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1
M.Sc. (Previous) Mathematics
Paper –V
Differential Equations
BLOCK- I
UNIT –I : Homogeneous linear differential equations with variable
coefficients, Simultaneous differential equations and Total
differential equations.
UNIT -II: Picard’s method of integration, Successive approximation,
Existence and uniqueness theorem.
2
BLOCK-INTRODUCTION
The study of differential equations is a wide field in pure and applied
mathematics, physics, meteorology, and engineering. All of these disciplines are concerned
with the properties of differential equations of various types. Pure mathematics focuses on
the existence and uniqueness of solutions, while applied mathematics emphasizes the
rigorous justification of the methods for approximating solutions. Differential equations
play an important role in modeling virtually every physical, technical, or biological process,
from celestial motion, to bridge design, to interactions between neurons. This block
contains two units:
Unit I: This unit deals the concept, different methods to solve the Homogeneous linear
differential equations with variable coefficients, Simultaneous differential
equations and Total differential equations with several solved examples followed by
exercise to check the progress of reader.
Unit II: This unit deals the concept of numerical problems and their solutions especially in
reference to Picard’s method of integration, Successive approximation, Existence
and uniqueness theorem with several solved examples followed by exercise to
evaluate reader by himself.
At the end a list of reference books are given for the convenience to the reader.
3
UNIT –I : Homogeneous linear differential equations with variable
coefficients, Simultaneous differential equations and Total
differential equations.
Unit Structure :
1.0 Object
1.1 Introduction
1.2 Linear differential equations and its kinds
1.3 Homogeneous Linear differential equations with variable coefficients
1.4 Algorithm to solve Homogeneous Linear differential equations with variable
coefficients with solved examples
1.5 Equations reducible to Homogeneous Linear differential equations
1.6 Simultaneous differential equations
1.7 Algorithm to solve Simultaneous differential equations by different methods with solved
examples
1.8 Total differential equations
1.9 Algorithm to solve Total differential equations by different methods with solved
examples
1.10 Unit brief review (Summary)
4
1.1 Object:
At the end of this unit students are in a position to find the solutions of Homogeneous
Linear differential equations with variable coefficients, Simultaneous differential
equations and Total differential equations in easy manner.
1.1 Introduction:
Present chapter is deal with the study of Homogeneous Linear differential equations
with variable coefficients, Simultaneous differential equations and Total differential
equations. For the sack of convenience to students a brief summary related to the topic
viz. linear differential equations, its kind and method to find the solutions in a simple
manner is given.
1.2 Linear differential equations and its kind:
1.2.1 Linear differential equations: A differential equation of the form
+
= X ………………. (1)
where and X are function of x or constants.
In terms of D- notations (1) can be written as f(D)y = X or
+
………………(2)
where D =
,
, ……………. and so on.
The general solution of equation (1) or (2) is given by
y= complementary function (C.F.)+ particular integral (P.I.)
Things to remember:
(i) If X=0 in equation (1), the general solution is C.F. itself.
5
(ii) If all are constant in (1) then equation (1) is called linear
differential equations with constant coefficients or otherwise it is known as linear
differential equations with variable coefficients.
1.2.2 Algorithm to find complementary function (C.F.):
Step I: find auxiliary equation (A.E.) f(m)=0 by writing D=m in f(D) of equation (2).
Step II: find the roots of the A.E. i.e. values of m. Let the roots are m1, m2 ,…… , mn .
Step III: required C.F. is obtained as per the roots stated below:
Roots of A.E. Complementary function (C.F.)
All roots m1 , m2 ,…… , mn are real and
different.
+
+……+
m1 = m2 , but other roots are real and
different.
+ +……+
If roots are imaginary (say)
If ( ), ( ) repeated twice.
cos sin )
or )
or )
corresponding part of C.F. is
[
Solved examples:
Example (1): Solve .
Solution: Here A.E. is
- 3m - 4 = 0
(m- 4)(m+1) = 0
Hence roots are 4 and -1, real and different. Therefore C.F. is y = +
,itself
general solution.
6
Example (2): Solve .
Solution: Here A.E. is
= 0
(m+ 2)(m- 2)2 = 0
Hence roots are 2,2 and -2, two are equal and one is different. Therefore C.F. is
y = + , itself general solution.
Example (3): Solve .
Solution: Here A.E. is
= 0
(m- 2)(m2
+ 2m + 4) = 0 m = -1
Hence roots are 2, and -1 , one is real and other is a pair of imaginary. Therefore
C.F. is
y = + cos sin ), itself general solution.
1.2.3 Particular integral (P.I.): If X 0 in equation (1) then
P.I. =
.
In previous classes we studied the method to find P.I. either by resolving the f(D) into linear
factors and partials fractions and then each factor of the form
solve by
∫ ………… (3)
or by shortcut methods given below:
Type of function X Corresponding P.I.
X = then put D= a in f(D) i.e.
.If
then (D-a) is one of the factor of f(D).This
7
factor is solve by formula (3) and rest is
solve by shortcut method given here.
X = sin ax (or cos ax) then put D2= -a
2 in
f(D) i.e.
sin ax (or cos ax), provided
or otherwise use following
formula:
sin ax = -
or
cos ax =
X = xm
then P.I. is [
Expand [ using binomial expansions
and solve corresponding terms treating D as
differentiation.
X = ,
, solve V as per format given
above.
Solved examples:
Example (1): Solve .
Solution: Here A.E. is
+4m +3 = 0
(m+3)(m+1) = 0
Hence roots are -3 and -1, real and different. Therefore C.F. is y = +
Now A.T.Q. P.I. =
[here X= ]
Since f(a) on putting D=-2, we get
P.I. y =
= -
= -
Hence the general solution is y = C.F.+ P.I.
i.e. y = +
- .
8
Example (1): Solve .
Solution: Here A.E. is
-4 = 0
(m+2)(m-2) = 0
Hence roots are 2 and -2, real and different. Therefore C.F. is y = +
Now A.T.Q. P.I. =
. [we know that cos 2x= 2 -1.]
Therefore P.I. =
y =
[ ]
y =
y =
y =
y =
=
Hence the general solution is y = C.F.+ P.I.
i.e. y = +
.
Example (3): Find particular integral (P.I.) .
Solution: Here f(D) = = (D- 2)(D- 3)
Therefore P.I. =
=
= (
)
=
=
=
9
=
[since (1-D)
-1 = 1+D+D
2+…….]
=
[ D means differentiation wrt. x]
= (
) +
=
.
Check your progress:
Solve the following differential equation:
Q.1 . [Ans. y = +
+ ]
Q.2 . [Ans. y = +
]
Q.3 . [Ans. y = + + +
]
Q.4
[Ans. y = + + +
]
Q.5 [Ans. y =
].
1.3 Homogeneous Linear differential equations with variable coefficients
1.3.1 Homogeneous Linear differential equations with variable coefficients:
An equation of the form
+
= X
+
…………. (1)
where are constant and X is function of x or constant, is called
Homogeneous Linear differential equations.
10
Note: (i) Equation (1) above is called homogeneous because in each term the power of x is
same as the order of the derivative in that term.
(ii) Since the coefficient in each term is variable, therefore it is known as Homogeneous
Linear differential equations with variable coefficients.
(iii) This equation is also known as Cauchy’s linear equation.
(iv) By the substitution of x = or z = log x, the above equation (1) is transferred into
the linear differential equation with constant coefficient changing the independent
variable x to z as below:
If x = or z = log x then we have
=
………… (2)
=
or
(say), ..……… (3)
..………. (4)
Now
(
) =
(
)=
(
) =
(
)
Hence
(
) = ,
Similarly we can obtain = and so on. On putting the
equivalent values of etc. in equation (1),it reduces to linear
differential equation with constant coefficients having independent variable z. This
equation can be solved easily by the method given in (1.2).
Important Note: Notation D stands for
and
.
1.4 Algorithm to solve Homogeneous Linear differential equations with variable
coefficients with solved examples
Algorithm:
11
Step I: Write the given equation in D-notation form.
Step II: Replace etc in the equation by
= , = and so on.
Step III: Obtain equations is linear differential equation with constant coefficients, find
C.F. and P.I. as per the method given in article 1.2 treating z as independent variable.
Step IV: Lastly put back z = log x to get the required result.
Solved examples :
Example 1: Solve
-
.
Solution: D-notation form of the given equation is
-
On putting x = and = , we have
[ ] = 2z (as log )
Here A.E. is = 0 gives m = 1,1. So C.F. is y =
Now P.I. =
[
= [1+2 +……….] =
Hence the general solution is y = C.F.+ P.I.
i.e. y =
or y = . ( putting z = log x) Ans.
12
Example 2: Solve
-
.
Solution: D-notation form of the given equation is
-
On putting x = and = , we have
[ ] y = 2 (as )
Here A.E. is = 0 gives m = 2, 2. So C.F. is y =
Now P.I. =
=
(here X is of the form , V=1,art.1.2)
P.I. =
=
(here
∬ )
=
=
=
Hence the general solution is y = C.F.+ P.I.
i.e. y =
or y = . (putting z = log x)
or y = . Ans.
Example 3: Solve
+
.
Solution: D-notation form of the given equation is
.
13
On putting x = and = , =
we have
[ ] y = .
Here A.E. is = 0
or
Hence, C.F. = .
Now P.I. =
=
First part of P.I. =
= 5 [On putting ] ………… (i)
Now second part of P.I. =
=
On putting because it becomes zero, a failure case of f(a)=0
which can be solved by the formula
∫ ,
=
=
= 2
= 2 ∫
= 2 ……….. (ii)
from (i) and (ii) P.I. = 5 + 2
Hence the general solution is y = C.F. + P.I.
14
i.e. y =
or y = .
(putting z = log x)
y = ( ) Ans
Check your progress:
Solve the following differential equation:
Q.1
+
. [Ans. y =
+
+
]
Q.2
[Ans. y =
-
]
Q.3
[Ans. y =
+ +
]
Q.4
+
[Ans. y = + +
]
(Hint: Divide whole equation by x)
Q.5
[Ans. y =
+ +
].
1.5 Equations reducible to Homogeneous Linear differential equations
An equation of the form
+
= X
+
…………. (1)
where are constant and X is function of x or constant, can be
reduced into Homogeneous Linear differential equations by putting
15
then we get b=
and so we have
= b
,
(
)
(
)
…………etc.
On putting the above values in (1), equation reduces to Homogeneous Linear
differential equations and easily solve by the same technique as per article (1.4).
Solved Examples:
Example 1:
Solution: on putting , then , the equation becomes
] y = , where
.
,
A.E. is ,
Hence, C.F. =
Now P.I. =
=
, (on writing )
=
=
Hence the general solution is y = C.F. + P.I.
i.e. y =
or y =
(back substitution ) Ans.
16
Example 2:
Solution: on putting , then , the equation becomes
[ ] y = , where
.
A.E. is ,
Hence, C.F. =
Now P.I. =
=
, (on writing , failure case)
= 4
) =2z (since
cos ax =
Hence the general solution is y = C.F. + P.I.
i.e. y = +2z
y = +2
(back substitution ) Ans.
Check your progress:
Solve the following differential equation:
Q.1
[Ans. y =
[ ]
Q.2
[Ans. y = }.
17
1.6 Simultaneous differential equations
System of differential equations in which number of differential equations
will be same as is the number of dependent variables, called the system of Simultaneous
differential equations and the differential equations are called Simultaneous differential
equations. Present chapter deal with two types of Simultaneous differential equations.
1.7 Algorithm to solve Simultaneous differential equations by different methods with solved
examples
1.7.1 Type I: Simultaneous differential equations of first order and of the first degree with
constant coefficients i.e.
…… (1)
and …… (2)
Now to solve the above equations we apply elimination method. We first eliminate x
(or y) with suitable operation and obtain linear differential equation with constant
coefficients in y and t (or in x and t) which can be solved by the method, discussed in
article (1.2). Then we find the value of other variable by substituting value of first
variable in (1) or (2).
Important note: The number of arbitrary constants in general solution is equal to the
degree of D in determinant |
| = 0
Solved Examples:
Example 1: Solve
.
Solution: On writing given equations in D-notation form, we have
18
(D- 7) x + y = 0 …. (i)
-2x + (D- 5) y = 0 …. (ii)
To eliminate y we operate (i) by (D- 5 ) and then subtracting, we get
(D- 7) x + 2x=0 i.e. (D2-12D+37) x = 0
A.E. (m2-12m+37) = 0 , m=6
x = e6t
(c1cos t +c2sin t)
On substituting the above value of x in (i), we get
y = -(D- 7) e6t
(c1cos t +c2sin t)
=7e6t
(c1cos t +c2sin t)-D e6t
(c1cos t +c2sin t)
=7e6t
(c1cos t +c2sin t)- 6e6t
(c1cos t +c2sin t)- e6t
(-c1 sin t +c2 cost)
= e6t
(c1cos t +c2sin t) - e6t
(-c1 sin t +c2 cost)
= e6t
[ (c1 - c2 ) cost + (c1+ c2 ) sin t]
Hence the general solution is
x = e6t
(c1cos t +c2sin t), y= e6t
[ (c1 - c2 ) cost + (c1+ c2 ) sin t] .
Example 2: Solve
.
Solution: On writing given equations in D-notation form, we have
(D+5) x + y = …. (i)
-x + (D+3) y = …. (ii)
To eliminate x we operate (ii) by (D+ 5 ) and then adding, we get
(D+5)(D+3)y + y =(D+5) i.e. (D2+8D+15) y = D
19
(D
2+8D+15) y = 2
(D2+8D+15) y =
A.E. (m2+8m+15) = 0 , m= -3, -5
C.F. y = (c1 e-3t
+c2 e-5t
) and
P.I. y=
( ) =
( )
( )
=
( )
{as per method given in art.1.2}
=
y = (c1 e-3t
+c2 e-5t
)+
On substituting the above value of y in (ii), we get
x = (D+3) { (c1 e-3t
+c2 e-5t
)+
} -
=D{(c1 e-3t
+c2 e-5t
)+
+3{(c1 e
-3t +c2 e
-5t)+
}-
= -3c1 e-3t
-5c2 e-5t
+
+3c1 e
-3t +3c2 e
-5t+
-
=
-2c2 e
-5t +
Hence the general solution is
x = -2c2 e-5t
+
, y = (c1 e-3t
+c2 e-5t
) +
.
1.7.2 Type II: Simultaneous differential equations of first order and of the first degree in
the derivatives. Here we are giving the technique to solve the equations containing three
variables. The general form of Simultaneous differential equations of first order and of
the first degree containing three variables is
20
P1 dx + Q1 dy + R1 dz = 0, P2 dx + Q2 dy + R2 dz = 0 ……… (1)
where the coefficients are function of x, y, z. Solving these equations simultaneously,
we get,
or
……… (2)
Thus simultaneous equations (1) can always be put in the form (2).
Methods to solve
1.7.3 First method:
Suppose that any two fractions, directly integrated then after integration we find an
integral. Same procedure we apply for other two fractions. The two integrals so
obtained form the complete solution. Sometimes first integral may be used to simplify
the other two fractions.
Solved Examples:
Example1: Solve
.
Solution: On taking first two fractions, we have
= y , on integration, we get + c or = c
Similarly, taking second and third fractions, we get,
,gives c’
Thus the complete solution is = c, c’
Example 2: Solve
.
Solution: On taking first two fractions, we have
= y , on integration, we get + c or = c ...... (i)
21
Similarly, taking second and third fractions,
( )
c y =
,
on integration, we get c
c’ or
c’ ii
(i) and (ii) form the complete solution.
Note: It will be noted that this technique to solve the equations applied to Simultaneous
differential equations containing any number of variables.
1.7.4 Second Method:
We have
. If l, m, n are such that ,
then we get . If it is exact differential equation (say) du = 0, then
u=c is one part of the complete solution. Here l, m, n are known as multipliers. This
method may be repeated to get another integral by choosing new multipliers l’, m’, n’.
Note: Sometimes one integral we find by using method first and second integral using
second method.
Solved Examples:
Example1: Solve
.
Solution: On choosing 1, 1, 1 as multipliers, we get
or
, gives x+ y+ z = c ………. (i)
Again, choosing x, y, z as multipliers, we get
, gives c ‘ ii
22
(i) and (ii) form the complete solution.
Example 2: Solve
.
Solution: Taking last two fractions, we have
, on integration, we get + or = zc ...... (i)
Now using x, y, z as multipliers, we get
or
On taking fractions
, on integration, gives
0r ii
(i) and (ii) form the complete solution.
Example3: Solve
.
Solution: On choosing 1, -1, 0 as multipliers, we get
or
... (i)
Similarly, taking 0, 1, -1 and – 1, 0, 1 as multipliers, we get
or
…(ii)
and
or
… (iii)
From (i),(ii),and (iii) we have
23
Taking first two fractions of (iv), we get
on integration gives
i.e.
= c ….(v)
Similarly on taking last two fractions, we get
= c’ ….. (vi )
(v) and (vi) form the complete solution .
CHECK YOUR PROGRESS:
Solve the following simultaneous differential equations:
Q.1
[Ans. √ ].
Q.2
[Ans.
].
Q. 3
[Ans. ].
Q.4
[Ans. ].
Q.5
[Ans. ].
1.8 Total differential equations:
An equation of the form P ….. (1)
where P, Q, R are the functions of x, y, z is called total differential equation.
24
Sometimes equation (1) can be directly integrable, if there exists a function S(x, y, z) whose
total differential dS is equal to L.H.S. (1), otherwise we find the condition for which
equation (1) be integrable.
1.8.1 Condition for integrability of Total differential equations P
The equation is P …. (1)
Let S(x, y, z) = c be a solution of equation (1). Then total differential dS must be equal
to P , but, we know that
…. (2)
On comparing (1) and (2), we have
= , say
so that
…. (3)
now from first two equations, we have
= P
and
= Q
Since
, we have P
= Q
Gives
) = Q
P
…. (4)
Similarly,
) = R
Q
…. (5)
and
) = P
R
…. (6)
Multiplying (4), (5) and (6) by R, P and Q respectively and adding, we get
) +
) +
) = 0. ….. (7)
This is the required condition for integrability of equation (1).
25
Note: (I) If an equation satisfies condition (7), has an integral. In this case given equation is
directly integrable.
(II) Equation (1) is exact, if
these conditions are
obtained by putting in (4), (5) and (6).
1.9 Algorithm to solve Total differential equations by different methods with solved
examples
Let The equation is P . …… (1)
Step I: First compare the given equation to P and find P, Q, and R.
Step II: Check the condition of integrability
) +
) +
) = 0. .….. (2)
Case I: If the given equation is exact then the coefficients of P, Q and R in (2) are zero. In
this case given equation is directly integrable after regrouping the terms of the equation.
Example 1: Solve (x - y)dx – x dy + z dz = 0
Solution: A.T.Q. , we have P = (x - y) , Q = – x, R = z. The condition of integrability is
) +
) +
) = 0
) – x ) + ) = 0
satisfied here.
Therefore equation is exact, so on regrouping the terms of the given equation ,we have
x dx –( y dx + x dy) + z dz = 0,
on integration, we get
, is the solution.
Example 2: Solve (y + z) dx + dy + dz = 0
Solution: A.T.Q. , we have P = (y + z) , Q = 1, R = 1.
26
Here
,
= 0,
,
The condition of integrability is
) +
) +
) = 0
A.T.Q., we have ) +1 ) + ) = 0, satisfied here.
Therefore equation is exact, so on rearranging the terms of the given equation, we have
=0, integrating, we get = c
Or , is the solution.
Case II: If P, Q, R are homogeneous functions then substitute to
separate one variable z (say) from the other variables. After separating variable z, equation
is directly integrable. Satisfaction of condition of integrability is must in each problem.
Students are advised to try the same in each question, and then go ahead as follows:
Example 1: Solve (x - y) dx – x dy + z dz = 0
Solution: Here P = (x - y) , Q = – x, R = z. Clearly P, Q, R are homogeneous function, so on
putting , we get and the
equation becomes
) – uz ) + = 0
Or +{ u - ) } = 0
Or
—
, integration gives, — )+ 2
Or — ) = c, solution is obtained by putting u =
.
Example 2: Solve ( + yz ) dx +( xz + dy + - xy) dz = 0
Solution: Here P = ( + yz ), Q = ( xz + , R = - xy). The equation satisfies
integrability condition. Also P, Q, R are homogeneous function, so on putting
27
, we get and the equation
becomes
( + v ) ( )+( + - ) dz = 0
( + v ( )+ + - ) dz = 0
Or
,
Or
-
, integration gives,
Or
, required solution is obtain by putting u =
i.e.
complete solution is
or (x + z) y = c (y + z).
Case III: Check first, which two terms can be readily solved then take remaining third
term variable constant viz for our convenience let two terms P readily
integrable, then we take the third variable z = constant , so we get . Put in
the given equation and then integrate the remaining and take (z) as integration constant.
Differentiate obtained relation with respect to x, y, z and then compare with the given
equation to find (z), which on integration gives the value of (z) to complete the solution.
NOTE: Satisfaction of condition of integrability is must in each problem. Students are
advised to try the same in each question, and then go ahead as follows:
Example 1: Solve (x - y) dx – x dy + z dz = 0
Solution: Rearranging the equation, we have
x dx –( x dy+ y dx) + z dz = 0 …… (1)
28
now, if we put z=constant, then we have dz = 0 and the equation becomes
x dx –( x dy+ y dx) = 0, integrating gives,
……. (2)
on differentiating, we get (x - y) dx – x dy = dz …….. (3)
comparing (1) and (3), we get gives
+ c
putting this value in (2),the solution is ,
+ c
or [ Ans.
Example 2: Solve ( ) dx + dy + 2 dz = 0
Solution: Taking x= constant, we have dx = 0 and the equation becomes
dy + 2 dz = 0, integrating give, ….. (1)
now on differentiating (1), we get
dy + 2 dz or dy + 2 dz ….. (2)
comparing it to given equation, we get
= = [ by (1)]
Or = - ( ), which is linear equation in , therefore
I.F. = ,
. ∫
= ∫ ( ) ∫
= ∫ ∫
=
= , putting this value of
the general solution is or
. Ans.
29
Note: If none of the above three cases are applicable, then we apply the following case.
Case IV: Compare (1) and (2) to obtain auxiliary equation (A.E.) in the form of
simultaneous equations as below:
.
Solve the above equations by the methods discussed in (1.7.3) and (1.7.4) earlier. If two
integrals are say, then by comparing with (1), we get
the values of A and B and then the complete solution.
Example 1: Solve ( + yz ) dx +( xz + dy + - xy) dz = 0.
Solution: here P = ( + yz ), Q = ( xz + , R = - xy)
,
Thus A.E. is
or
or
….. (1)
Taking last two fractions, we get
Also taking 1, 0, 1 as multipliers, we get
Or or = v (say ).
Then A du + B dv =0 …. (2)
gives A ) + B + ) =0
i.e. [
Comparing with the given equation, we get
( + yz ) i.e. ( + z ) = u
30
And ( xz + i.e.
Or =
Hence (2) becomes du + u dv =0 i.e. on integration, we get
Or
Ans.
CHECK YOUR PROGRESS:
Solve the following total differential equation:
Q.1 - - 4 = 0. [Ans. ]
Q.2 ( +2x) + + + + = 0. [Ans. ]
Q.3 . [ ]
Q.4 . [ ]
Q.5 . [Ans. ].
1.10 Unit brief review (Summary):
31
UNIT –II: Picard’s method of integration, successive approximation,
Existence theorem Uniqueness theorem, Existence and
uniqueness theorem (All proof by Picard’s method)
Unit Structure:
2.0 Object
2.1 Introduction
2.2 Picard’s method of integration
2.3 Algorithm to solve initial value problem by Picard’s method with solved examples
2.4 Existence theorem and its proof
2.5 Uniqueness theorem with proof
2.6 Existence and Uniqueness theorem with proof.
2.7 Unit brief review (Summary)
2.0 Object:
This unit is presented in such an easy manner with solved examples that after going
through the whole unit students is easily solved the problems related to I.V.P.
32
2.1 Introduction:
A differential equation along with the subsidiary conditions on dependent variable
and its derivatives at the some value of the independent variable is known as initial
value problem (I.V.P.) while subsidiary conditions on dependent variable and its
derivatives given at more than one value of the independent variable is called Boundary
value problems (B.V.P.). In many of the engineering field and practical problems we
are often confronted with the differential equation whose exact solution cannot be
found by standard methods. Such problems cannot solve exactly and we find an
approximate solution only. This process of finding approximate solution is called
iteration method or successive approximation method.
2.2 Picard’s method of integration:
Given differential equation
……... (1)
= ……... (2)
From (1), we have
∫
∫
, [using (2)]
Or ∫
, ……... (3)
Now the integral on the right hand side can only evaluated if we known the value of y in
terms of x. This is not known. As an initial approximation, we replace y by in R.H.S. of
(3) and call the value of y on L.H.S. as first approximation of y, so that
∫
,
In the same manner if we substitute y by in R.H.S. of (3) and call the value of y on
L.H.S. as second approximation of y, we get
∫
,
33
Continuing in this way, we get better approximation in each step than the preceding one.
At the nth step we get
∫
,
This process is called Picard’s method of successive approximation.
2.3 Algorithm to solve initial value problem by Picard’s method with solved examples:
Algorithm to solve initial value problem by Picard’s method:
Step I: Find and by comparing given problem to equation (1) and (2).
Step II: Find by putting n=1, 2, 3….. in
∫
and corresponding values given of and
Solved Examples:
Example 1: Apply Picard’s method up to third approximation to solve
.
Solution: Here and . We know that the nth approximation by
Picard’s method is given by ∫
So, first approximation is obtained by putting n=1
∫
or ∫
[since ]
A.T.Q. ∫
or
Similarly, second approximation is obtained by putting n=2
∫
∫
or
And, third approximation, ∫
. Ans.
34
Example 2: Apply Picard’s method up to third approximation to solve
.
Solution: Here and . The nth approximation by Picard’s
method is given by ∫
………. (A)
So, first approximation is obtained by putting n=1 in (A), therefore we have
∫
or ∫
∫
[since ]
Or
Now the second approximation given by
∫
∫
,
-
+
The third approximation given by
∫
∫
,
-
Or ∫
+
+
. Ans.
Example 3: Apply Picard’s method up to third approximation to solve
.
Solution: Here and . The nth approximation by Picard’s
method is given by ∫
………. (A)
35
So, first approximation is obtained by putting n=1 in (A), therefore we have
∫
or ∫
∫
[Since ]
Or
Now the second approximation given by
∫
∫
∫
+
The third approximation given by
∫
∫
+
Or ∫
+
+
+
. Ans.
Example 4: Apply Picard’s method to approximate y up to second approximation and
corresponding to for that particular solution of
.
Solution: Here and .
The nth approximation by Picard’s method is given by
∫
and ∫
… (A)
36
So, first approximation is obtained by putting n=1 in (A),
therefore we have
∫
or ∫
∫
[Since ]
Or
and …. (i)
∫
or ∫
…. (ii)
The second approximation given by
∫
∫
∫
∫
Or
and …. (iii)
∫
or ∫
Or ∫ (
)
or
.... (iv)
On putting x= 0.1in (i), (ii), (iii) and (iv), we get particular solution
Ans.
CHECK YOUR PROGRESS:
Solve by Picard’s method up to three approximations (or iteration):
37
Q.1
. [Ans.
].
Q.2
. [Ans.
].
Q.3
.
[Ans.
].
Q.4
.
[Ans. ].
Q.5
.
[Ans.
].
2.4 Existence theorem and its proof:
Existence theorem: The I.V.P.
, = ……... (1)
has at least one solution y(x), provided the function f(x, y) is continuous and bounded for all
values of x in a domain D and there exist positive constants M and K such that
| | ………. (2)
and satisfies the Lipschitz condition
| | | | ……….. (3)
for all points in the domain D.
Proof: As per Picard’s method, the iterative sequence is given by
∫
, ……….. (4)
38
The initial value problem has a solution if necessarily the sequence {yn(x)} converges to y(x)
is either a solution of (1) or of an equivalent integral solution
∫
, ……….. (5)
ensure the existence of the limiting function
………… (6)
Let be the sum of successive difference
∑ …………. (7)
Clearly sequence { } converges if the infinite series ∑ converges.
From (4) ∫
,
∫
,
= ∫ [
………. (8)
The relation (8) is true for k = 1, 2, 3, ………. ,
Also from (4), we have ∫
∫
……… (9)
Now | | ∫ | || |
∫ | |
, from (2)
Or | | | | ……….. (10)
Again from (8) = ∫ [
| | ∫ | || |
∫ | || |
, from (3)
∫ | || |
=
| |
from (10) ………(11)
39
Similarly | | | |
=
| |
On continuing, we have | | | |
……… (12)
By Mathematical induction we easily establish that (12) holds true for (k+1) i.e.
| | | |
....…… (13)
From (13) it is quite clear that the value of each term of the series (7) is getting smaller
term by term and we get
| | | |
| |
Or
[ | | ], which converges for all values ( ) and so,
.
Taking limit , we get from (4)
∫
,
∫
= ∫
Thus the iterative sequence (4) converges to a solution, hence the theorem.
2.5 Uniqueness theorem with proof:
Uniqueness theorem: The I.V.P.
, = has a unique solution y(x),
provided the function f(x, y) is continuous and bounded for all values of x in a domain D
and there exist positive constants M and K such that
| | ………. (1)
and satisfies the Lipschitz condition
40
| | | | ……….. (2)
for all points in the domain D.
Proof: Let if possible the I.V.P.
, = have two solutions y(x) and v(x).
Then ∫
and ∫
∫ [
| | ∫ | ( )| | ( )|
| |
∫
= 2M| |, from (1) …….. (3)
Again using (2), we have
| | ∫
| || | ……..(4)
On combining (3) and (4), we get
| | ∫
| || | =
| |
…… (5)
In view of (5) and (6), we have
| | = | |
Continuing in this way, we get
| | = | |
, n= 1, 2, 3,…… …..... (6)
Right hand side of (6) tends to zero as n tends to infinity, thus
| | = 0 gives, , solution is unique.
2.6 Existence and Uniqueness theorem with proof:
41
Existence and Uniqueness theorem: Let function f(x, y) is continuous and bounded for all
values of x in a domain D and there exist positive constant M such that
| | . ………. (1)
Let the function f(x, y) satisfy the Lipschitz condition
| | | | ……….. (2)
Where the constant K is independent of . Let the rectangle R defined by
| | | | ……… (3)
lie in D, where Mh . Then the I.V.P.
, = has a unique solution
y(x).
Proof: By Picard’s successive approximation, we have
∫
∫
… … … … … … … …….. (4)
∫
∫
First we claim that the function lies in R. We have
∫
or | | ∫ | || |
∫ | |
, from (1)
Or | | | | Mh by (3).
This proves the desired result for n= 1.By Mathematical induction we assume that
lies in R. Then from (4), we have
| | ∫ | || |
∫ | |
, from (1)
42
Or | | | | Mh by (3).
Shows that lies in R.
Secondly we claim that | | | |
……. (5)
Since, we have | | | |
| | ∫ | || |
∫ | || |
, from (2)
∫ | || |
=
| |
Similarly | | | |
=
| |
On continuing, we have | | | |
, conclude that (5) is true for
n= 1, 2, 3, ….. .
By (3) and (5), we have | |
, for n= 1, 2, 3, ….. . ……. (6)
Using the infinite series
……. (7)
Mh+
+
+…………+
+ ….
[ ] which is convergent and so series (7) is
convergent. Thus .
Now we show that satisfies the differential
, = . Since
tends to uniformly in R and by Lipchitz conditions, we have on taking limit
, we get from (4)
∫
,
43
∫
= ∫
…… (8)
The integrand on right hand side of (8) being a continuous function, we come to conclusion
that the integral has a derivative and thus y(x) satisfies the I.V.P. Hence the solution of
given I.V.P. is exist. Uniqueness of solution already proved in (2.5), shall left to the reader.
Solved Examples:
Example 1. If rectangle R is defined by | | | | show that the function
f (x, y)= , satisfy the Lipchitz condition. Find the Lipchitz constant.
Solution: Here f (x, y)=
= . Since f is real valued
function on R ,
is exist and being continuous and bounded in R.
|
| = | | | | | | | | | | | | | | +1} +1}.
As we know that | | and | | , where is positive constant. Hence f satisfies
Lipchitz condition. Thus |
| where K= +1.
Example 2: Examine existence and uniqueness theorem for the I.V.P.
, =
Solution: Here f (x, y)=
and
. Clearly f and
are both continuous for all
(x , y). We consider rectangle R as | | | | about the point (1, -1).
Obviously in this rectangle R
| | |
| | || |
= (2+2b) | |
Clearly implies that Lipchitz condition is satisfied in R. Now let M= max. | | and
h = min.{a, b/M}, then the problem posesses a unique solution in | | .
44
CHECK YOUR PROGRESS:
1. State and prove Picard’s theorem.
2. Show that if the solution of the I.V.P. is exist, is unique.
3. If rectangle R is defined by | | | | show that the function f(x, y)= ,
satisfies the Lipchitz condition. Find the Lipchitz constant.
2.7 Unit Summary:
45
46
M.Sc. (Previous) Mathematics
Paper –V
Differential Equations
BLOCK- II
UNIT –III : Dependence on initial conditions and parameters;
Preliminaries, Continuity, Differentiability, Higher order
Differentiability. Poincare-Bendixson theory –Autonomous
systems, Index of a stationary point, Poincare-Bendixson
theorem, Stability of periodic solutions, rotation point, foci,
nodes and saddle points.
UNIT -IV: Linear second order equations-Preliminaries, Basic facts.
Theorems of Sturm, Sturm-Liouville Boundary value problems.
Numbers of zeros, Nonoscillatory equations and principal
solutions, Nonoscillation theorems.
UNIT -V: Partial differential equations of first and second order, linear
partial differential equation with constant coefficient.
47
BLOCK-INTRODUCTION
Differential equations are mathematically studied from several
different perspectives, mostly concerned with their solutions —the set of functions that
satisfy the equation. An Ordinary Differential Equation is a differential equation in which
the unknown function known as the dependent variable is a function of an independent
variable. On the other hand a Partial Differential Equation is a differential equation in
which the unknown function is a function of multiple independent variables and the
equation involves its partial derivatives. The order of partial differential equations defined
similarly to the case of ordinary differential equations. Both ordinary and partial
differential equations are broadly classified as linear and nonlinear homogeneous and non
homogeneous. This block contains three units:
Unit III: This unit deals the concept of Dependence on initial conditions and parameters,
Continuity, Differentiability, Higher order Differentiability. Poincare-Bendixson theorem,
Index of a stationary point, Poincare-Bendixson theorem, Stability of periodic solutions,
rotation point, foci, nodes and saddle points with several solved examples followed by
exercise to check the progress of reader.
Unit IV: This unit deals the concept of linear second order equations-Preliminaries, Basic
facts, Theorems of Sturm, Sturm-Liouville Boundary value problems, Numbers of
zeros, Nonoscillatory equations and principal solutions, Nonoscillation theorems
with several solved examples followed by exercise to evaluate reader by himself.
Unit V: This unit deals the solutions of Partial differential equations of first and second
order, linear partial differential equation with constant coefficient by usual and
shortcut methods followed by several solved examples and exercise to check the
progress of reader.
At the end a list of reference books are given for the convenience to the reader.
48
UNIT –III: Dependence on initial conditions and parameters; Preliminaries,
Continuity, Differentiability, Higher order Differentiability.
Poincare-Bendixson theory –Autonomous systems, Index of a
stationary point, Poincare-Bendixson theorem, Stability of
periodic solutions, rotation point, foci, nodes and saddle points.
Unit Structure :
3.0 Object
3.1 Dependence on initial conditions and parameters; Basic facts
3.2 Continuity, differentiability theorems
3.3Poincare- Bendixson theory-Autonomous systems
3.4 Stability of periodic solutions, rotation point, foci, nodes and saddle points
3.5 Unit brief review (Summary).
3.0 Object:
The main object of the present unit is to provide all the important contents in easy manner
with several solved examples.
3.1 Dependence on initial conditions and parameters; Basic facts:
3.1.1 Dependence on initial conditions and parameters:
If is defined on an open interval with the property that then
initial value problem (I.V.P.)
….. (1)
has a unique solution
49
defined on maximal interval of existence where depends on initial
conditions .
In general the I.V.P. depends upon a set of parameters. If , then I.V.P.
…. (2)
Where for each fixed z, then I.V.P. (2) has a unique solution
Now condition (1) can be reduced to the system (2) by change of variables say
and . Clearly I.V.P. (1) changes into the form
…. (3)
In which (
) considered as parameters.
(3.1.2) Lower and Upper Semi continuity:
The lower semi-continuity of function at ( ) is defined as
( ) ( ) as ( ) ( )
Similarly upper semi-continuity of function at ( ) is defined as
( ) ( ) as ( ) ( )
3.2 Continuity and differentiability theorems:
3.2.1 Continuity theorem: Let function be continuous on an open set E with the
property that for every ( ) E then I.V.P.
…..(1)
Let be the maximal interval of existence of the solution
, then
50
or is a lower(upper) semi continuous
function of and is continuous on the set , ( ) E
Proof: The I.V.P. (1) can be replaced by
, ,
i.e. by the I.V.P.
and ( ) by ….. (2)
Without loss of generality assume that function does not depend on z. Thus is
defined on an open interval with the property that Then
I.V.P. (2) has a unique solution
on the maximal interval of existence ( ), where
Now we choose a sequence of points ( ) (
) ( )
( )
And ( ) where as
.
Since the solution of (2) is unique, we can apply the theorem if and
be a sequence of continuous function defined on open set E such that
as
holds uniformly on each compact subset at E. Let of
(
)
And let ( ) be the maximal interval of existence. Also ( )
51
, then there exist a solution and a sequence of
positive integers with the property that
then
and as uniformly for .
In particular
lim lub
as n = n(k) .
Applying the result of this theorem we get
Implies is lower semi continuous.
Similarly , thus is upper semi continuous.
Now we claim that is continuous on ( ).
In the above theorem we have as , therefore
is considered for a fixed x.
We know that is continuous for fixed x, so in the above reference
is uniformly continuous i.e. for a given depending upon
s.t.
| |
If | | , then | |
Hence is continuous on set ( ).
52
3.2.2 Differentiability Theorem: Let f(x, y, z) be a continuous on an open (x, y, z) set E and
possess continuous first order partial derivatives
with respect to the components y
and z. Then
(i) The unique solution of …. (1)
Is the class on its open domain ( ) where .
(ii) Furthermore if J(x)= J is the Jacobian matrix
of f(x, y, z) with respect to
y at J(x)= J
…. (2)
Then
is the solution of the I.V.P. …. (3)
Where
and
{
And
is the solution of ... . (4)
Where = is the vector.
at and
is given by
∑
…. (5)
Proof of this theorem is beyond our scope. Reader may find the proof of the above theorem
in various reference books given at the end of this block.
3.3 Poincare- Bendixson theory-Autonomous systems:
3.3.1Basic concepts -
(I) Autonomous system:
A system of two first order equations of the form
…. (1)
53
is said to be autonomous, when the independent variable t does not appear explicitly.
The system (1) defines as
…. (2)
which is indeterminate only at points where vanish simultaneously.
Such points of course represent singular solution of (1).
(II) Phase Plane :
Let and be continuously differentiable functions in some region R in
the xy-plane. Then xy-plane is called Phase plane for the system
…. (1)
(III) Trajectory or The Orbit: A unique solution of the system
…. (1)
Is called trajectories or the orbit if it is exists in the phase plane for some open interval
and any point ( ) of region R and satisfies the initial condition
(IV) Critical Point: A critical point of the system
…. (1)
54
is a point ( ) obtained by solving
and
and such that
( ) ( )
On the other hand if ( ) is a critical point of the system, then the constant valued
function satisfying system (1) and are called equilibrium solution
of the system (1). A point which is not critical is called regular point.
(V) Index: An index is a significant property of a critical point P of the system
…. (1)
This is an integer which represents the net rotation of the direction field along a simple
closed curve.
(VI) Centers: A center (or vertex) is a critical point which is surrounded by a family of
closed paths. It is not approached by any path as .
(VII) Spirals : A spiral (or sometimes called a focus) is a critical point which is approached
in a spiral like manner by a family of paths that wind around it an infinite number of times
as
(VIII) Almost Linear System: The non linear system is of the form
….(1)
where are constants. The matrix form of above system is written as
( ) (
) (
) (
) …. (2)
By ignoring the non linear terms in (2), related linear system is
55
…. (3)
If we assume that
|
| , for system (2) then clearly (0,0) is the critical point to related
linear system (3).
are continuous and have continuous partial derivatives for
all (x,y).
√ and
√
Then (0, 0) is said to be simple critical point of the system (2) and the system is called
almost linear.
3.4 Stability of periodic solutions, rotation point, foci, nodes and saddle points:
3.4.1 Stable and Unstable critical Point:
A critical point ( ) of the almost linear system (2) is said to be stable provided
that if the initial point ( ) is sufficiently close to ( ) then the point
remains close to ( ) t 0. In other words for given if
| | |( ) |
Then critical point ( ) is called stable. Further a point which is not stable is called
unstable.
The critical point ( ) is called asymptotically stable if it is stable and every trajectory
that begins sufficiently close to ( ) also approaches to ( ) as
56
| | .
3.4.2 Nature and Stability of the critical point (0,0) :
The linear autonomous system is
…. (1)
If we assume that
|
| , then clearly (0,0) is the critical point to related linear system (1).
The above system can be written in matrix form as
( ) (
) (
)
Then the eigen values of the coefficient matrix (
) are the roots of the eigen
equation | |
Or .
Let the roots (eigen values) of the above eigen equation are . Then the nature of
the critical point (0,0) is depend on the values of and describes as below:
57
Nature of roots Linear system
Type Stability
Almost linear system
Type Stability
Node Unstable Node Unstable
Node Asymptotically
stable
Node Asymptotically
stable
Saddle Unstable
point
Saddle Unstable
point
Node Unstable Node or Unstable
Spiral
Node Asymptotically
stable
Node or Asymptotically
Spiral stable
Spiral Unstable Spiral Unstable
Spiral Asymptotically
stable
Spiral Asymptotically
stable
Centre Stable Centre Indeterminate
or Spiral
Solved Exampes based on Nature and Stability of the critical point (0,0) :
Example 1: For the system of the equations
Verify that (0,0) is a critical point . Show that the system is almost linear and discuss the
type and stability of the critical point (0,0).
Solution: The given system can be written as
58
Where and .
Now for critical point we must have
and
Solving the equations, we get . Thus (0,0) is a critical point.
Also
√ =
√ = 0
and
√ =
√ = 0
Therefore the system is almost linear. Again the related linear system for the given system
is
,
whose matrix form is,
( ) (
) ( )
Now the eigen values are the roots of the eigen equation
| | or |
|
or
Therefore
{Here
}
So from the table (3.4.2) it is clear that the critical point is a spiral and asymptotically
stable.
59
Example 2: For the system of the equations
Verify that (0,0) is a critical point . Discuss the type and stability of the critical point (0,0).
Solution: For critical point we must have
and
Solving the equations, we get . Thus (0,0) is a critical point.
Since ine given system , therefore we have
√ =
√ = 0
and
√ =
√ = 0
and the system is almost linear. Again the related linear system for the given system is
,
whose matrix form is,
( ) (
) ( )
Now the eigen values are the roots of the eigen equation
| | or |
|
or
Therefore , here have
same sign. So from the table (3.4.2) it is clear that the critical point is a node and unstable.
60
CHECK YOUR PROGRESS:
Q.1 For the system of the equations
(i) Show that (0,0) and(1,1) a critical point for the given system.
(ii) Show that (0,0) is a saddle point and (1,1) is the centre of above system.
Q.2 Determine the nature and stability of the critical point (0,0) for each of the following
systems of the equations
(i)
,
(ii)
,
(iii)
,
(iv)
,
(v)
,
.
3.4.3 Periodic Solutions:
Let
…. (1)
is a non linear autonomous system. Then a solution is said to be
periodic if neither function is constant, if both are defined for all and if such
that
61
Then the number T with the above property is called periodic solution with period T.
3.4.4 Closed Paths: A non linear system of equation has a closed path if it has a periodic
solution. On the other hand a linear system has closed path if the roots of its auxiliary
equation are purely imaginary. We can easily understand the concept of closed path
with the following example:
…. (2)
By using polar co-ordinates and taking we have and
which gives
x
+
and x
.... (3)
By (2) and (3) we get
or
.... (4)
and
.... (5)
The system (2) has a single critical point at To find the path consider
Separate the variables and integration we get
√ and …..(6)
The corresponding general solution of (2) is
√ and
√ … . (7)
Geometrically if we analyze (6) we come to the
conclusion that for c = 0 , we have
62
as and
If c 0 then . Also if c 0 then r 1 and again .
The above observations clearly shows that there is only a single closed circular path
for .
3.4.5 Poincare- Bendixson Theorems:
Theorem 1: A closed path of the system
necessarily surrounds at least one critical point of this system.
Proof: Let C be a simple closed curve and assume that C does not pass through any critical
point of the given system. If P= (x, y ) is a point on C then
V(x, y) =
is a non zero vector and have a definite direction given by If P moves once around C
then change by 2 , n is an integer and it is called the index of C.
If C shrinks continuously to a smaller simple closed curve without passing over any
critical point, then index change continuously but index is an integer which cannot change.
Theorem 2: Let R be a bounded region of the phase plane together with its boundary, and
assume that R does not contain any critical points of the system
If C = [ x(t), y(t)] is a path of (1) that lies in R for some T and remains in R for all t ,
then C is either itself a closed path or it spirals toward a closed path as . Thus in
either case system has a closed path in R.
The proof of this theorem can be easily illustrated by an example we discuss in 3.5.4.
63
UNIT –IV: Linear second order equations-Preliminaries, Basic facts.
Theorems of Sturm, Sturm-Liouville Boundary value problems.
Numbers of zeros, Nonoscillatory equations and principal
solutions, Nonoscillation theorems.
Unit Structure :
4.0 Object
4.1 Linear second order equations-Preliminaries, Basic facts
4.2 Algorithm to solve linear second order differential equations with solved examples
4.3 Sturm-Liouville Boundary value problems
4.4 Algorithm to solve Sturm-Liouville Boundary value problems and solved examples
4.5 Number of zeros
4.6 Nonoscillatory equations and principal solutions.
4.7 Nonoscillation theorems
4.8 Unit brief review (Summary)
4.1 Object:
4.1 Linear second order equations-Preliminaries, Basic facts:
4.1.1 Linear second order equations: An equation of type
,
64
Where P, Q, X are functions of x alone, is called the linear equation of second order. If
the coefficients P and Q are constants, the equation can be solved by the methods which we
have discussed in unit II, block I, otherwise there is no general method to solve such
equations. We are giving a procedure in which integral belongs to the complementary
function can be found.
4.1.2 Complete solution of linear equation of second order, when one integral belonging to
the C.F. is known:
The given equation is
, ….. (1)
Let y = be a known part of the C.F. i.e. a solution of equation
, ….. (2)
Let y =u be the complete solution of equation (1), where u is the function of x.
Now y =u gives
=
and
Substituting these values in (1), we get
(
) (
) ,
(
) (
)
,
Or
=
….. (3)
Putting
in (3), we get
,
- =
….. (4)
Which is linear in t, therefore I.F. = ∫{
}
∫
Hence solution of (4) is
65
t (I.F.) = ∫
or t (
∫ ) = ∫
∫
Solving and putting back
, and integrating we get
u = ( ∫ ) [∫ ∫ ]
Complete solution of (1) is y =u , where u is given by (5).
4.1.3 Special integral part of C.F. to
: The integral part given below
is obtain by putting and .. in (2).
Condition Integral part is
1 + P + Q = 0
1 - P + Q = 0
+ mP + Q = 0
P + Qx = 0 X
2+2Px + Q + = 0
m(m-1) + mPx + Q + = 0
4.2 Algorithm to solve linear second order differential equations with solved examples:
4.2.1 Algorithm to solve linear second order differential equations:
Step I: Put the equation in standard form i.e.
coefficient of
is unity.
Step II: Test for integral part of C.F. as per table given in 4.1.3 above.
Step III: Put y = u in step I equation and simplify as simplified in 4.1.2.
The following method fully illustrates the method.
Solved Examples:
Example 1: Solve
66
Solution: Given differential equation can be written as
Here P =
, Q =
and R = 0;
Clearly 1 + P + Q = 1
+
= 0 is a part of C.F.
Now we assume that the Complete solution is y =u , then the given equation reduces to
= 0, Putting
, we get x
= or
=
Integrating both sides, we get
log t + log x = log c or t . x = c
Putting back
, we get
. x = c or =
Integrating both sides, we get = c log + .
Hence complete solution is y =(c log + ) . Ans.
Example 2: Solve
Solution: Here P = , Q = and R = x;
Clearly P + Q x = = 0 is a part of C.F.
Assume that the Complete solution is y =u then the given equation reduces to
= x, Putting
, we get x
=
or
(
) = , ….. (I)
a linear equation in t whose I.F. = ∫(
)
Solution of (I) is ∫
67
. (
)/ ∫ .
(
)/
= - (
) [Putting
]
This gives
.
/
or
.
/
[
]
Or (
.
/
)
Integrating both sides, we get =
.
/
dx + .
Hence complete solution is y = .
/
dx + . Ans.
CHECK YOUR PROGRESS:
Q.1 Solve
[Ans.
dx + ].
Q.2 Solve
[Ans.
+ ].
Q.3 Solve
[Ans.
+ ].
4.2.2 Reduction of equation
to its normal form:
When we fail to obtain a part of C.F. we cannot apply method 4.2.1. In such cases the
equation may be solved by reducing equation
….. (1)
to its normal form i.e. by removing first derivative as below:
68
Let y =u gives
=
and
Substituting these values in (1), we get
(
) (
) ,
(
)
(
) ,
Now we choose in such a way that the coefficient of
i.e. (
)
Or
∫
.. (2)
Or
(
) =
….. (3)
Again from (2) we have
∫
, differentiating both sides gives
And
=
=
Putting (3), we get
(
–
) =
∫
Or
( –
) .
∫
Or
, where ( –
) = .
∫
or
….. (4)
Equation (4) is the required normal form of equation (1), can be easily integrated.
4.2.3 Algorithm to solve linear second order differential equations by reducing to its
Normal form:
Step I: Put the equation in standard form i.e.
coefficient of
is unity.
69
Step II: Find P, Q, X,
∫
, ( –
) = .
∫
.
Step III: Put the values of in normal form
.
Step IV: Obtained equation is linear differential equation with constant coefficient and
solve by finding C.F. and P.I. as given in Unit I, Block I.
Step V: Required Solution is obtain by putting the values of
Solved Examples:
Example 1: Solve
Solution: Here P = Q = to reduce in normal form we choose
∫
= , =
And ( –
) ( –
)= 2
Equation reduces to
.
Here A.E. is C.F. =
And P.I. =
=
( )
, therefore
,
And the complete solution is
Ans.
Example 2: Solve
Solution: Here P = Q =
To reduce in normal form we choose the complete solution of the given equation as
, where
∫
= = , =
70
And ( –
) ( –
)= 6
Equation reduces to
.
Here A.E. is C.F. =
And P.I. =
=
( )
( )
,
Therefore
,
And the complete solution is
Ans.
CHECK YOUR PROGRESS:
Q.1 Solve
[Ans.
].
Q.2 Solve
[Ans. ].
Q.3 Solve
[Ans. ].
Q.4 Solve
[Ans. ].
4.2.4 Transformation of the equation
by changing the independent
variable:
Sometimes the equation is transformed into an integrable form by changing the
independent variable. Let the equationbe
….. (1)
Let the independent variable x be changed to z by taking z as the function of x.
and
(
)
(
)
,
71
Substituting these values in (1), we get
[
(
)
,
Or
(
)
[
]
+ ,
Or
…. (2)
Where
(
) ,
(
) and
(
) .
After obtaining equation (2) we like to choose z in such a way that (2) can be easily
integrated.
Case I: .
We choose z to make the coefficient of
in (2), equal to zero i.e.
(
) or
or
Integrating, we get log
or
integrating again, we get , this value of z reduces (2) to
which can be easily solved provided comes out to be a constant or a constant multiplied
by
.
Case II:
We choose z such that
(
) (constant),
i.e. (
)
or
, integrating gives ∫ .
72
The above value of z reduces (2) to
Which can be solved easily, if comes out to be a constant.
Things To Remember:
To apply these methods student are advised to remember equation (2) and the values of
.Since we have assumed that z is f(x),so to get the required the required
result replace z with its corresponding value in terms of x.
Solved Examples:
Example 1: Solve
.
Solution: Writing given equation in standard form, we have
…. (i)
Here P = , Q = , X =
Changing independent variable from x to z, equation becomes
….. (ii)
Where,
(
) ,
(
) and
(
) .
Let us choose z such that
(
) = - 2 (constant)
i.e. (
)
or
, integrating gives . ….. (iii)
Then
(
)
and
(
)
=
73
Hence equation (ii) transferred to
Or
Or
By (iii)
Or ….. (iv)
Now C.F. = +
and
P.I. =
=
,
Hence the solution of the given equation is
y = +
+
Or y = +
+ [ As z = sin x ] Ans.
Example 2: Solve
Solution: Writing given equation in standard form, we have
…. (i)
Here P =
, Q =
, X =
Changing independent variable from x to z, equation becomes
….. (ii)
Where,
(
) ,
(
) and
(
) .
Let us choose z such that
A.T.Q.
( )
, putting
then
( )
Separating the variables and integrating, we get
74
+ or
( )
Since
gives
( ) Separating the variables and integrating, z =
(
) =
(
( )) = 4 and
(
( )) ,
Hence the transformed equation is
its solution is y = +
or y = + [As z = ] Ans.
CHECK YOUR PROGRESS:
Q.1 Solve
[Ans. y = + ].
Q.2 Solve
[Ans. y = + ].
Q.3 Solve
[Ans. y =
+
(
) ].
Q.4 Solve
[Ans. y = -
].
4.3 Sturm-Liouville Boundary value problems:
4.3.1 Boundary value problems:
A differential equation along with some conditions on the unknown function and its
derivatives given for one specific value of the independent variable is called initial value
75
problem (I.V.P.). If conditions on the unknown function and its derivatives given for more
than one value of the independent variable is called boundary value problem (B.V.P.).
Ex. (1) y is an I.V.P. because for one value of
independent variable we have conditions on the unknown function and its
derivatives .
Ex. (2) y is a B.V.P. because for two values
of independent variable we have conditions on the unknown function and
its derivatives .
4.3.2 Sturm-Liouville Boundary value problem:
The boundary value problem given by the second order differential equation of the form
[ [ ….. (1)
on some interval [a , b] satisfying the conditions of the form (i)
and (ii) , where are a real parameter, and
the real valued continuous functions of x . Also are real constants
at least one in each is non zero of conditions (2) is called the Sturm Liouville Problem. Here
equation (1) is known as Sturm Liouville equation and equation (2) is called boundary
conditions.
Note (1): Two real functions f(x) and (x) are called orthogonal function on the interval
[a , b] , if ∫
(2) If are two eigen functions of Sturm Liouville Problem (1)
corresponding to two distinct eigen values respectively and their derivatives are
continuous function in the same interval [a, b], then orthogonal .
4.3.3 Theorem: Eigen values of the Sturm Liouville Problem are all real.
Proof: Sturm Liouville Problem is [ [ …….. (i)
On the interval [a, b] satisfying the conditions
76
and
We assume that p(x) 0 when . Let y(x) is an eigen function corresponding to an
eigen value , then this function satisfies equations (i), (ii), and (iii), and it may
be a complex function. Taking conjugate throughout in equations (i), (ii), and (iii), we get
[ ] [ ] ….(iv)
.… (v)
and …. (vi)
The above equations shows that is eigen function corresponding to the eigen value
. Multiplying (i) by and (iv) by and subtracting, we get
) = [ ] [
= [ ] [
Integrating both sides from a to b, we get
)∫
= * [ ] [ + -
* [ ] [ +
)∫
= 0, in view of (ii), (iii), (v) and (vi).
Thus we have )= 0 or as ∫
.Hence the theorem.
4.4 Algorithm to solve Sturm-Liouville Boundary value problems and solved examples:
Algorithm to solve Sturm-Liouville Boundary value problems: Standard form of the
equation is
Step I: Consider three different cases for in each problem.
Step II: solve each case separately by finding C.F.
77
Step III: Use boundary condition to find the values of constants taken in C.F.
Conclusion: Non zero value of constant gives eigen value and corresponding eigen vectors.
Solved Examples:
Example 1: Findthe eigen values and eigen functions of the Sturm Liouville problem
Solution: Case I. If , the given equation reduces to
Integrating twice gives
Using given boundary conditions, we have B = 0 and or
Therefore , which is not an eigen function.
Case II. If ,let the given equation reduces to which is a
linear differential equation with constant coefficient, whose solution is
Using given boundary conditions, we have A + B = 0 and or
Therefore , which is not an eigen function.
Case III. If ,let he given equation reduces to which is a
linear differential equation with constant coefficient, whose solution is
Using given boundary conditions, we have A = 0 and
or
If B = 0then we have, , which is not an eigen function. Therefore taking
, we have
78
Therefore the eigen function are taking B = 1 and the eigen values are
……. Ans.
Example 2: Find the eigen values and eigen functions of the Sturm Liouville problem
Solution: Case I. If , the given equation reduces to
Integrating twice gives and
Using given boundary conditions, we have B = 0 and .
Therefore , which is not an eigen function.
Case II. If ,let the given equation reduces to which is a
linear differential equation with constant coefficient, whose solution is
and
Using given boundary conditions, we have A + B = 0 and or
Therefore , which is not an eigen function.
Case III. If ,let he given equation reduces to which is a
linear differential equation with constant coefficient, whose solution is
and
Using given boundary conditions, we have A = 0 and
or
If B = 0then we have, , which is not an eigen function. Therefore taking
, gives
79
Therefore the eigen function are taking B = 1 and the eigen values
are ……. Ans.
CHECK YOUR PROGRESS:
Q.1 For the eigen values and eigen functions of the Sturm Liouville problem
[Ans. eigen function eigen values are …….]
Q.2 Find the eigen values and eigen functions of the Sturm Liouville problem
[Ans. eigen function eigen values are …….]
Q.3 Find the eigen values and eigen functions of the Sturm Liouville problem
[Ans. eigen function eigen values are …….]
4.5 Number of zeros:
We shall now discuss the problem of determining the number of zeros of non trivial
solutions of the general second order differential equation
+ q(t)y = 0 ….. (1)
Where the functions and q(t) are continuous on some interval Before
moving to the main result, we first understand the Prufer’s transformation.
4.5.1. Cor. I: let be a non trivial solution of ….. (i)
existing on the interval , then the transformation
, reduces (i) to ….. (ii)
80
….. (iii)
and (
) ….. (iv)
Proof: The proof of the above cor. is quite usual and it can be obtained by differentiating
(iii) with respect to t and using relation y = and .
4.5.2 Let the coefficient function P(t) and q(t) in be
continuous on the interval and let u(t) be a nontrivial solution of
,
Suppose u(t) has exactly n ( ) zeros at t = .
[a, b]. If is a function defined by (ii) then ) for
and for , k=1 , 2 , 3…… .
Proof: Since t = . are the zeros of y(t), it follows from the second relation of (ii),
that at t = .Thus from(iii), we have
From the continuity of this implies that is increasing in the nbd of the points
t = , gives the result.
4.5.3 The Sturm Theory:
Let u(t) be a solution of
….. (A)
having first derivative u (t) on Also suppose that u(t) has an infinite number of
zeroes on then for all t on
Proof: Since u(t) has an infinite number of zeros on then by Bolzano-weirstrass
theorem the set of zeroes has a limit point l [ . Therefore there exists a sequence ⟨ ⟩
of zeroes which converges to l. Now since u is continuous
,
81
where through any sequence of points on [ Let through the sequence
of zeroes ⟨ ⟩. Then
.
Now since u (t) exists, then by the definition of differentiability, we have
u (l) =
,
where through any sequence of zeroes ⟨ ⟩.
For such points
u (l) = 0.
Therefore is a solution of equation (A) such that u (l) = 0. Since l [ ,
therefore . This concludes the theorem.
4.5.4. Sturm Separation Theorem: Let u and v be real linearly independent solutions of
….. (A)
on t [ . Then between any two consecutive zeroes of u, there is precisely one zero of v.
Proof: Let are two consecutive zeroes of u on [ . Then by theorem 4.5.3 we
have v( ) v( ) . Now we assume that v has no zero in the open interval
. Then since solutions u and v have continuous derivatives on [ , the quotient u/v
has a continuous derivative on the interval [ .
Also u/v is zero at the end points of this interval, then by Rolle’s theorem point
c such that
*
+ , at t = c . But
*
+
and therefore u and v
are linearly independent on [a, b].
Thus
*
+ , which is a contradiction. Hence v has a zero in Now to claim that
v has precisely one zero in we assume that v has more than one zero in
be two consecutive zeroes of v. Then by interchanging the role of
82
u and v and proceeding as above, we can easily claim that u must have at least one zero in
This concludes the theorem.
4.6 Nonoscillatory equations and principal solutions:
4.6.1 Oscillatory and Nonoscillatory equations and principal solutions:
Equation is of the form
where is a real valued and continuous function on is called oscillatory
equation if it’s all the non trivial solutions have an infinite number of zeros on
These nontrivial solutions are known as oscillatory solutions. If some of them are
oscillatory and remaining are non oscillatory, then such equations are known as non-
oscillatory equation and their solutions are known as non-oscillatory solutions. Clearly the
equation
,
is non oscillatory equation, if is a real valued and continuous function in the
interval . In other word an equation is called non oscillatory in
interval , if no solution of the equation can change its sign more than once in
the interval.
4.7 Nonoscillation theorems:
Theorem: If all the non trivial solutions of are oscillatory, is
continuous, and , then some non trivial solutions of
(1)
are oscillatory. On the other hand, if some non trivial solution of equation (1) are non
oscillatory and then some non trivial solutions of must be non oscillatory.
83
Proof: Let and are the non trivial solution
of (1). Multiplying (1) by y and the equation by z,
and subtracting we get
[
Or [ ……..(2)
Let and be two consecutive zeros of and let and that y(t)
on the interval [ ]. By integrating equation (2) ranging from , we get
∫ [
….(3)
Since and be two consecutive zeros of , we have = 0 with
and Therefore from (3), we get
∫ [
….. (4)
Now we claim that has a zero in the interval [ ]. On the contrary we assume that
has no zero in [ ].Then does not change its sign. Since and
and and are non negative in the interval [ ] leads a
contradiction. Thus has a zero in [ ] and changes its sign in the interval
[ ]. This concludes that between any two consecutive zeros of there is a
Second part follows with the similar argument as we used in first part.
4.8 Unit brief review:
84
UNIT -V: Partial differential equations of first and second order, linear
partial differential equation with constant coefficient.
Unit Structure :
5.0 Introduction
5.1 Partial differential equations of order one and two with examples
5.2 Linear and non linear Partial differential equations
5.3 Linear partial differential equation with constant coefficient
5.4 Algorithm to solve linear partial differential equations with constant coefficient and
solved examples.
5.5 Unit brief review (Summary)
5.0 Introduction:
In the theory of partial differential equations, a variable (say) is a function of
more than one independent variable. Here we are restrict ourselves to assume that is a
function of two variables x and y; we write than
the partial
derivatives of with respect to of order one. Similarly the second order partial
derivatives of are denoted by
. A partial differential
equation is a relation between dependent variable, independent variables and partial
derivatives of dependent variables.
5.1 Partial differential equations of order one and two with examples:
5.1.1 Partial differential equations of order one:
The order of partial differential equation is determined by the highest order partial
derivative in it. Thus a Partial differential equation is called of order one if it contains
highest derivatives of order one i.e. the equation contains
only.
Symbolically denoted as . For examples,
(1) ,
(2) ,
85
are partial differential equations. Clearly order of partial differential equation (1) is one
and order of (2) is two. We already studied the formation of partial differential equations
by eliminating arbitrary functions and arbitrary constants in earlier classes. Thus in the
present chapter we restricted ourselves to discuss the solution of linear partial differential
equations with constant coefficients.
5.2 Linear and non linear Partial differential equations:
5.2.1 A partial differential equation involving partial derivatives p and q and no higher
derivatives is called of order one . In addition, the degree (or power) of p and q is unity
(one), then it is a linear partial differential equation of order one. For example,
2xp+ 5yq= z and p are both linear partial differential equation of order
one. On the other hand + 3q = z and x + y are both partial differential
equation of order one but non linear.
The solutions of linear partial differential equation of order one we already been
studied in the previous classes using Lagrange’s and Charpit’s method. So we left this
to the reader as practice exercise.
5.3 Linear partial differential equation with constant coefficients:
5.3.1 A partial differential equation is called linear partial differential equation with
constant coefficient if it consist higher partial derivatives of z with respect to x and y
but the power of each derivatives that occurs and variable z is one and the coefficient of
various terms are constant quantities. The general form of such equation is denoted as
{[
]
+ [
]
+ ………. +[ ] + N}z = ....(1)
In (1) the notations and taken for the operators
respectively and the
coefficients are all constants.
In brief (1) can be written as , where
86
{[
] + [
]
+ …+[ ] + N}.
5.3.2 Homogeneous linear partial differential equation: Equation (1) in 5.3.1 is called
homogeneous linear partial differential equation if in (1) is in the form
{[
]}
i.e. {[
]}z = ….(2)
The general solution of homogeneous linear partial differential equation consist two parts
namely complementary function (C.F.) and particular integral (P.I.).ion Thus the general
solution of above equation is
z = C.F. + P.I.
Here C.F. is the general solution of while P.I. is the any particular solution
of .
5.3.3Solution of homogeneous linear partial differential equation:
Complementary function(C.F.): We know that C.F. is the general solution of
Let one of the linear factor of is say ( - ).Then we have
( - ) or
or p -
The above equation is clearly of Lagrange form . Therefore the auxiliary
equations of Lagrange’s are
First two relations give (on integration). And the last
relation gives Hence the required solution corresponding to factor
( – ) is
In the same fashion solutions corresponding to all factors can be obtained.
87
Particular solution (P.I.): We know that P.I. is the solution of free
from arbitrary constants. P.I. of any equation is taken as
. Here the symbolic
function can be treated as algebraic function of and and can be factorized or
expanded in ascending powers of or to obtain the required P.I.
Note: In solving the problem treat and as partial operator w.r.t. x and y while
as integration w.r.t. x and y respectively. In the next section we will discuss the
shortcut method to find P.I. in detail.
5.4 Algorithm to solve linear partial differential equations with constant coefficient and
solved examples:
5.4.1. Algorithm to find complementary function (C.F.):
Let the given equation is . We follow the steps given below to
obtain the required C.F:
Step I: Take and put to get auxiliary equation (A.E.).
Step II: Find the roots of the equation i.e. values of m. Let the values of m are
. Then
Case I: If values of m i.e. all are distinct, then C.F. is given by
.
Case II: If values of m are repeated, say then, C.F. is given by
SOLVED EXAMPLES:
Example 1: Solve ( ) z = 0
Solution: The auxiliary equation is {put and =1}
88
Or gives
Therefore required C.F. is
Or
Ans.
Example 2: Solve ( )z = 0.
Solution: The auxiliary equation is {put and =1}
Or gives ,
Therefore required C.F. is
Ans.
Example 3: Solve
Solution: Given equation is
Here A.E. is or gives
[repeated roots]
Therefore required C.F. is
(
)
(
) Ans.
Example 4: Solve (
Solution: A.E. is (
Or ) )= 0
Or or
Or [repeated roots] ,
therefore required C.F. is
Ans.
89
5.4.2 Shortcut Methods To Obtain Particular Integral (P.I.):Let the given equation is
, then
Case I: When is a function of , then to get the particular solution of the
equation , proceed as below [where F is a homogeneous
function of of degree n]:
Algorithm:
Step I: Put and integrate , n times with respect to t.
Step II: Put a for D and b for to get .
Step III: Required P.I. =
[n
th integral of obtained in step I], where .
SOLVED EXAMPLES:
Example 1: Solve ( ) z =
Solution: The auxiliary equation is {put and =1}
Or gives
Hence C.F. is
Now P.I. =
Here is a function of the form and is a homogeneous
function of degree 2 in .
We first put , and then integrating twice with respect to t, we get
i.e. [as ]
Also putting 2 for and 3 for in
We get P.I. =
=
.
Therefore the complete solution is
Ans.
90
Example 2: Solve ( ) z =
Solution: The auxiliary equation is {put and =1}
Or gives
Hence C.F. is
Now, P.I. =
( ) =
( )
[
=
( )
( )
Here is a function of the form and is a
homogeneous function of degree 2 in .
We first put , and then integrating twice with respect to t, we get
i.e. [as ].
Now on putting for and in , we get
P.I1. =
( ) …(A)
The same procedure we apply for second part i.e. for , we get
and putting m for and for in
We get P.I2. =
( ) .
Therefore the complete solution is
P.I1.+ P.I2.
( )
( ) Ans.
5.4.3 Exceptional case when F(a,b) = 0:
If On Putting a for D and b for in , the above method gets
fails. Then to evaluate P.I., proceed as below:
91
Step I: Differentiate with respect to D partially and multiply the expression by x,
so that
Step II: If is also zero, Differentiate with respect to D partially and
multiply the expression by x again, so that
Repeat the above procedure till the derivative of vanishes. If , then
, solved by shortcut method above.
Example 1: Solve ( ) z = 4
Solution: The auxiliary equation is {put and =1}
Or gives
Hence C.F. is
Or
Now P.I. =
4 ,
the denominator becomes zero when 2, therefore differentiating the
denominator with respect to D partially and multiplying by x,
P.I. =
4 ,
The denominator again vanishes when 2, therefore again differentiating the
denominator with respect to D partially and multiplying again by x,
P.I. =
4 ,
The denominator does not vanishes when 2, therefore
92
P.I. =
[ cos = ,
Therefore the complete solution is
.
CHECK YOUR PROGRESS:
Q.1 [Ans.
].
Q.2 Solve ( ) z = [Ans.
].
Q.3 Solve ( ) z = [Ans.
].
Q.4 Solve ( ) z = ( ) [Ans.
].
5.4.4 General method of finding the P.I.
Consider the equation .
This can be written as , which is of Lagrange’s form. Therefore
A.E. is
The first two relations give . …(1)
Taking
On integration we have
z =∫ or z =∫ , from (1)
Thus z =∫ ,
Where constant c is replaced by , after integration.
The above method be repeated for all the factors of .
93
Solved Examples:
Example 1: .
Solution: Given equation can be written as
( ) z =
The auxiliary equation is {put and =1}
Giving
Hence C.F. is
For finding P.I., we use general method.
P.I. =
Or =
Or =
∫ ,
[Because corresponding to the factor we have solution .
Or =
[ ,
Or =
[ , replacing
Or =
[ ,
Or = ∫[
[Because corresponding to the factor we have solution .
Therefore P.I. = [
Or = [
94
Or = [ ,
Thus the complete solution is
CHECK YOUR PROGRESS:
Q.1 [Ans.
].
Q.2 ( ) z =
.
[Ans.
].
5.4.5 Non homogeneous linear equations: A linear partial differential which is not
homogeneous is called non homogeneous linear equation. We consider the differential
,
Where is not necessarily homogeneous. We Classify mainly in two types,
which shall be treated separately as :
(i) is reducible i.e. can be expressed as product of linear factors of the
form , where a and b are constants.
(ii) is irreducible i.e. is not reducible as above.
Case I: is reducible:
Complementary function (C.F.) :
Let be the factor of , then C.F. is easily obtain by applying
Lagrange’s equation concept and write as , where is an arbitrary
function. We now give the different cases here:
(I) have distinct linear factors: If
Where all the factors are distinct, then corresponding C.F. is given by
95
(
)
(
)
Things to remember(1): If the linear factor is , then the corresponding C.F.
is .
(II) have repeated linear factors:
Let a factor occur twice in . Then corresponding C.F. is
given by
(
)[
In general if occur n times in , then the corresponding C.F. is
given by
(
)[
Things to remember(2): If the linear factor is and occurs
twice, then the corresponding C.F. is [
.
Particular integral (P.I.): Particular integral of non homogeneous partial differential
equation can be found in a way similar to those of ordinary differential equations. Here we
are giving some standard form of and their corresponding P.I. format.
Form I: If = , then corresponding P.I.
is obtain by putting
’ in .e.
Required P.I. =
, [Provided ].
Form II: If = , then corresponding P.I.
is
obtain by putting and in .e.
Required P.I. =
,
If the terms of D and D’ remains in the denominator, then
96
(i) Rationalize the denominator and replace and
(ii) Treat D and D’ as partial differential operator to get required P.I.
Form III: If = , then corresponding P.I.
is obtain by expanding
in terms of .
Form IV: If = , then corresponding P.I.
is obtain by
putting ’ in then solve according to the form of V
Required P.I. =
.
SOLVED EXAMPLE:
Example 1: Solve ( ) ( ) .
Solution: According to 5.4.5 case I (I) of complementary functions, here are two distinct
linear factors ( ) and ( ). After comparing to we have
and . Therefore
C.F. =
. [See things to remember(I),
5.4.5]
Now P.I. =
, writing 2 for D, -1 for
=
,
Hence the complete solution is
=
+
.. Ans.
Example 2: Solve ( ) .
Solution: The given equation is ( ) ( ) z = ,
97
According to 5.4.5 case I (I) of complementary functions,
C.F. =
. [See things to remember (I), 5.4.5]
Now P.I. =
( )
( ) , writing for , (-1).2 for
=
,
Rationalizing the denominator, we get
=
, writing .
=
=
=
[
Hence the complete solution is
=
[ …Ans.
Example 3: Solve ( ) .
Solution: The given equation is ( ) ( ) z = ,
According to 5.4.5 case I (I) of complementary functions,
C.F. =
. [See things to remember (I), 5.4.5]
Now P.I. =
[
= [
98
= [
= [
= [
= [ .
[Taking D and means partial differentiation w.r.t. x and respectively]
Hence the complete solution is
=
[ . …Ans.
CHECK YOUR PROGRESS:
Q.1 Solve
Ans. [
[ ].
Q.2 Solve ( )z =
Ans. [
].
Q.3 Solve ( )z = xy Ans. [
].
Case II: is irreducible i.e. is not reducible into linear factor:
In this case we follow the few steps to find C.F. as below:
Step I: Take trial solution =
Step II: Put the above value in the given equation.
Step III: Find the value of k in terms of h.
Put the value of h in trial solution to get the required solution.
NOTE: To find Particular Integral we adopt the same technique as in case I.
Solved Examples:
Example 1: Solve ( )z = 0.
99
Solution: Let the solution of the above equation be = ….. (1)
Then from the equation
( ) = 0 ( ) = 0
( )
Putting in (1), we get the required solution is
= … Ans.
Example 2: Solve ( ) z = .
Solution: Let the solution of the above equation be = ….. (1)
Then from the equation
( ) = 0 ( ) = 0
( )
Putting in (1), C.F. is
=
Now P.I.=
( )
( ) , writing for
=
,
After rationalizing the denominator, we get
=
100
= ( )
= ( )
, writing for
=
( )
=
( )
Hence the complete solution is
=
( ) … Ans.
CHECK YOUR PROGRESS:
Q.1 Solve ( ) z = 0 [Ans. = ]
Q.2 Solve ( ) z = 0 [ = ].
101
LIST OF REFERENCES:
1. Evans, Lawrence C.(1-CA) Partial differential equations. (English. English summary)
Graduate Studies in Mathematics, 19. American Mathematical Society, Providence, RI,
1998.
2. Folland, Gerald B.(1-WA) Introduction to partial differential equations. (English.
English summary) Second edition. Princeton University Press, Princeton, NJ, 1995
3. Andrews, Larry C. Elementary Partial Differential Equations with Boundary Value
Problems New York, NY: Academic Press, 1986.
4. Balachandra Rao S. & H.R. Anuradha Differential Equations
5. Sankara Rao, Introduction to Partial Differential Equations
6. Raisinghania, M D Ordinary And Partial Differential Equations book
7. Raisinghania, M D , Advanced Differential Equations book
8. Zafar Ahsan, Differential Equations and Their Applications, Prentice Hall of India, New
Delhi 1999.
9. B. D. Sharma, Differential Equations, K. N. R. N. Delhi 1983
10. Kapoor, N. M. Differential Equations, P.P. Company Ltd. Delhi 1999.
11. Renardy, Michael Rogers, Robert C.(1-VAPI) An introduction to partial differential
equations, Springer-Verlag, New York, 1993