) ( m + 2) ( m + 3) ( m + 4) úû
2
m+2
é
( m2 +1) x 2 ( m2 +1) ( m2 + 4m + 5) +1 x 4....ùú
1
= A0 x m ê1x3 + 2
2 ( m + 2) ( m + 3)
2 . ( m +1) ( m + 2) ( m + 3) ( m + 4)
êë 2 ( m +1) ( m + 2)
úû
2
é
ù
1
( m +1) +1 x 3 + A1 x m ê x x 4...ú
2 ( m + 3) ( m + 4)
êë 2 ( m + 2) ( m + 3)
úû
Putting m  0 and taking A0  a and A1  b, we have
1
5 4 
1
1 4 
 1

y  a 1  x 2  x 3 
x .....  b  x  x 3 
x .....
12
96
6
24
 4



…(5)
This gives two arbitrary constants, so it may be taken as the complete primitive.
If we take m  1, then from (3) we have A1  0 .
1
1 4 

x .....
Therefore y  A0 x  x  x 3 
6
24


This is a constant multiple of the second in the solution by (5)
Hence (5) is the complete primitive.
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
24
Lecture notes on Ordinary Differential Equations
Exercise: - Solve 1  x 2 
Solution:
-:
Khanday M. A.
d2y
dy
 2x  y  0
2
dx
dx
…(1)
Let us assume that the solution of the given equation (1) is

y   Ar x m  r
r 0
So that

dy
  Ar (m  r ) x m  r 1
dx r 0
and
d2y 
  Ar (m  r )( m  r  1) x m r 2
2
dx
r 0
Substituting these values in the given equation, we have,
 A [1  x ((m  r )(m  r  1) x

2
mr 2
r
r 0

or
 A [ (m  r )m  r  1  2m  r   1x
r 0
r
¥
or
 2 x(m  r ) x m  r 1  x m  r ]  0
å A [-{( m + r ) ( m + r - 3) -1} x
r
m+r
m r
 m  r m  r  1x m  r  2 ]  0
+ ( m + r ) ( m + r -1) x m+r-2 ] = 0
r=0
…(2)
Now this being an identity, we can equate to zero the coefficient of various powers of x ,
Therefore, equating to zero the coefficients of lowest powers of x i.e., x m  2 , we have
A0 mm  1  0
Now A0  0 , as it is the coefficient of the first term of the first term with which we start
the series,
Therefore m  0 or 1
Thus the difference of the roots of the indicial equation is an integer.
Equating to zero the coefficient of the next higher power of x m 1 , we have
A1 m  1m  0
The coefficient of A1 vanishes when m  0.
i.e., A1 is indeterminate.
Again equating to zero the coefficient of the general term i.e., of x m p , we have
 Ap m  p m  p  3  1  m  p  2m  p  1Ap  2  0
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
25
Lecture notes on Ordinary Differential Equations
Ap  2 
Khanday M. A.
m  p m  p  3 A
p
(m  p  1)m  p  2
…(3)
Putting p  0,1,2,3,.. in (4) , we have,
A2  
mm  3  1
A0
(m  p  1)m  p  2
m  1m  1  1 A
1
(m  2)m  3
m  2m  1  1 A  m  2m  1  1mm  3  1 A
A4 
2
m  1m  2(m  3)m  4 0
(m  3)m  4
m  3m  1 A  mm  3  1m  1m  2  1 A
A5 
3
m  2(m  3)m  4m  5 1
(m  4)m  5
A3 


y   Ar x m  r = A0 x m  A1 x m1  A2 x m  2  A3 x m3  .....
r 0

mm  3  1 2 m  2 m  1  1mm  3  1 4 
 A0 x m 1 
x 
x .....
m  1m  2m  3m  4
 m  1m  2 


m  1m  2  1 x 3  mm  3  1m  1m  2  1 x 5 ...
 A1 x m  x 

m  2m  3
m  2m  3m  4m  5


Putting m  0 and taking A0  a and A1  b, we have
1
1
1 5
 1



y  a 1  x 2  x 4 .....  b  x  x 3 
x .......
8
2
40
 2



…(4)
This gives two arbitrary constants, so it may be taken as the complete primitive.
If we take m  1 , then we shall get series which is constant multiple of the second series
in (4)
Hence (4) is the complete primitive of (1).
HOME ASSIGNMENT
1  x  d
2
1.
1  x  d
2
2.
3.
d2y
 x2 y  0
2
dx
2
dx
2
dx
y
2
y
2
 2x
dy
 2y  0
dx
 2x
dy
 nn  1 y  0
dx
(Legendre’s equation of order n)
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
26
Lecture notes on Ordinary Differential Equations
Khanday M. A.
The simultaneous equations
dx dy dz


P Q
R
…(1)
Those equations express that the tangent to a certain curve at any point (x, y, z) has
direction cosines proportional to (P, Q, R). If P, Q and R are constants, we thus get a
straight line; or rather a double infinite system of straight lines, as one such line goes
through any point of space. If however, P, Q and R are functions of x, y and z, we get a
similar system of curves, any one of which may be considered as generated by a moving
point which continuously alters its direction of motion. The lines of forces of electro
statistics form such a system.
The following methods are used for the solution of simultaneous equation in the form (1)
First Method. First take any two members of (1)
dx dy

P
Q
(say)
Integrating, we obtain an equation. Again take other two members of equation (1)
dy dz

Q
R
(Say)
Integrating this, we obtain another equation. Thus two equations, so obtained from the
complete solution.
Second Method. We may be able to find l , m.n and L, M , N , such that one of the
equations
dx dy dz dx  mdy  ndz Ldx  Mdy  Ndz
=



P Q
R
P  mQ  nR
LP  MQ  NR
can be easily integrated
If l , m, n are such that P  mQ  nR  0 . Then we get dx  mdy  ndz  0 which
gives another equation on integration. These two equations, so obtained form the
complete solution.
Geometrical Interpretation of the equations.
dx dy dz


P Q
R
…(1)
From solid geometry, we know that the direction cosines of the tangent to a curve are
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
27
Lecture notes on Ordinary Differential Equations
Khanday M. A.
dx dy dz


ds ds ds
dx  dy  dz
i.e., in the ratio
Hence geometrically, the equation (1) represents a system of curves in space such that the
direction ratios of the tangent to it at any point (x, y, z) are proportional to P, Q and R.
Examples for solution
Obtain the system of curves, defined by two equations with arbitrary constants in each,
satisfying the following simultaneous differential equations.
(1)
dx dy dz


x
y
z
(2)
dx
dy
dz


mz  ny nx  z y  mx
(3)
dx
dy
dz


2
2
 2 xy  2 xz
y z x
(4)
dx dy dz


yz zx xy
(5)
dx
dy
dz


yz zx x y
(6)
xdx
dy
dz


2
yz yz
z  2 yz  y
2
2
dx dy dz


x
y
z
Solution: (1)
…(1)
Taking first two members of (1), we have
log x  log y  log a
or x  ay
Also taking last two members, we have
log y  log z  log b
or
y  bz
Now the complete solution is given by
x y
  z , which represents straight lines through the origin
a b
(2)
dx
dy
dz


mz  ny nx  z y  mx
Solution. Using multiplies ( x, y, z ) , we have
xdx  ydy  zdz
0
Therefore
xdx  ydy  zdz  0
Integrating, we have
x2  y2  z 2  a
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
28
Lecture notes on Ordinary Differential Equations
Khanday M. A.
Also using multiplies l , m, n , we have
dx  mdy  ndz  0
On integrating, we get  x+ my+ nz = b,
3.
Solution:
dx
dy
dz


2
2
 2 xy  2 xz
y z x
2
Using multiplies (x, y, z) we have
xdx  ydy  zdz
dy

2
2
2
 x( x  y  z )  2 xy
2 xdx  2 ydy  2 zdz dy

y
x2  y2  z2

Integrating, we have
log(x2 + y2 +z2) = log ay
or
x 2  y 2  z 2  ay
Also, taking last two members of (1), we have
dy dz

y
z
Now integrating, we have
log y  log z  log b
(4)
or
y  bz
dx dy dz


yz zx xy
Solution:. Taking first two members, we have
dx dy

y
x
or
xdx  ydy , integrating, we get
x2  y2  a ,
Also, taking last two members, we have
dy dz
Or

z
y
ydy  zdz
On integrating, we get
y2  z2  b
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
29
Lecture notes on Ordinary Differential Equations
Khanday M. A.
This is clearly the intersection of two families of rectangular hyperbolic cylinders.
5.
dx
dy
dz


yz zx x y
Solution: Using multiplies (1, 1, 1), we have
dx  dy  dz
2( x  y  z )
Also using multipliers (1,1,0) and (0,1,1) , we have
dx  dy
dy  dz
&
yx
zy
Integrating, we have
dx  dy 
dx  dy  dz
 2
x  y  z 
x y
log( x  y  z )  2 log( z  y )  log a
or
a  ( x  y) 2 ( x  y  z )
and
dx  dy
dy  dz

 ( x  y)  ( y  z)
or
log( x  y )  log b  log( y  z )
( x  y )  b( y  z )
or
6.
xdx
dy
dz


2
yz yz
z  2 yz  y
Solution.
Using multipliers (1, y, z), so that
2
xdx  ydy  zdz  0

x2  y2  z 2  a
Also taking last two members, we have
( y  z )dy  ( y  z )dz
or
ydy  zdy  ydz  zdz
or
ydy  zdz  ydz  zdy
or
y 2  z 2  2 yz  b
or
y 2  2 yz  z 2  b .
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
30
Lecture notes on Ordinary Differential Equations
Exercise.
Khanday M. A.
Find the radius of the circle of
dx dy dz
that passes through the point (0, – n, m).


x
y
z
Solution: Clearly the solution of the given system of equations is
x2  y2  z 2  a
Now
p
b
2  m2  n2
lx  my  nz  b
and

b
2  m2  n2
Also we know that if
x 2  y 2  z 2  2ux  2vy  2wz  d  0 .
is the equation of a sphere
passing through origin.
Then
R  u 2  v 2  w2  d
Here u  v  w  0 and

d  a ,
R a
Further the circle passes through (0, -n, m)

x  my  nz  b
gives b = 0,
and
x2  y 2  z 2  a
gives m2 + n2 = a
Now
r 2  a  m2  n2

 p


or
r  m 2  n 2 is the required radius of the circle.
(8)

0 

2  m2  n2

b
Find the surface generated by the curves
dx dy dz
that intersects


yz zx xy
the circle y 2  z 2  1, x  0
Solution. We know that x2 – y2 = a and x2 – z2 = b is the solution of the given system of
simultaneous equations,

As x = 0, We have
a = -y2
&
b = -z2
so that a + b = -(y2 + z2) = -1
Now adding x2 – y2 = a and x2 – z2 = b, we have 2x2 – y2 – z2 = a + b.
or
y2 + z2 – 2x2 = 1 is the required surface.
Ex .9. Find the surface generated by the curves of
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
31
Lecture notes on Ordinary Differential Equations
Khanday M. A.
dx dy dz


x
y
z
…(1)
that intersects the helix x2 + y2 = r2, z = K tan-1(y/x).
Solution. Since, we have already proved that the curves of the simultaneous differential
equation
dx dy dz


x
y
z
x y
 z
a b
are
Now
x
z
a=
a2  b2 
Clearly
and
b=
y
z
x2  y2
z2
Since z  k tan 1  y / z , therefore, we have x  ak tan 1  y / x  and y  bk tan 1  y / x 
Thus
x
2


  k tan  y / x  x
k tan  y / x  r
 y 2  r 2 gives ak tan 1  y / x 
a
or
2
 b2
1
2
2
1
2
2
 y2

2
Substitute the value of (2) in (3), we have
x
2


 y 2 k tan 1  y / x   r 2 z 2 is the required surface
2
Exercise: Find the curve which passes through the point (1, 2, -1) and is such that at any
point the direction-cosines of its tangent are in the ration of the squares of the coordinates of that point.
Solution. We have given that the curve is such that at any point, the direction cosines of
its tangent are in the ratio of the squares of the co-ordinates of that point.
Therefore, the simultaneous differential equation is given by
dx dy dz


x2 y2 z2
Taking first two members, we have
1 1
 a
x y
&
1 1
 b
x z
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
32
Lecture notes on Ordinary Differential Equations
Since the curve passes through (1. 2. -1) therefore from
Khanday M. A.
1 1
  a gives
x y
1
1
1  a  a 
2
2
Also
1  1  b  b  2
Thus, the required curve is
1 1 1 1
   z
x y z z
Home Assignments
Solve the following simultaneous differential equations
(1)
dx dy dz


1
1
1
(2)
dx dy dz


z
0 x
(3)
dx
dy
dz

 2
z( x  y) z( x  y) x  y 2
(4)
dx
dy
dz


1 y 1 x
z
A second integral found by the help of the first
Solve the following.
(1)
dx dy
dz
 2 
2
nxy
x
y
(2)
dx dy
dz


1
3 5  tan( y  3x)
(3)
dx dy
dz

 2
z
 z z  ( y  x) 2
(4)
dx
dy
dz

 4
2
2
xz( z  xy  yz ( z  xy) x
(5)
dx dy
dz
 2 
xy y
zxy  2x 2
(6)
dx
dy
dz
 2
 2
x  yz y  zx z  xy
2
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
33
Lecture notes on Ordinary Differential Equations
Solution. (1).
Khanday M. A.
dx dy
dz
 2 
2
nxy
x
y
Using first two members, we have
dx dy

x2 y2
1 1
 a
x y
or

 1 1 a
Again, using multipliers  , ,
 x y n
y  x  axy
…(1)

 , we have

dx dy z
- + dz
dx dy
dz
x
y n
 2 
=
2
nxy
x
y
x - y + axy
dx dy a

 dz  0.
x
y n

 from
(1) 
Integrating, we get
log x  log y 
az
b
n
or
log
x
a
b z
y
n
or
z

n
x  nxy 
 b  log

 b  log x y 
a 
y  y  x 

…(2)

y x
 a 

xy 


(1) and (2) form the solution of the given equation.
2.
dx dy
dz


1
3 5 z  tan( y  3x)
…(1)
Solution. We have from first two members
dx dy

 3dx  dy
1
3
or
3x  a  y
or
3x  a  y  3x  a
Also from first and last member of (1), we have
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
34
Lecture notes on Ordinary Differential Equations
Khanday M. A.
dx
dz

1 5 z  tan a
dz

 dx   5 z  tan a
or
1
x  log 5 z  tan a   b
5
or
5z  tan  y  3x   ce 5 x
3.
dx dy
dz

 2
z
 z z  ( y  x) 2
Solution. We have
dx dy

 dx  dy
z
z
or
x = -y +a
Also
dx
dy
 2
z
z  a2
or
dx 
or
x+y=a
zdz
,
z  a2
2
Integrating, we get
x
1
log( z 2  a 2 )  b
2
x2 + y2 + z2 + 2xy = ce2x
or
dx
dy
dz

 4
2
2
xz( z  xy)  yz ( z  xy) x
4.
Solution. We have
dx dy
a
, so that log x = log

x y
y
or
xy  a

dx
dy
 4
2
xz( z  xy) x
or
dx
dz
 3
2
z ( z  a) x
------------- (1)
or
x 3 dx  z ( z 2  a)dz
Integrating, we get
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
35
Lecture notes on Ordinary Differential Equations
Khanday M. A.
x4 z4 z2

 ab
4
4
2
or
x 4  z 4  2az 2  c.
or
x 4  z 4  2 xyz 2  c
…(2)
(1) and (2) gives the required solution.
5.
Solution.
dx dy
dz
 2 
xy y
zxy  zx 2
We have
dx dy
so that x = ay

x
y
Also
dx
dz

xy x( zy  zx )
or
dx
dz

y
zy  2ay
or
dx
dz

1
z  2a
x  log z  2a   b

z  2a  ce x
zy  2 x  cye x is the required solution.
or
dx
dy
dz
 2
 2
x  yz y  zx z  xy
6.
2
Solution. Using multiplies (1, -1, 0) (0, 1. -1), (-1. 0. 1) so that
dx  dy
dy  dz
dz  dz


( x  y )( x  y  z ) ( y  z )( x  y  z ) ( z  x)( x  y  z )
or
dx  dy dy  dz dz  dz


.
x y
yz
zx
Therefore integration, we get
x y
a
yz
and
x y
b
zx
which is the complete solution of the given differential equation.
Home Assignments
Solve the following
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
36
Lecture notes on Ordinary Differential Equations
(1)
dx dy
dz

 2
1  2 3 x sin( y  2 x)
(2)
dx dy
dz
 2 
xy y
zxy  2x 2
(3)
adx
bdy
cdz


(b  c) yz (c  a) zx (a  b) xy
(4)
dx
dy
dz

 2
z( x  y) z( x  y) x  y 2
(5)
 dx
dy
dz


1 y 1 x
z
(6)

Khanday M. A.
dx
dy
dz


x( x  y ) y ( x  y ) ( x  y )( 2 x  2 y  z )
TOTAL DIFFERENTIAL EQUATION
An equation of the form
Pdx  Qdy  Rdz  0
where P, Q, R functions of x, y, z is called the total differential equation or single
differential equation.
If we are given a relation of the form
F x, y, z   cons tan t
…(1)
Then, we have
f
f
f
.dx  .dy  .dz  0
x
y
z
i.e., we can express this in the form
Pdx  Qdy  Rdz  0
where P, Q, R functions of x, y, z
Geometrical Interpretation of the Equation
Pdx  Qdy  Rdz  0
This is differential equation expresses that the tangent to a curve is perpendicular to a
certain line the direction cosines of this tangent and line being proportional to dx, dy, dz 
and ( P, Q, R ) respectively.
But we say that the simultaneously equations
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
37
Lecture notes on Ordinary Differential Equations
Khanday M. A.
dx dy dz


P Q R
expressed that the tangent to a curve was parallel to the line ( P, Q, R ).
We thus get two sets of curves. If two curves, one of each set, intersect, hey must
intersect at the right angles.
Now two cases arise.
It may happen that the equation
Pdx  Qdy  Rdz  0
is integrable. This means that a family of surfaces can be found, all curves on which are
perpendicular to the curves represented by the simultaneous equations at all points where
these curves cut the surface.
Examples for Solution
(1)
xdx  ydy  zdz  0
(2)
y
(3)
yzdx  zxdy  xydz  0
(4)
 y  z dx  z  xdy  x  ydz  0 .
(5)
z  ydx  xdy  y 2 dz
(6)
xdx  zdy   y  2z dz  0
Sol. (1).
2

 z 2  x 2 dx  2 xydy  2 xzdz  0
xdx  ydy  zdz  0
Integrating, we get
x2  y2  z 2  a
Ex (2).
y
2

 z 2  x 2 dx  2 xydy  2 xzdz  0
Dividing throughout by x , we have
y2  z 2  x2
2 xy
2 xz
dx  2 dy  2  0
2
x
x
x
or
y 2 dx  2 xydy z 2 dx  2 xzdz

 dx  0
x2
x2
Or
  y2    z2 

  
  x  a
x
x

 

M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
38
Lecture notes on Ordinary Differential Equations
or
Ex (3).
Sol.
Khanday M. A.
x 2  y 2  z 2  bx
yzdx  zxdy  xydz  0
Dividing throughout by xyz  a
dx dy dz


0
x
y
z
or
log xyz  log z log xyz = log a
 xyz  a
or
yzdx  zxdy  xydz  0 is the exact differential equation.

d xyz  0 or xyz  a is the required solution.
Ex (4).
 y  z dx  z  xdy  x  ydz  0
Solution.
We can write
 ydx  xdy  zdx  xdz  zdy  ydz   0
Thus xy  xz  yz  a is the required solution of the equation.
Ex. (5).
z  ydx  xdy  y 2 dz
Solution. We can write the differential equation as
ydx  xdy dz

z
y2

x
d    d (log z ),
 y
Integrating, we get
or
x
 log az
y
x  y log az
Ex (6).
xdx  zdy   y  2z dz  0
Solution.
xdx  zdy  ydz   2zdz  0
Integrating, we get
x2
 yz  z 2  a
2
or
x 2  2 yz  2 z 2  b
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
39
Lecture notes on Ordinary Differential Equations
Khanday M. A.
Method of Integration, when the solution is not obvious.
When an integrable equation of the form Pdx  Qdy  Rdz  0 can not be solved by
inspection, we seek for a solution by considering first the simpler case where z is
constant, so that dz  0 .
Examples for solution.
(1)
yz log zdx  zx log zdy  xydz  0
Solution. Treating z-constant, We have dz  0 ,
So that
yz log zdx  zx log zdy  0
or
yz log zdx  zx log zdy

ydx  xdy
This gives
log x  log ay or x  ay
a
or
x
y
Since z was treated as constant, therefore ‘a’ is a function of z , So let f ( z )  a ( say ) ,
x
 f (z )
y
i.e.,
This leads to
ydx  xdy  y 2
df
dz  0
dz
which being identical with the original equation.
Therefore, we have
y
x


yz log z zx log z
i.e.,
dz
df ( z )

yz log z
f ( z)
or
dz
df ( z )

z log z
f ( z)
or
df ( z )
dz

f ( x)
z log z
df
dz
xy
 y2
Integrating, we get
or
log f ( z )   log(log z )  log b
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
40
Lecture notes on Ordinary Differential Equations
or
f ( z) 
Khanday M. A.
b
 b  f ( z ) log z
log z
or
x
b
1

 y  x log z  cx log z
y log z
b
i.e.,
y  cx log z is the required solution.
2.
2 yzdx  zxdy  xy(1  z )dz  0
Solution.
Taking z constant, so that dz  0 ,
We have
2 yzdx  zxdy  0
or
2dx
dy
  , integrating we get
x
y
log x 2   log y  log a or a  x 2 y
Since ‘a’ is a constant and z was treated as constant, therefore it follows that ‘a’ will the
function of z.
say
a  f (z )
Thus
x 2 y  f ( z)
This gives
2 xydx  x 2 dy 
df
.dz  0
dz
which is identical with the given equation

df
2 xy x
dz


2 yz zx  xy(1  z )
or
dz
df
 2
z
x y (1  z )
or
(1  z )
df
dz 
z
f ( z)
or
2

df
1 
   1dz
f ( z)  z 
Integrating we get
log f ( z )  log z  z  b
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
41
Lecture notes on Ordinary Differential Equations
f ( z)
 zb
z
f ( z)
 e z b
z
or
log
or
f ( z )  cze z
i.e.,
x 2 y  ce z is the required solution.
3.
2x
2
or
Khanday M. A.

 2 xy  2 xz 2  1 dx  dy  2 zdz  0
Solution. Assume x constant, so that dx  0 , we have
dy  2 zdz  0
or
y  z2  a
Since x is treated as constant, therefore ‘a’ will be a function of x . Say
a  f ( x), therefore., f ( x)  y  z 2  0
which gives
Therefore
df
.dx  dy  2 zdz  0 which is identical to the given equation.
dx
df
 1  2z
dx


2
2
1
2z
2 x  2 xy  2 xz  1

df
dx
 1
2
2 x  2 x( y  z 2 )  1
or

or


df ( x)
 (2 x 2  1)  2 xf ( x)
dx
df ( x)
 2 xf ( x)  (2 x 2  1)
dx
This is a linear differential equation.
Now
I.F = e  2 xdx  ex 2
2
Multiplying both sides by e x , we have
2
2
d
( f ( x)e x )  (2 x 2  1)e x
dx
This on integration gives
f ( x)e x   (2 x 2  1)e x dx
2
2
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
42
Lecture notes on Ordinary Differential Equations
or
f ( x)ex 2    (2t  1)e t .

 

 t
 


 




t





1
2 t
Khanday M. A.
put x 2  t
dt
1  t 
e dt 
2 t

t
2 xdx  dt
 t
e dt
2 t 





1
 t e  b
x=
t
dx 
dt
2 t

   ( f ( x)  f / ( x)e x dx  f ( x)e x  c

t



  xe x  b
2
i.e.,
Exercise 4.
Solution.
x  y  z e  b . Is the required solution.
y  yz dx  zx  z dy  y  xydz  0
( f  x)e x  b or
2
2
2
2
Assume y as constant, so that dy  0
y y  z dx  yx  y dz  0

or
x2
2
dx
dz

y( x  y) y y  z 
Integrating, we get
log x  y   log  y  z   log a
a
x y
yz
Since y is treated as constant, therefore the constant ‘a’ is a function of y. say a  f ( y )
x  y  
i.e.
This gives
or
or
yz
f ( y)
 y  z dx  dy   x  y dy  dz    y  z 2 df
dy
dy
 y  z dx   y  x  y  z   y  z 2 df dy   y  x dz  0

dy 

df 
dy  ( y  x)dz  0
( y  z )dx   x  z  ( y  z ) 2
dy 

This is identical with the given equation
Therefore, we have
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
43
Lecture notes on Ordinary Differential Equations
Khanday M. A.

df 

  x  z  ( y  z ) 2
dy 
yz
yx



y( y  z )
z( x  z)
y ( y  x)
1  ( x  z ) ( y  z ) 2 df


y z ( x  z ) z ( x  z ) dy
i.e.,
or
1 1  ( y  z ) 2 df
 
y z
z ( x  z ) dy
yz
( y  z ) 2 df

yz
z ( x  z ) dy
or
or
xz
df

y( y  z)
dy
df
x y yz

dy
y( y  z )


or

x y
yz

y( y  z) y( y  z)
df
f 1
 
dy y y
df
( f  1)

dy
y
or
df
dy
 ,
f 1
y
or
Integrating, we have
log  f  1   log y  log b
or
f 1 
b
y
b  y f  1
or
x y 
b  y
 1
 yz 
or
 xz

 y
 yz
Therefore
Exercise 5:
yx  z   b y  z  is the required solution.
x
2





y  y 3  y 2 z dx  xy 2  x 2 z  x 3 dy  xy 2  x 2 y dz  0
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
44
Lecture notes on Ordinary Differential Equations
Khanday M. A.
Dividing throughout by x 2 y 2 , we have
Solution:
1 z
1 y
1 1
z 
x 
  2  2 dx    2  2 dy    dz  0
x 
y 
y x
x y
x y
or
1
 1

x
y  1
z
z
 1
 dx  2 dy    dy  2 dx    dz  2 dx    dz  2 dy   0
y
x
x
y
 x
 y
y
 x

or
ydx  xdy xdy  ydx xdz  zdx ydz  zdy



 0.
y2
x2
x2
y2
Integrating, we have
or
x y z z
   a
y x x y
or
x 2  y 2  yz  xz  axy is the required solution.
Exercise 6:
mz  nydx  nx  lz dy  ly  mxdz
Solution:
Consider z = Constant,
…(1)
so that dz  0 ,
mz  ny dx  lz  nxdy  0
This implies
dx
dy

lz  nx (mz  ny)
Integrating both sides, we have
1
1
log a
 log( lz  nx)   log( mz  ny ) 
n
n
n
or
or
 lz  nx 
  log a
log 
 mz  ny 
nx  lz
 a.
ny  mz
Since ‘a’ is constant, therefore ‘a’ is a function of z.
i.e.,
(say)
a = f(z)
nx  lz
 f (z )
ny  mz
we have ny  mz ndx  ldz   nx  lz ndy  mdz   ny  mz 
2
or
df
dz
dz

2 df 
nny  mz dx  nnx  lz dy   l ny  mz   mnx  lz   ny  mz 
dz
dz 

This is identical with (1).
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
45
Lecture notes on Ordinary Differential Equations
Khanday M. A.

 df

mnx  lz   ny  mz  ny  mz   l 
 nmz  ny 
nnx  lz 
 dz


Therefore


nx  lz 
mz  ny
ly  mx
mnx  mlz  nly  mlz  ny  mz 
2
 n  n 
ly  mx
nmx  ly   ny  mz 
2
n 
or

ly  mx
 n  n 
df
dz
df
dz
ny  mz 2
df
ly  mx dz
or
df
 0 implies that df  0
dz
or we have
f  cons tan t  b ( say )
i.e.,
nx  lz
b
ny  mz
or
nx  lz  c(mz  ny )
This is the required solution of the given differential equation.
Exercise: Solve the following differential equation.
y 2 z x cos x  sin x dx  zx 2  y cos y  sin y dy
 xy y sin x  x sin y  xy cos z dz  0
Solution: Choose
z  cons tan t so that dz  0
Therefore,
y 2 zx cos x  sin x dx  zx 2  y cos y  sin y dy  0
Clearly on integrating both sides, we get
y sin x  x sin y
 a  f ( z) ,
xy
because “a” is a constant and ‘z’ is treated to be constant.
sin x sin y

 f ( z ),
x
y
Differentiating, we get
x cos x  sin x
y cos y  sin y
df
dx 
dy  .dz  0
2
2
dz
x
y
which being identical with the given equation, gives
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
46
Lecture notes on Ordinary Differential Equations
Khanday M. A.
df
x cos x  sin x
y cos y  sin y
dz
 2 2

2 2
x y z ( x cos x  sin x) zx y ( y cos y  sin y ) xy( y sin x  x sin y  xy cos z )

df
1
1
dz
 2 2 
2 2
xy( y sin x  x sin y  xy cos z )
x y z zx y

or
or
xy( y sin x  x sin y  xy cos z )
df

2 2
dz
zx y
or
y sin x  xSiny xy cos z
df

 z
xy
xy
dz
f ( z)  z
df
 cos z  0.
dz
dt f
cos z
 
dz z
z
or
which is a linear differential equation,
Therefore, we have
I.F. = e
1
 dz
z
 z.
Multiplying both sides by I.F, we get.
z
or
df
 f   cos z
dz
d
( fz)   cos z ,
dz
Integrating, we get
fz   Sinz  b
or
fz  Sinz  b
or
 sin x sin y 
  sin z  b.
z

x
y


or
yzSinx  zxSiny  xySinz  bxy
Exercise.
( z  z 3 ) cos x
dx
dy
dz
 ( z  z 3 )  (1  z 2 )( y  sin x)  0
dt
dt
dt
Solution. Put z  cons tan t so that dz  0.

( z  z 3 ) cos xdx  ( z  z 3 )dy
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
47
Lecture notes on Ordinary Differential Equations
Khanday M. A.
Sinx  y  a
or
As z = treated to be constant, therefore ‘a’ is a function of z, (say) a = f (z)
Sinx  y  f (z )
so that
cos xdx  dy 
df
.dz  0
dz
This is identical with (1)

or
or
or
 df
cos x
1
dz


( z  z 3 ) cos x  ( z  z 3 ) 1  z 2  y  Sinx 
df
1
dz

( z  z 3 ) (1  z )(1  z )(sin x  y )
df
1
dz

z (1  z 2 ) (1  z 2 ) f
(1  z 2 )
df
dz 
2
f
z (1  z )
df 1
2z
 
f
z (1  z 2 )
( By partial fraction)
Integrating, we get
Error! Objects cannot be created from editing field codes.
or

z
f  b
 1 z2 )





i.e.,
Sinx  y 1  z 2   bz
Exercise.
(2 x  y 2  2 xz)
Solution.
Multiplying both sides by dt, we have
dx
dy
dz
 2 xy  x 2
1
dt
dt
dt
(2x +y2 +2xz)dx + 2xy dy+x2dz =dt.
or
2xdx + (2xzdx + x2dz )+ (2xydy + y2dx) = dt.
or
x2 + x2z + xy2 = t + a
or
x(x + xz + y2) = t + a.
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
48
Lecture notes on Ordinary Differential Equations
Khanday M. A.
Necessary and Sufficient condition for Pdx  Qdy  Rdz  0 to be integrable is that
 Q R 
 P Q 
 R P 
  Q
  0 .
p



  R

z

y

x

z

y

x






Proof. Suppose that the total differential equation
Pdx  Qdy  Rdz  0 be integrable.
Let
x, y, z   c be the integral of Pdx  Qdy  Rdz  0
…(1)
which on differentiation gives



.dx 
.dy 
.dz  0 ,
x
y
z
which is identical with (1), therefore, we have
  
x  y  z   (say )
P
Q
R
which implies



 P ;
 R .
 Q ;
x
z
y

     2
    
Now consider
( P )    
    (Q)
y
y  x  yx x  y  x
P

Q

P

Q
y
y
x
x
i.e;

or
 
 P Q 


  Q

P
x
y
 y x 
…(2)
 Q R 


  R

Q
y
z
 z y 
…(3)


 R P 

R
P
z
x
 x z 
…(4)
Similarly, we have
 
and

Multiplying (2), (3) and (4) by R, P and Q respectively, we get
 P Q 
 Q R 
 R P 
  0 .



  Q
  R
 x z 
 x z 
 y x 
P 
That proves the necessary part of the result.
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
49
Lecture notes on Ordinary Differential Equations
Conversely,
suppose
that
(1)
is
satisfied
Khanday M. A.
for
the
differential
equation
Pdx  Qdy  Rdz  0
Here, we use a lemma, which states that if P, Q, R satisfy the condition, so also does
P1  P, Q1  Q, R1  R , where λ is any function of x, y and z.
Now, to find out the solution of
Pdx  Qdy  Rdz  0
Consider z  Cons tan t so that dz  0 .
We have
Pdx  Qdy  0
Let the solution be
F ( x, y , z )  a
Since ‘a’ is constant, so ‘a’ will be a function of z.
a  f ( z ) ( say )
i.e;
Thus,
F ( x, y , z )  f ( z )
This gives
F
F
F
dx 
dy 
dz  df
x
y
z
or
F
F
 F df 
dx 
dy  
 dz  0 .
x
y
 z dz 
…(5)
This will be identical with Pdx  Qdy  Rdz  0 if
F F F df

x  y  z dz   ,
P
Q
P
F
 P ,
x
if
F
 Q ,
y
Put
P1  P, Q1  Q, and R1  R
i.e;
if
F df

 R1
z dz
i.e;
if
df F

 R1 .
dz z
F df

 R
z dz
…(6)
 F

 R1  involves x and y only as a function of z.
We now show that 
 z

This will be the case if
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
50
Lecture notes on Ordinary Differential Equations
Khanday M. A.
 F   F

F   F
 R1  
. 
 R1   0

x dy  z
 y x  z

…(7)
Identically
Now, by the lemma, the relation P, Q, R leads to the similar relation
 Q1 R1 
 R1 P1 
 P1 Q1 
P1 



  Q1 
  R1 
0
y 
x 
 z
 x z 
 z
…(8)
Also, since equation (5) is integrable.
 Q1   f df  
   F df  P1 
   Q1 


P1 
 




z

y

x
dz

z

x
dz


 


 z 
…(9)
 F df  P1 Q1 

 


0
x 
 z dz  y
Subtracting (8) from (9), we have
P1
  F df


 R1
y  z dz

  F df
  Q1 

 R1
x  z dz




 F df
 P1 Q1 


 R1 

0
x 
 z dz
 y
P1 
But
F
F
, Q1 
,
x
y
and
  f    f 

 
  0 as f is a function of z alone.
x  z  y  z 
F  
 F
. 
 R1
x dy 
 z
Hence

 F  
 F
. 
 R1


 y x 
 z


0  0


This is equivalent to (7).
i.e.,
F
 R1 can be expressed as a function of F and Z, say  (F, z)
z
Hence we have
f
  ( f , z)
z
If the solution of this is f   (z ), then
F ( x, y, z )   ( z ), is a solution of
Pdx  Qdy  Rdz  0 . Hence the proof is completed.
Exercise: Find
f ( y ) if f ( y )dx  zxdy  xy log ydz  0 is integrable. Find the
Corresponding integral.
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
51
Lecture notes on Ordinary Differential Equations
Khanday M. A.
Solution: We know that the condition for Pdx  Qdy  Rdz  0 to be integrable is
 Q R
P

 z y



  Q R  P

 x z


Q = -zx,
and


  R P  Q

 y x


Here
P = f (y),
Now


Q
R
1
  x,
  x y.  log y    x(1  log y )


z
y
y



0


…(1)
R = - xy log y
R
P
P
Q
  y log y,
 0,
 f / ( y),
 z
x
z
y
x
Therefore From (1)
f ( y)( x  x  x log y)  zx( y log y  0)  xy log y( f / ( y)  z)  0

xf ( y) log y  xyz log y  xyz log y  xy log yf / ( y)  0
or
f ( y)  yf / ( y)
or
f y
or
dy df

y
f
df
dy
Integrating we get
log k  log y  log f
or
f  yk
i.e.,
f(y) = yk.
Substitute in the given equation, we have
ykdx  zxdy  xy log ydz  0
…(2)
Treating ‘x’ as constant, so that dx  0
Therefore
 zxdy  xy log ydz  0
or
- zxdy  xy log ydz
or

dy
dz
 ,
y log y
z
integrating we get
log(log y )  log a  log z
or
log y 
a
z
or
a  z log y
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
52
Lecture notes on Ordinary Differential Equations
Khanday M. A.
Now
z log y  f ( x) , as ‘a’ should be the function of x,
Therefore
df
z
.dx  dy  log ydz  0
dx
y
which is identical with (2) gives
df
dx   z / y   log y
yk
 zx
 xy log y
or
1
df

xy kydx
or
k log x  f  log b
or
k log x  z log y  log b
or
 xk
l o g
 b

or
dx df

x
k

l o g
y z or x k  by z is the corresponding integral of the given


differential equation.
Method of Auxiliary equations
If
Pdx  Qdy  Rdz  0
…(1)
is integrable, then we have,
 Q R 
 P Q 
 R P 
  Q
  0
P



  R
 x z 
 z y 
 y x 
…(2)
Comparing (1) and (2) we get,
dx
 Q R 



 z y 

dy
 R P 



 x z 

dz
 P Q 



 y x 
Known as auxiliary equations. These can be solved like Simultaneous equations.
Let   a and v  b be its integrals. We can find the value of Ad  Bdv  0 and
compare with the equation (1) to find A and B. substituting these values of A and B in
Ad  Bdv  0 and solving, we get the required solution.
Note:- The method fails if the equation is exact.
Solve the following differential equations.
1. zz  y dx  zz  xdy  xx  y dz  0
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
53
Lecture notes on Ordinary Differential Equations
Khanday M. A.
2. xz3 dx  zdy  2 ydz  0
Solution:- Comparing the differential equation with
 Q R 
 P Q 
 R P 
  Q
  0
P



  R
 x z 
 z y 
 y x 
Here
P  z ( z  y ), Q  z ( z  x), R  x( x  y )
 Q R 

  2 z  x   x  2 z

 z y 
 R P 


  2 x  y   2 z  y   2 x  2 y  2 z
 x z 
Therefore,
 P Q 

   z   z  2 z

 y x 
Clearly the condition of integrability is satisfied. Hence the auxiliary equations are
dx
dy
dz


2z 2x  2 y  2z  2z
dx
dy
dz


z
x yz z
or
dx dz

 dx  dy  0
From the two members
z
z
 x  z   ( say )
Again, we have
dx  dy dz

x y
z
log ( x + y) + log z = logv
Þ
log ( x + y) z = log v
or
( x + y) z = v
(say)
Let the given equation be Ad  Bdv  0
or
Adx  dz   Bx  y dz  dx  dy z  0
A  zBdx  Bzdy  A  Bx  y dz  0
Comparing this with the given equation, we have
Therefore
A  zB  zz  y; Bz  zz  x; A  Bx  y  xx  y
B  x  z    and A  x  y x  B  x  y z  v
Hence the given equation is Ad  Bdv  0 or
 vd  dv  0
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
54
Lecture notes on Ordinary Differential Equations
d
Khanday M. A.
dv
 0   log   log v  log c

v
i.e., v  c or x  y z  cx  z 

or

This is the required solution.
Homogeneous Differential equations
In case P, Q, R are homogeneous functions of x, y, z then one variable say z may be
reported from other two by substituting x  zu
dx  zdu  vdu ;
So that
and
y  zv (say )
dy  zdv  vdz.
Now two cases arise in which the given equation will be reduced to the form which either
the coefficient of dz in zero or not zero. In both cases the new equation may easily be
integrated.
Example: Solve by Homogeneous method,
z
Solution:-
2





 zy dx  z 2  xz dy  x 2  xy dz  0
The condition of integrability can be easily verified. Here P, Q, R are
homogeneous functions. We take x  zu
dx  zdu  vdu ;
So that
and
y  zv
dy  zdv  vdz.
Substituting these in the given differential equation, we have
z
2





 z 2 v zdu  udz   z 2  uz 2 zdv  vdz  z 2 u 2  z 2 uv dz  0
or
z 2 1  v zdu  udz   1  u zdv  vdz  uu  v z 2  0
or
z1  u du  dv  u  vdu  u  v1  u dz  0
or
du  dv
du
dz


0
uv
1 u z
On integration we get,
log u  v  log 1  u   log z  log c
log
uv
z  log c
1 u
or we have
Or
x y
 c is the required solution.
xz
zx  y   cz  x
This is the required solution.
Home Assignments
1.
x
2





y  y 2  y 2 z dx  xy2  x 2 z  x 2 dy  x 2 y  xy 2 dz  0
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
55
Lecture notes on Ordinary Differential Equations
Khanday M. A.
2.
xz3 dx  zdy  2 ydz  0
3.
dy
dz
z  z cos x dx
 z  z   1  z  y  sin x   0
dt
dt
dt
3
2
2


4.
zdz  x  a dx  b 2  z 2  x  a 
5.
zz  y dx  zz  xdy  xx  y dz  0
6.
x
7.
8.
2
2 1/ 2

 y 2  z 2  2 xy  2 xz dx 
y
2xyz  y
2
2
dy  0



 z 2  x 2  2 yz  2 yx dy  z 2  x 2  y 2  2 zx  2 zy dz  0





z  yz 2 dx  x 2 z  2 xyz  xz 2 dy  x 2 y  xy 2  2 xyz dz  0
Check the integrability of the equation
x
2





y  y 3  y 2 z dx  xy 2  x 2 z  x 3 dy  x 2 y  xy 2 dz  0 and solve it i)
By
the method of inspection; ii) By treating one variable constant iii)
By Homogeneous method iv)
9.
Find
By using the method of Auxiliary equation.
f (z) such that
y2 + z 2 - x 2
dx - ydy + f (z)dz = 0
xz
is integrable. Hence solve completely.
M.A./M.Sc. 3rd Semester Department of Mathematics, University of Kashmir
56