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.
1
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.
2
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)
3
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
+ where
………………. (1)
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.
4
(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
If (
), (
cos
(say)
) repeated twice.
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 = general solution.
+
,itself
5
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 Hence roots are 2, and -1 C.F. is y=
+
m = -1
, one is real and other is a pair of imaginary. Therefore
cos
sin
1.2.3 Particular integral (P.I.): If X
), itself general solution.
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 X=
then put D= a in f(D) i.e.
Corresponding P.I. .If then (D-a) is one of the factor of f(D).This
6
factor is solve by formula (3) and rest is solve by shortcut method given here. X = sin ax (or cos ax) then put D2= -a2 in
sin ax (or cos ax), provided
f(D) i.e.
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. Since f(a)
P.I. =
[here X=
+ ]
on putting D=-2, we get P.I. y =
=-
=-
Hence the general solution is y = C.F.+ P.I. i.e.
y=
+
-
.
7
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. =
Therefore
P.I. =
.
+
[we know that cos 2x= 2
y =
[
-1.]
]
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. =
= =
.
=(
)
=
= 8
=
[since (1-D)-1 = 1+D+D2+…….]
=
[ D means differentiation wrt. x]
=(
)
=
+ .
Check your progress: Solve the following differential equation: Q.1
.
Q.2
[Ans. y =
+
+
. [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. 9
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
..……… (3)
(say),
..………. (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:
10
Step I: Write the given equation in D-notation form. etc in the equation by
Step II: Replace =
,
=
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
[
Here A.E. is
]
, we have = 2z
(as log
)
= 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. or
y= y=
.
( putting z = log x)
Ans.
11
Example 2: Solve
-
.
Solution: D-notation form of the given equation is On putting x =
=
and
[
Here A.E. is
, we have
]y=2
(as
)
= 0 gives m = 2, 2. So C.F. is y =
Now P.I. = = P.I.
=
(here X is of the form
= =
∬
(here
, V=1,art.1.2)
)
=
= Hence the general solution is y = C.F.+ P.I. i.e.
y=
or
y=
or
y=
Example 3: Solve
. .
+
(putting z = log x)
Ans.
.
Solution: D-notation form of the given equation is . 12
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
P.I. = 5
∫ ……….. (ii)
=2 from (i) and (ii)
=2
+2
Hence the general solution is y = C.F. + P.I. 13
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 14
then we get b= =b
and so we have
( )
,
(
)
…………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 ]y=
, the equation becomes .
, 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.
15
Example 2: Solution: on putting
, then
[
]y=
A.E. is Hence,
, the equation becomes .
, where
, C.F. =
Now P.I. = = =
, (on writing 4
) =2z
, failure case)
(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 =
}.
16
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) …… (2)
and
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 17
(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
18
(D2+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)+ = -3c1 e-3t -5c2 e-5t +
}+3{(c1 e-3t +c2 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
19
……… (1)
P1 dx + Q1 dy + R1 dz = 0, P2 dx + Q2 dy + R2 dz = 0
where the coefficients are function of x, y, z. Solving these equations simultaneously, we get, ……… (2)
or 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
Similarly, taking second and third fractions, we get, Thus the complete solution is Example 2: Solve
= c,
+ c or ,gives
=c c’
c’
.
Solution: On taking first two fractions, we have =y
, on integration, we get
+ c or
= c ...... (i)
20
Similarly, taking second and third fractions, (
cy
)
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 21
(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)
or
… (iii)
and
From (i),(ii),and (iii) we have
22
Taking first two fractions of (iv), we get on integration gives i.e.
….(v)
=c
Similarly on taking last two fractions, we get
….. (vi )
= c’
(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.
23
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 …. (1)
The equation is P
Let S(x, y, z) = c be a solution of equation (1). Then total differential dS must be equal to P
…. (2)
, but, we know that
On comparing (1) and (2), we have
=
, say …. (3)
so that now from first two equations, we have =P
Since
and
, we have P
Gives
)=Q
=Q
=Q …. (4)
P
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).
24
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 obtained by putting
these conditions are
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 )+
)+ )–x
)=0 )+
)=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. 25
Here
,
= 0,
,
The condition of integrability is )+
)+
A.T.Q., we have
)=0
) +1
)+
) = 0, satisfied here.
Therefore equation is exact, so on rearranging the terms of the given equation, we have =0, integrating, we get Or
=
c
, 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 Or Or Or
+{
u-
, integration gives,
—
—
Example 2: Solve ( Solution: Here P = (
)+
=0
)}
=0 —
)+ 2
) = c, solution is obtained by putting u = + yz ) dx +( xz + + yz ), Q = ( xz +
dy + ,R=
.
- xy) dz = 0 - xy). The equation satisfies
integrability condition. Also P, Q, R are homogeneous function, so on putting 26
, we get
and the equation
becomes ( (
+v +v
)( (
)+( )+
+
-
+
Or
-
) dz = 0
) dz = 0
,
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 integrable, then we take the third variable z = constant , so we get
readily . 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) 27
now, if we put z=constant, then we have dz = 0 and the equation becomes
x dx –( x dy+ y dx) = 0, integrating gives, on differentiating, we get (x - y) dx – x dy = comparing (1) and (3), we get
……. (2) …….. (3)
dz
gives
+c
putting this value in (2),the solution is ,
+c [
or Example 2: Solve (
Ans.
) 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 = Or
I.F. =
,
=-(
=
the general solution is
∫
[ by (1)]
), which is linear equation in
∫
. =
=
∫(
= ∫
)
, therefore
∫
=
, putting this value of or
.
Ans. 28
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 +
Solution: here P = (
dy +
+ yz ), Q = ( xz +
,R=
- xy) dz = 0. - xy)
, Thus A.E. is or
….. (1)
or
Taking last two fractions, we get Also taking 1, 0, 1 as multipliers, we get Or
or
= v (say ). …. (2)
Then A du + B dv =0
gives
A
)+B
+
) =0
[
i.e.
Comparing with the given equation, we get (
+ yz )
i.e.
( +z) =u 29
And
( xz +
Or
i.e. =
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 Q.2 (
+2x)
-4 +
+
= 0. +
Q.3
[Ans. +
= 0.
[Ans.
.
] [
Q.4 Q.5
]
. .
[ [Ans.
] ] ].
1.10 Unit brief review (Summary):
30
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.
31
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 ∫ Or
∫
,
∫
[using (2)] ……... (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 ∫
, 32
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). by putting n=1, 2, 3….. in
Step II: Find ∫
and corresponding values given of
and
Solved Examples: Example 1: Apply Picard’s method up to third approximation to solve . Solution: Here
and ∫
Picard’s method is given by So, first approximation ∫ A.T.Q.
. We know that the nth approximation by
is obtained by putting n=1 ∫
or ∫
[since
]
or
Similarly, second approximation
is obtained by putting n=2
∫ And, third approximation,
∫ ∫
or . Ans.
33
Example 2: Apply Picard’s method up to third approximation to solve . Solution: Here
. The nth approximation by Picard’s
and ∫
method is given by So, first approximation ∫
………. (A)
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 method is given by
and ∫
. The nth approximation by Picard’s ………. (A)
34
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)
35
So, first approximation
is obtained by putting n=1 in (A),
therefore we have ∫
or
∫
∫
[Since
]
Or
….
and ∫
or
(i)
∫ …. (ii)
The second approximation
given by
∫
∫
∫
Or
∫
∫
Or
…. (iii)
and 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):
36
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) 37
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 ∑
Clearly sequence {
…………. (7)
} converges if the infinite series ∑ ∫
From (4)
converges.
, ∫
,
= ∫ [
………. (8)
The relation (8) is true for k = 1, 2, 3, ………. , ∫
Also from (4), we have ∫
……… (9)
Now
|
|
∫ |
Or
|
|
|
|
|
∫
|, from (2)
|
……….. (10)
= ∫ [
Again from (8) |
||
|
∫ | ∫
|| |
||
| =
|
|
∫ |
|
|| from (10)
|, from (3) ………(11)
38
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 39
|
|
|
|
……….. (2)
for all points in the domain D. Proof: Let if possible the I.V.P.
,
∫
Then
=
have two solutions y(x) and v(x). ∫
and
∫ [ |
|
∫ | (
)|
∫
)| |
| (
|
|, from (1)
= 2M|
…….. (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:
40
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 |
|
|
|
Where the constant K is independent of | lie in D, where Mh
|
|
……….. (2)
. Let the rectangle R defined by |
……… (3)
. 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
or |
∫ Or
|
|
lies in R. We have
|
|
|
∫ |
Mh
||
|
∫
|
|, from (1)
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) 41
Or
|
|
Shows that
|
|
Since, we have | |
by (3).
|
|
lies in R.
Secondly we claim that |
|
Mh
|
|
Similarly |
| |
|
On continuing, we have
……. (5)
|
∫ | ∫
|
|
||
|
||
| =
|
|
∫
||
|
|
|
=
|
|
|
|, from (2)
|
, 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 tends to
+…………+
satisfies the differential
,
=
. Since
uniformly in R and by Lipchitz conditions, we have on taking limit
, we get from (4) ∫
,
42
∫
∫
=
…… (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 | | f (x, y)=
| |
, satisfy the Lipchitz condition. Find the Lipchitz constant.
Solution: Here f (x, y)= function on R ,
show that the function
=
. Since f is real valued
is exist and being continuous and bounded in R.
| |=|
|
As we know that |
| |
Lipchitz condition. Thus | |
|
|
|
and | |
| ||
, where
where K=
|
|
|
| | +1}
is positive constant. Hence f satisfies
+1.
Example 2: Examine existence and uniqueness theorem for the I.V.P. Solution: Here f (x, y)=
and
. Clearly f and
(x , y). We consider rectangle R as |
+1}.
|
|
,
=
are both continuous for all
|
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. | h = min.{a, b/M}, then the problem posesses a unique solution in |
|
| and .
43
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:
44
45
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.
46
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.
47
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 48
defined on maximal interval of existence conditions
where
depends on initial
.
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 ( (
) as (
Similarly upper semi-continuity of function (
) is defined as
)
at (
(
)
(
)
) is defined as ) as (
)
(
)
3.2 Continuity and differentiability theorems: 3.2.1 Continuity theorem: Let function property that for every (
be continuous on an open set E with the
) E then I.V.P. …..(1)
Let
be the maximal interval of existence of the solution , then 49
or function of
and
is a lower(upper) semi continuous ,(
is continuous on the set
) E
Proof: The I.V.P. (1) can be replaced by
,
,
i.e. by the I.V.P. and ( Without loss of generality assume that function defined on an open interval
….. (2)
) by does not depend on z. Thus
with the property that
is 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 ( And let (
of )
) be the maximal interval of existence. Also (
) 50
, 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
Now we claim that
is upper semi continuous.
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 Hence
|
| |
, then |
is continuous on set (
| ).
51
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 Is the class
on its open domain (
(ii) Furthermore if J(x)= J y at
…. (1)
of ) where
.
is the Jacobian matrix
…. (2)
J(x)= J
Then
…. (3)
is the solution of the I.V.P.
Where
and
And
{
is the solution of
Where
= at
of f(x, y, z) with respect to
... . (4)
is the vector. 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)
52
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)
53
is a point (
) obtained by solving (
)
(
On the other hand if (
and
and such that
) ) 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 ( )
(
)( )
By ignoring the non linear terms
(
)
…. (2) in (2), related linear system is
54
…. (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 ( that if the initial point ( remains close to (
) of the almost linear system (2) is said to be stable provided ) is sufficiently close to (
) t 0. In other words for given
| Then critical point (
) then the point
|
|(
if )
|
) 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
55
|
|
.
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 equation | Or
(
) are the roots of the eigen
| .
Let the roots (eigen values) of the above eigen equation are the critical point (0,0) is depend on the values of
. Then the nature of and describes as below:
56
Nature of roots
Linear system Type Node
Node
Stability Unstable
Almost linear system Type Node
Asymptotically Node stable
Saddle
Unstable
point Node
Stability Unstable
Asymptotically stable
Saddle
Unstable
point Unstable
Node or
Unstable
Spiral Node
Spiral
Spiral
Asymptotically Node or Asymptotically stable
Spiral
Unstable
Spiral
Unstable
Spiral
Asymptotically
Asymptotically stable
Centre
Stable
stable
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
57
Where
and
.
Now for critical point we must have and Solving the equations, we get
. Thus (0,0) is a critical point.
Also =
√
and
=
√
=0
√
√
=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. 58
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 =
√
and
=
√
=0
√
=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.
59
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 periodic if neither function is constant, if both are defined for all
is said to be and if
such
that
60
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 √
…..(6)
and
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 61
as If c
0 then
. Also if c
and 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 then
change by
2
If P moves once around C
, 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. 62
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 ,
63
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 Let y =
….. (1)
be a known part of the C.F. i.e. a solution of equation ,
Let y =u Now y =u
….. (2)
be the complete solution of equation (1), where u is the function of x. gives
=
and
Substituting these values in (1), we get (
) (
)
Or
(
)
(
)
, , ….. (3)
=
Putting ,
in (3), we get - =
Which is linear in t, therefore I.F. =
….. (4) ∫{
}
∫
Hence solution of (4) is 64
t (I.F.) = ∫
or
Solving and putting back u=(
∫
)=∫
, and integrating we get ∫
∫
) [∫
Complete solution of (1) is y =u 4.1.3 Special integral part
∫
t(
]
, where u is given by (5). : The integral part given below
of C.F. to
is obtain by putting
and Condition
.. in (2). Integral part
is
1+P+Q=0 1-P+Q=0 + mP + Q = 0 P + Qx = 0 2+2Px + Q
X
+=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. Step II: Test for integral part Step III: Put y = u
coefficient of
is unity.
of C.F. as per table given in 4.1.3 above.
in step I equation and simplify as simplified in 4.1.2.
The following method fully illustrates the method. Solved Examples: Example 1: Solve 65
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 = 0, Putting
, we get x
=
, then the given equation reduces to 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;
=0
Clearly P + Q x =
is a part of C.F.
Assume that the Complete solution is y =u = x, Putting or
(
) =
, we get x
= ….. (I)
,
a linear equation in t whose I.F. = Solution of (I) is
then the given equation reduces to
∫(
)
∫ 66
.
(
)
/
=.
(
)
/
)
[Putting
/
.
This gives
(
[
)
=
/
dx + .
Hence complete solution is y =
]
/
.
Integrating both sides, we get
]
/
or .
Or
(
.
∫
.
/
dx +
.
Ans.
CHECK YOUR PROGRESS: Q.1 Solve
[Ans.
Q.2 Solve
[Ans.
dx +
].
+
].
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: 67
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 (
Or
) =
And
, differentiating both sides gives
=
Putting
.. (2) ….. (3)
∫
Again from (2) we have
=
(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.
68
∫
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=
∫
=
,
( –
And
to reduce in normal form we choose
)
= (
–
)= 2
Equation reduces to
. C.F. =
Here A.E. is 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
∫
=
=
,
=
69
( –
And
)
( –
)= 6
Equation reduces to
. C.F. =
Here A.E. is 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
( )
(
)
,
70
Substituting these values in (1), we get [
( )
,
( )
Or
[
]
+
, …. (2)
Or
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
(
or
)
integrating again, we get
or
Integrating, we get log
or
which can be easily solved provided by
in (2), equal to zero i.e.
, this value of z reduces (2) to comes out to be a constant or a constant multiplied
.
Case II: We choose z such that
i.e.
( )
or
(
)
(constant),
, integrating gives
∫
.
71
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,
(
)
,
(
Let us choose z such that
i.e.
Then
( )
(
or
(
)
and
)
)
(
)
.
= - 2 (constant)
, integrating gives
and
(
.
)
….. (iii)
=
72
Hence equation (ii) transferred to Or
Or
By (iii) ….. (iv)
Or Now C.F. =
+
and
P.I. =
=
,
Hence the solution of the given equation is y=
+
+
Or y =
+
+
Ans.
[ As z = sin x ]
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 73
+ Since
or
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 Q.4 Solve
+
[Ans. y =
[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
74
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
independent variable we have conditions on the unknown function derivatives
of and its
.
Ex. (2) y
is a B.V.P. because for two values of independent variable we have conditions on the unknown function
its derivatives
and
.
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,
the real valued continuous functions of x . Also
and 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
75
and We assume that p(x) 0 when eigen value
. Let y(x) is an eigen function corresponding to an
, 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) …. (vi)
and 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 )∫
*[ )∫ Thus we have
[ ]
+[
+
= 0, in view of (ii), (iii), (v) and (vi). )= 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. 76
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 Therefore Case II. If
or
, which is not an eigen function. ,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 Therefore Case III. If
or
, which is not an eigen function. ,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
77
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 Case II. If
.
, which is not an eigen function. ,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
Therefore Case III. If
or
, which is not an eigen function. ,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
78
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 ….. (1)
+ q(t)y = 0 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
….. (i)
be a non trivial solution of
existing on the interval
, then the transformation , reduces (i) to
….. (ii)
79
….. (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 = 4.5.2 Let the coefficient function P(t) continuous on the interval
and
.
and q(t) in
be
and let u(t) be a nontrivial solution of ,
Suppose u(t) has exactly n ( [a, b]. If
.
is a function defined by (ii) then
and .
t=
for
are the zeros of y(t), it follows from the second relation of (ii),
at t =
From the continuity of
)
, k=1 , 2 , 3…… .
for
Proof: Since t = that
) zeros at t =
.Thus from(iii), we have
this implies that
is increasing in the nbd of the points
, gives the result.
4.5.3 The Sturm Theory: Let u(t) be a solution of ….. (A) having first derivative u (t) on zeroes on
then
Also suppose that u(t) has an infinite number of for all t on
Proof: Since u(t) has an infinite number of zeros on theorem the set of zeroes has a limit point l
[
then by Bolzano-weirstrass
. Therefore there exists a sequence 〈 〉
of zeroes which converges to l. Now since u is continuous , 80
through any sequence of points on [
where
Let
through the sequence
of zeroes 〈 〉. Then . Now since u (t) exists, then by the definition of differentiability, we have u (l) =
,
through any sequence of zeroes 〈 〉.
where
For such points Therefore
u (l) = 0. is a solution of equation (A) such that
therefore
[
u (l) = 0. Since l
,
. 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. are two consecutive zeroes of u on [
Proof: Let have v( )
v( )
. Then by theorem 4.5.3 we
. Now we assume that v has no zero in the open interval
. Then since solutions u and v have continuous derivatives on [ has a continuous derivative on the interval [
.
Also u/v is zero at the end points of this interval, then by Rolle’s theorem c
such that
*
+
, at t = c . But
, the quotient u/v
*
+
point
and therefore u and v
are linearly independent on [a, b]. Thus
*
+
, which is a contradiction. Hence v has a zero in
v has precisely one zero in
Now to claim that
we assume that v has more than one zero in be two consecutive zeroes of v. Then by interchanging the role of 81
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 nonoscillatory 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 continuous, and
are oscillatory,
is
, 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.
82
Proof: Let
and
are the non trivial solution
of (1). Multiplying (1) by y and the equation by z, and subtracting we get [ [
Or Let
and
……..(2)
be two consecutive zeros of
on the interval [
and let
and that y(t)
]. By integrating equation (2) ranging from
, we get
∫ [ Since
and
be two consecutive zeros of and
….(3)
, we have
= 0 with
Therefore from (3), we get ∫ [
Now we claim that has no zero in [ and contradiction. Thus [
….. (4)
has a zero in the interval [ ].Then and
]. On the contrary we assume that
does not change its sign. Since
and
are non negative in the interval [
has a zero in [
] and
] leads a
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:
83
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 derivatives of
with respect to
derivatives of
are denoted by
the partial
of order one. Similarly the second order partial . 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 Symbolically denoted as (1) (2)
only.
. For examples, , , 84
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 {[
] +[
] + ………. +[
In (1) the notations
and
coefficients In brief (1) can be written as
] + N}z =
taken for the operators
....(1)
respectively and the are all constants.
, where 85
{[
]+[ + …+[
]
] + N}.
5.3.2 Homogeneous linear partial differential equation: Equation (1) in 5.3.1 is called homogeneous linear partial differential equation if
i.e.
{[
]}
{[
]}z =
in (1) is in the form
….(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 of
while P.I. is the any particular solution
.
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 ( -
)
or
The above equation is clearly of Lagrange form
is say ( -
).Then we have
or p . Therefore the auxiliary
equations of Lagrange’s are
First two relations give relation gives ( –
) is
(on integration). And the last Hence the required solution corresponding to factor
In the same fashion solutions corresponding to all factors can be obtained.
86
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 function
can be treated as algebraic function of
expanded in ascending powers of
or
Note: In solving the problem treat
and
. Here the symbolic and
and can be factorized or
to obtain the required P.I. 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 ( Solution: The auxiliary equation is
)z=0 {put
and
=1} 87
Or
gives
Therefore required C.F. is
Or Example 2: Solve (
Ans.
)z = 0.
Solution: The auxiliary equation is Or
{put
gives
and
=1}
,
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
)
Or Or
)= 0
or [repeated roots] ,
therefore required C.F. is
Ans.
88
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
, proceed as below [where F
equation
is a homogeneous
of degree n]:
function of Algorithm: Step I: Put
and integrate
Step II: Put a for D and b for
, n times with respect to t.
to get
.
[nth integral of
Step III: Required P.I. =
obtained in step I], where
.
SOLVED EXAMPLES: Example 1: Solve (
)z=
Solution: The auxiliary equation is Or
Now
=1}
P.I. =
Here
is a function of the form
function of degree 2 in We first put
and
[as
is a homogeneous
. , and then integrating
twice with respect to t, we get
]
Also putting 2 for We get
and
gives
Hence C.F. is
i.e.
{put
and 3 for
in
P.I. =
=
Therefore the complete solution is
.
Ans.
89
Example 2: Solve (
)z=
Solution: The auxiliary equation is Or
{put
and
=1}
gives
Hence C.F. is Now,
P.I. = =
(
=(
)
(
)
Here
)
(
[
)
is a function of the form
homogeneous function of degree 2 in We first put
and .
, and then integrating
i.e.
[as
Now on putting
twice with respect to t, we get
].
for
and
P.I1. =
is a
(
in
, we get …(A)
)
The same procedure we apply for second part i.e. for and putting m for We get
P.I2. =
(
and
for
, we get in
.
)
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:
90
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 Or
{put
and
=1}
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 91
P.I.
=
cos
[
=
,
Therefore the complete solution is
.
CHECK YOUR PROGRESS: Q.1 Q.2 Solve (
)z=
[Ans.
[Ans.
Q.3 Solve (
)z=
[Ans.
Q.4 Solve (
)z=(
) [Ans.
]. ].
].
].
5.4.4 General method of finding the P.I. Consider the equation
. , which is of Lagrange’s form. Therefore
This can be written as A.E. is The first two relations give
…(1)
.
Taking On integration we have z =∫ Thus
z =∫
or z =∫
, from (1)
,
Where constant c is replaced by
, after integration.
The above method be repeated for all the factors of
.
92
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 Or
=
[
Or
=
[
Or
=
[
Or
=
we have solution
.
, , replacing ,
∫[
[Because corresponding to the factor
we have solution
.
Therefore P.I. = [ Or
=[
93
=[
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.
form
can be expressed as product of linear factors of the
, where a and b are constants.
(ii)
is irreducible i.e.
Case I:
is not reducible as above.
is reducible:
Complementary function (C.F.) : Let
be the factor of
Lagrange’s equation concept and write as
, then C.F. is easily obtain by applying , 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 94
(
)
(
)
Things to remember(1): If the linear factor is is
, then the corresponding C.F.
.
(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 twice, then the corresponding C.F. is
and occurs
[
.
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 Form I: If
= ’
, then corresponding P.I. in
Required P.I. = Form II: If
and their corresponding P.I. format.
=
obtain by putting
is obtain by putting
.e. ,
[Provided
].
, then corresponding P.I. and
is
in
Required P.I. =
.e. ,
If the terms of D and D’ remains in the denominator, then 95
(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.
in terms of Form IV: If
.
=
, then corresponding P.I. ’
putting
is obtain by expanding
in
is obtain by
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
and
. Therefore
C.F. =
we have
.
[See things to remember(I),
5.4.5] Now
P.I. =
, writing 2 for D, -1 for
=
,
Hence the complete solution is
=
+
Example 2: Solve (
)
Solution: The given equation is (
)(
.. Ans. . )z=
, 96
According to 5.4.5 case I (I) of complementary functions,
C.F. = Now
P.I. =
(
.
[See things to remember (I), 5.4.5]
)
(
, writing
)
=
for
, (-1).2 for
,
Rationalizing the denominator, we get =
, writing
.
= = [
=
Hence the complete solution is
=
[
Example 3: Solve (
)
Solution: The given equation is (
)(
…Ans.
. )z=
,
According to 5.4.5 case I (I) of complementary functions, C.F. = Now
P.I. =
.
[See things to remember (I), 5.4.5]
[
= [
97
= [ =
[
= [ =
[
. [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 ( Case II:
].
)z = xy is irreducible i.e.
Ans. [
].
].
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. 98
….. (1)
Solution: Let the solution of the above equation be = Then from the equation (
)
=0
( (
Putting
)
=0
)
in (1), we get the required solution is … Ans.
= Example 2: Solve (
)z=
. ….. (1)
Solution: Let the solution of the above equation be = Then from the equation (
)
=0
( (
Putting
)
=0
)
in (1), C.F. is =
Now
P.I.=
(
(
=
)
, writing
)
for
,
After rationalizing the denominator, we get = 99
=
=
(
)
(
)
=
(
=
(
, writing
for ) )
Hence the complete solution is =
(
)
… Ans.
CHECK YOUR PROGRESS: Q.1 Solve (
)z=0
[Ans. =
Q.2 Solve (
)z=0
[
=
] ].
100
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
101
