UNIT 8 METHOD OF SYMBOLIC
OPERATORS
Structure
8.1
Introduction
Objectives
8.2
8.3
8.4
8.5
8.6
5.7
Differential Operator
General Method of Finding Particular Integral
Short Methods of Finding Particular Integrals
Euler's Equations
Summary
Solutions/Answers
8.1 INTRODUCTION
In Unit 5, we saw that the general solution of a non-homogeneous linear
differential equation consists of two parts, namely, complementary function and
particular integral. In Unit 6, we developed the method of undetermined
coefficients for finding the particular integral of non-homogeneous linear
differential equations with constant coefficients for certain particular forms of the
non-homogeneous terms. We dealt with the method of variation of parameters in
Unit 7. This method provides the particular integral of non-homogeneous linear
differential equations with constant as well as variable coefficients provided, all the
linearly independent solutions of the corresponding homogeneous equation are
known. In the case of non-homogeneous equations with constant coefficients, the
constraints on the use of both these methods namely, the method of undetermined
coefficients and the method of variation of parameters call be overcome to a large
extent when we use the method of differential or symbolic operators. The notion
of symbolic operator can be traced back t o Brisson (1808) and its use was carried
out by Cauchy (1789-1857).
You may recall that in Sec. 5.2.1 of Unit 5 we ,defined a linear differential
operator L of order n in the form
L
=
aoDn + alDn-'
where D =
+ ....+
an-lD + a,,
d
denotes differentiation with respect to x.
dx
---
In this unit we shall make use of the method of differential operator to find a
particular integral of non-homogeneous linear differential equation with constant
coefficient. The determination of a particular integral of a non-homogeneous
equation depends upon the properties of the operators inverse to D, that is, D - I .
Although the problem of inverse operator was taken up by Lobatto in 1837 dnd
Boole in 1841, the full justification of inverse operators has not been given to
date. Inverse operators are treated only as symbolic operators but, when applied,
this technique normally yields particular integral by simple and short methods.
We have also given the properties of differential operators and inverse operators in
this unit.
Objectives
After going through this unit you should be able to
define an operator and symbolic inverse operator;
state properties of operators and symbolic inverse operators;
obtain particular integral of a given equation using general method of symbolic
operators:
I
Sccond and Higher Order
Ordinary Differential Equations
use shorter operator methods of finding particular integral when nonhomogeneous term is of the form exp(ax), sin(ax + b) or cos(ax + b), polynomial
in x, exp (ax).V(x) (where V(x) is a polynomial in x).
8.2 DIFFERENTIAL OPERATORS
In calculus, we often use the symbol Dn to denote the nth order derivative of a
function, i.e., Dny
=
-.dny
dxn
Using this notation, a linear differential equation with constant coefficients, viz.,
aoY(")+ sly("-') + a2y(n-2)+ .... + any = b(x)
...(I)
can be written as
(a,,Dny + a , ~ " - 'y + a , ~ " - ~ +
y ....+ any) = b(x)
or, (a,,Dn + a l ~ " - '+ .... + a,) y = b(x).
The expression
a,,Dn
+
al~'-l
+ .... +
a , - 2 ~ 2+ an-lD
+
a,
%
...(2)
as we have already mentioned in Sec. 5.2.1 of Unit 5 is a linear nth order
differential operator. Since expression (2) is a polynomial in the symbol D, we
often abbreviate it by f(D). Then Eqn. (1) can be written in the form
...(3)
f(D)y = b(x)
and is read as "f(D) operating on y equals b(x)". It may be noted that f(D) has
meaning only when applied to some operand y.
We say that two operators L1 and L2 are equal, if and only if, the same result is
obtained when each acts upon the function y. That is, LI = L2 if and only if,
Ll(y) = L2(y) for all functions y possessing the derivatives necessary for the
operations involved.
The product L1L2of two operators L1 and L2 is defined as that operator which
produces the same result as is obtained by using the operators L2 followed by the
operator L1. Thus L1L2(y) = LI(L2(y)). The product of two differential operators
always exists and is a differential operator. Moreover, if L I and L2 are operators
with constant coefficients then L1 Lz = L2 L1, but it is usually not true for
operators with variable coefficients. For instance, if
L1 = D
+
2 and L2 = 3D-1, then
Similarly, we have L2(Ll(y)) = ( 3 + ~5D ~- 2) y, which shows that LlL2 = L2L1.
But if L1 = xD + 2 and L2 = D .- 1. then it can be checked that L1L2 = X D +
~
(3 - x) D - 2, whereas L2Ll = xD2 + (1 - X) D - 2. This is because L1 is an
operator with variable coefficients whose product is dependent on the order of the
factors. In this unit we shall be dealing with operators with constant coefficients.
The sum of two differential operators is obtained by adding their corresponding
coefficients. For instance, if L1 = 3D2 - D + x - 2 and L2 = X ~ +D 4D~ + 7
t h e n L , + L2 = (3 + x ~ ) D
+ 3D
~ + x + 5.
Differential operators are linear operators, that is, if L is any operator, cl and c2
are constants and yl and y2 are any functions of x each possessing the required
number of derivatives, then
L ( C ~ Y+I ~ 2 ~ 2= ) c1L (Y,) + c2L(y2)
Thus, on the basis of the above discussion and definitions of addition and
multiplication, we can say that differential operators satisfy the following
fundamental laws of operation.
Fundamental Laws of Operation
If L1, Lz and L3 be any three differential operators then,
L1 + Lz = Lz + L1 (addition is commutative)
i)
ii) (L1 + Lz) + L3 = L1 t. (L2 + L3) (addition is associative)
iii) (L1L2)L3 = LI(L2L3)(multiplication is associative)
iv) L1(L2 + L3) = LILz + L1L3(multiplication is distributive
with respect to addition)
v)
If Ll and L2 are operators with constant coefficients then
L1L2 = L2L1 (multiplication is commutative)
From the definition of differential operations, it also follows that if m and n are
any two positive integers, then
Note that under the operation of addition and multiplication the operators with
constant coefficients behave like algebraic polynomials. We can, therefore, use the
tools of elementary algebra. In particular, multiplication may be used to factor
operators with constant coefficients. For instance, we can write D3 - 3DZ + 4 =
(D+1) ( D Z - 4D + 4) = (D + 1) (D - 2)'and (D - 2) (D + 1)' = D3 - 3D - 2.
You may now try this exercise.
El) Factor each of the following operators:
a) 2 ~ +' 3D - 2
b) D3 - 2 ~ -' 5D
c) 2D4
+
+
6
11D3 + 18DZ + 4D - 8
d) D~ - 11D - 20
Operators (i) - (v) above are very useful for obtaining certain properties of
differential operators and inverse differential operators, which in turn are useful in
finding solutions of homogeneous linear equations. Before discussing the properties
of the operators let us define inverse differential operators.
Consider Eqn. (3), namely,
f(D) Y
=
b(x),
where f(D) is an operator with constant coefficients only. In order to find a
particular solution we write it as
'I
and then try to define an operator -so that, the function y in relation (4) will
f(D)
have a meaning and will satisfy Eqn. (3). In other words, what all we require is that
In particular, if we have f(D)
y = D-' b(x)
and b(x) = DD-' b(x),
=)
so that DD-' = 1.
=
D then Eqn. (3) reduces to Dy = b(x)
Method of Symbolic O p e ~ t o ~
W e n d and Higher Order
Ordinary Differential Equations
Thus, D-' represents such an operation on any quantity that, if the operation D is
subsequently performed, the quantity is left unaltered. Thus, D-' is an operator
inverse to D. Moreover, we know that differentiation and integration are inverse
operations. Therefore, D-' -is an operation of simple indefinite integration.
Similarly, D-P is the operation of p-times integration. You may note here that the
main object of these inverse operations is to find an integral but not the complete
integral, and therefore we can omit the arbitrary constant which arises in
integration.
1
From relations (4) and (5) we can thus say that -b(x) is that function of x
f(D)
which when operated upon by f(D) gives b(x).
1
(6 + 6 x 3 = x3, because (D' + 2D) x3 = 6 + 6x2.
D~ + 2D
1
Thus operator -is the inverse of operator f(D) and we can write the
f (Dl
particular integral of relation (3) as
For example,
We know that the general solution of Eqn. (3) is y
form of y,, it reduces to
=
y,
+
y,. With the above
The sum and difference of inverse operators is defined as
I
When we apply two or more inverse operators in succession to an operand, then
the operator immediately next to the operand in the left is applied first, then the
next and so on. Thus,
I
Moreover, any two operators,
1
1
and - -are equivalent, that is, they yield the same result
g(D) h(D)
g(D) ' h(D)
when applied to the same function b(x).
1
We now give some general properties of linear operators and inverse operators in
the form of theorems.
Theorem 1 : If yl and y2 are two particular integrals of'the equation f(D) y =
b(x), then their difference is a solution of the corresponding homogeneous
equation.
Proof : Since y, and y2 are particular integrals of f(D) y = b(x), we have
f(D) yl = b(x) and f(D) yZ
=
b(x)
Now, f(D) is a linear operator, therefore
f(D) (YI - ~ 2 =
) ~ ( D ) Y-I f(D)yz,
= b(x) - b(x),
= 0.
Hence in equations of form (3) we can use any particular integral of the given
equation to obtain the general solution.
Let us now consider an equation
64
Method of Symbolic Operators
or [D' - (a + P) D + aP1 y = X,
where a , P are constants.
Its particular integral is obtained as
We can, equivalently, write it as
i
Applying the operator (D - a ) (D - 0) to Eqn. (9) and using the iundamental laws
of operations for an operator with constant coefficients, we have
1
From this reduction you might have observed that we can as well write particular
integral (9) as
1
Therefore, we can say that the inverse operators with constant coefficients are also
commutative. We now give this property of inverse operation in the form of the
following theorem:
-1
1
1
1
1
a
Proof : We first show the equality between first and second operator, i.e.,
1
We know, from the definition of an inverse operator,' that
b(x) is a
g(D) h(D)
particular integral of g(D) h(D) y = b(x).
1
1
-b(x) is also a particular integral of g(D) h(D)y
If we can show that
g(D) ' h(D)
= b(x), then by Theorem 1, the two operators are equal.
-
which proves the equality between first and second operators. Similarly, we can
prove the equality between first and third operators and complete the proof of the
theorem. We are leaving it as an exercise for you.
>now tnat
=--
g(D) h(D)
h(D) ' g(D)
operators with constant coefficients.
, where g and h are linear ditterential
65
Second and Hlgbcr Order
Ordinmy MIfmnthl Equations
We can also extend Theorem 2 to any numbel of successive inverse operators.
That is, if g, , g2 , ..., g, are n linear differential operators with constant
coefficients, then
.
In the analogy of algebra, we call the replacement of a single operator by such a
succession of operators as factorisation.
I
i
I
Let us now sumose that & are dealing with two eauations. narnelv
i
d 5
and --T - y = 5x
dx
Since ( D -~ 1) (- 2)
=
2, therefore P.I. ylp =
!
1
2 = -2.
D ~ 1-
1
Also ( D -~ 1) (- .5x) = 5x, thus y 2 ~= -5x = - 5x.
D~ - 1
:.
P.1 of Eqn. (11) is
=
- 2 and P.1 of
Eqn. (12) is y2,
=
- 5x.
Now if we add the non-homogeneous terms of Eqns. (11) and (12) and consider an
equation
Then, ( D -~ 1) (- 2 - 5x) = 2
:.
+
5x
P.1 of Eqn. (13) is y, = - 2 - 5x
=
y,,
+ y2,.
Thus we see that if y,, and y2, are particular integrals of Eqns. (11) and (12) then
P.1 of Eqn. (13) is y,, + y2,. This is known as superposition of solutions.
We now take up the superposition theorem, namely,
.
Theorem 3 : If y,, y2, .. , y, are the solutions of the respective equations
f(D)y = bl(x), f(D)y = b2(x) ,...., f(D) y = b,(x), then y = y, + y2 + ...+ y, is
a solution of f(D)y = bl(x) + b2(x) + ...+ b,(x).
Proof : We know that a derivative of a sum is the sum of derivatives. Therefore,
it follows that
We are given that y,, y2 ,..., y, are the solutions of the equations f(D)y
f(D)y =b2x ,..., f(D)y = b,(x) respectively.
=
Hence under this hypothesis, the right-hand side of Eqn. (14) equals b,(x)
+ ...+ b,(x), which proves the required result.
.
bl(x),
+ b2(x)
This theorem allows us the distribution of inverse operator over a sum or
difference of functions.
E3) Using Theorem 3, find the particular integral of the equation
Let us consider
( D ~+ 3D
+ 4) eruy
It can be equivalently written as
+ 3D [ e y ] + 4e"y
D [eaXDy+ aeruy] + 3 [eaXDy+ aeaXy]+ 4eaxy
D [De"y]
=
I
+
+ ae"Dy + a 2 e 9 + 3e"Dy +
= eax [ D S + 2aDy + a? + 3Dy + 3ay + 4y]
= e" [(DS + 2aDy + a2y) + (3Dy + 3ay) + 4y]
= e" [(D + a)2 + 3(D + a) + 41 y
Thus, if f(D) = D~ + 3D + 4, then
f(D) eaXy = em f(D + a) y
= e " ~ ' ~ ae"Dy
3ae"y
+ 4e"y
Method of Symbolic Opmton
This is known as shift formula for operator polynomial f(D).
Relation (15) shows us how to shift an exponential factor from the right to the left
of a differential operator. The formula is very useful in finding the solutions of
differential equations. We shall be illustrating it in Sec. 8.3. However, in the next
theorem we shall prove this formula in general.
Theorem 4 : Suppose f(D) is a polynomial in D. If the first n derivatives of y w.r.
to x exist and are finite and a is any constant, then
Proof : We have
Suppose that for some integer k, we have
Differentiating both sides of Eqn. (17) w.r. to x, we get
D
+
(eay) = D [eax (D + a) 'y]
= e"D(D + a)k y + aea(D
= e" [(D + a). (D + alkyl
= eax (D + alk+ly
+
alky
Thus, if relation (17) is true for k, it is also true for k + 1. We have already
verified it for k = 1. Hence by induction, we conclude that relation (17) is true
for every positive integer k.
Since f(D) is a polynomial in D, using the superposition Theorem 3, and relation
(17), result (16) is proved.
Thus, in general, we have
So far, we have given in Theorems 1 - 4 certain properties of linear differential
operators and inverse operators. We have not discussed the methods of finding a
particular integral using differential operator. In the next section we shall give the
general method of finding the particular integral.
8.3 GENERAL METHOD OF FINDING PARTICULAR
INTEGRAL
Let us consider Eqn. (I), which can be written as
%Dny + a , ~ " - ' y + a 2 ~ n - 2+...+
y
any = b(x),
or f(D)y = b(x),
where, f(D) = a,,Dn + a l ~ " - '+ ...+ a,
If ml, m2 ,..., m, are the roots of the auxiliary equation corresponding to
differential Eqn. (18), then we can write Eqn. (18) in the form
f(D) Y = (D
- ml) (D - m2) ... (D - m,)
y = b(x)
...(19)
Second and Higher Order
Ordinary Differential Equations
Putting (D - m,) (D - m3) ... (D - m,)
Eqn. (19) reduces to
= 7,
...(20)
(D - mi) 711 = b(x)
Eqn. (20) is a linear differential equation of the first order and we can write its
solution in the form
J
Since we are looking for a particular integral, we may simplify the expression (21)
by putting cl = 0. Next, we put
(D - m3) (D - m4) ... (D - m,) y
so that,
= q2,
P
Solving this linear differential equation, we get
Continuing this process n-times, we get
Relation (25) yields the particular integral of Eqn. (18).
As an illustration of the above theory, consider the following example.
Example 1 : Find the particular integral of
( D -~ 5D
+ 6) y
= e3'
Solution : Here P.1 =
--
1
D, - 5D
+
6
e 3x
Let us consider the case of repeated roots.
Example 2 : Find the particular integral of the differential equation
...(22)
Method of Symbolic Operators
Solution = Writing Eqn. (26) in operator form, we get
( D ~+ 4D
:.
+
4) y = - x - e-2X
~
Particular Integral of Eqn. (26) is
D + 2
D + 2
Sometimes the actual integrations as in form (25) prove to be extremely tedious. In
such situations we use a symbolic method in which repeated integration can be
avoided. We call this method as symbolic partial fractions. In this method the
operator --- is resolved into partial fractions. If the factors of f(D) are distinct
(corresponding to the distinct roots of auxiliary equation), then we can write the
particular integral in the form
+...+
A)
D - mn
...(27)
b(x)
where a , , a2 ,..., a, are constants and for a particular problem we can obtain
these constants by simple algebraic manipulations.
r
Now suppose that we want t o solve the equation
...(28)
Solving, Eqn. (28) is equivalent to finding u such that
(D - m) u
= a
...(29)
b(x)
A particular integral of linear Eqn. (29) is
If we apply the same method t o solve each term of Eqn. (27), then the particular
integral can be obtained in the form
e-"'2" b(x) dx
+ ...
...(30)
In case a root m l o f auxiliary equation corresponding to differential Eqn. (27) is
1
repeated r-times, then corresponding partial fractions of -will be of the form
f(D)
a1
D
- ml
+ 2L
(D - m1)2
+ ....+
ar
(D-m,)'
+
ar+1
(D-mr)
+ .... +
an
(D - m,)
I
Second and Higher Order
Ordinary Differential Equations
and particular integral of Eqn. (27) will be
To have a better understanding of what we have discussed above, let us do a few
examples.
1
Example 3 : Find the particular integral of
( D ~ - ~+D6) y
=
Inx
...(31)
Solution : We have
1
o2- 5D +
1
6
(D - 3) (D - 2)
Hence the particular integral of Eqn. (31) is
--
1
1
lnx - --- Inx
D-3
D-2
=
e3"
S
e-3x (lnx) dx - e2'
S
e-" (lnx) dx
We now consider the case of repeated roots.
Example 4 : Find the particular integral of
(D - 1)' (D
+
1)' y
=
ex
Solution : Particular integral is
We now take up an example in which the right hand side is a trigonometric
function.
d2y
Example 5 : Solve + y = sec2x
dx
I
1
,Metbod of Symbolic Operators
Solution : The given differential .equation can be written as
Auxiliary equation is
m 2 + 1 = 0
* m = f i
:.
C.F.
=
y,
c, cosx
=
1
P.1 = y, =
D~
+
1
secz x =
(D
+
1
sec2x
i) (D - i)
cosx + i sinx
dx
cosZx
cosx - i sinx
-
2i
1
2i
-
-
[
-'[
-
eix
1
(secx - i secx tam) dx -
(2 i sinx) In I(secx
2i
= sinx In [(secx
:.
+ czsinx
(secx
I
+ i secx tam) dx
+ tam)) - (2 i cosx). secx
+ tanx)l - 1
General solution of the given differential equation is
+ y,
y = y,
= clcosx
+
c2sinx
+ sinx In I(secx + tanx)l
- 1
You may now try the following exercise.
E4) Find the particular integrals of the following equations.
,
+ n2) y = secnx.
(DZ - 3D + 2) y = sin x e-'
(DZ + 2D + 1) ) = 2e2'
a) (D2
b)
c)
The general method of computing a particular integral 'as discussed in Sec. 8.3
requires a lot of calculations. In certain cases, the P.I. can be obtained by methods
which are shorter than the general method. We shall discuss such methods in the
next section.
8.4 SHORT METHODS OF FINDING PARTICULAR
INTEGRALS
Consider the general nth order linear equation of the form (18) namely,
f(D) y
=
(a,,Dn
+
a1Dn-I + ...
+
a,-, D
+
a,) y = b(x),
where the coefficients a,,, a, ,...., a, are constants. For certain particular forms of
the non-homogeneous term b(x) in the above equation, there do exists shorter
methods of finding particular integrals.
elx = cosx -I- i sinx
e-IX = cosx - i sinx
k o n d and Iilgher Order
Ordinary Differential Equations
Let us take up these methods for various forms of b(x) one by one.
I. b(x) = eax, a constant
We know that
.........................................
Dn
eax
= an
+ al D"-' +...+ an-'D + a,,)eax
(an + alan-' + ...+ an-,a + a,J eax
.: f(D)eax = (Dn
=
= f(a) eax
Further, if f(a) # 0, then
- -
1
. f(a) eax (using Eqn. (32))
f(a)
=
1
That is, -- eax is the particular integral of f(D) y
f(a)
Thus,
1
e ax
,
f(a)
f (a)
beax
= -
=
eax, whenever f(a) # 0.
0
f
Now suppose that f(a) = 0. Then f(D) contains the factor (D - a). Suppose that
the factor occurs p times in f(D), that is, let
f(D) = (D -
alp4(D), 4 ( 4
...(34)
Z 0
By shift formula for operator polynomials (Theorem 4), we know that
f(D) [eaXV(x)]
=
1
Now, -eax
f(D)
eaXf(D
=
1
(D - dP4(D)
--
eaX(using Eqn. (34))
(using Eqn. (33))
-. 1 (using relation (39, when V(x) = 1)
G
4Q)
.
...(35)
1
-
Also consider, f(D)
+ a ) V(x)
[
'
DP
-. x
=
(D - alp
= 4(D)
I)%(
c.
DP
4(4
(using Theorem 4)
Method of Sy
(using Eqn. (32)
eax
I
-1
xP
P!
where f(D) = (D -
is the particular integral of f(D) y = em,
+(D) and +(a) # 0 .
That is,
I
We now illustrate this case with the help of the following examples.
+
Example 6 : Solve (D' - 4D
3) y = eZX
Solution : A.E. is
I
1
mZ-4m + 3 = O
=, (m - 3) (m - 1) = 0
m = 1, 3.
.: C.F = cleX + ~~e~~
1 .
P.1 =
eZX(here 2 is not a root of A.E)
(D - 1) (D -3)
-
:.
y
-eZX(using (33)).
Complete solution is
=
cleX + ~~e~~- ezx
Let us consider another example.
Example 7 : Solve - + y = 3 + e-' + 5ezx
dx
Solution : In the symbolic form, Eqn. (37) reduces to
( D ~+ 1) y = 3
+ e-' +
5eZx
A.E is
Hence C.F = c, e-'
[
+ eXI2
c2 cos
("):-
I)$(
+ c3 sin
Second and Higher Order
Ordinary Differential Equations
=
1
3 --eox +
1
O + I
1
e-'
D ~ - D+ I
-
(D
+
('.'
- 1 is a root of A.E)
I)
'
+
5
1
e2x
----
23
+
1
Hence the complete solution of Eqn. (37) is
y
+
=
[
ex"
e-'x
3
c2 cos ( xf)+c3~in($)]+3+-
+
5
9 eZx
And now an exercise for you.
E5) Solve the following differential equations.
+
a) (D2 - 2D
b) (D' - 1) y
=
1) y
=
3e(5'2)X
(ex
+
1)'
+ .7D - 3) y = eZXcoshx
(D3 - 6D2 + 11D - 6) y = eZX
c) (D3 - 5D2
d)
-
11. b(x) = cos(ax
- -
-
+ b) or sin (ax + b)
+ b) gives
Successive differentiation of cos (ax
D cps (ax
+ b)
D2 cos (ax
+
b)
D3 cos (ax
+
+
b)
D4 cos (ax
+ b)
= - a 2 cos (ax + b)
= a 3 sin (ax + b)
= (D')' cos (ax + b)
=
b)
- a sin (ax
..,...............*..........
a 4 sin (ax
=
+
b)
= (-
a2)' cos (ax
+
Therefore, in general,
(D2)" cos (ax
+
b)
=
(-a2)" cos (ax
+
b)
Thus, if +(D') is a rational integral function of D2, then
9(D2) cos (ax + b) = $(-a2) cos (ax + b)
Similarly, +(D2) sin (ax + b) = 4 (-a2) sin(ax
+
b)
Now two cases arise.
Case I : +(-a2) # 0.
In this case +(D')
cos(ax
+
b)
1
-
m 2 ) [COS(ax + b)l
1
+
-+(-a2) C O S ( ~ X b) (using (38))
+(-a2)
=
Similarly, +(D2)
sin(ax
+
b)
=
cos (ax
sin(ax
+
+ b)
b)
b)
Hence,
Method of Symbolic Operators
1
+
cos (ax
@(-a2)
and
+
b) is a P.I. of q 5 ( ~ ~=) cos
. ~ (ax
sin(= + b)isaP.I.ofq5(D2)y = sin(=
4(-a2)
whenever q5(-a2) # 0.
+
b)
b),
That is, we have
for 4 (-a2) # 0
1
1
cos (ax + b) = -cos(ax
4(D2)
4(-a 2,
1
sin (ax
+ b)
=
1
-sin(ax
4(-a2)
+
b)
+ b)
I
I
Case 11: q5(-a2) = 0
+
Let q5(D2) = (D2
1
-.
a2)' $(D2), where $(-a2) # 0.
+
cos (ax
q5(D2)
b)
1
=
(D2 + a2)' $(D2)
L
- -
1
$(-a2)
1
cos(ax
(D2 + a2)p
where
Similarly,
1
sin(= +
1
( D ~
+ a2)p sin(=
(D2
+ a2)p
+ b),
cos (ax
+ b),
+ b) can then be evaluated by general method.
1
1
b) = --sin (ax
'
(D2
+
a2)'
$(-a2)
q5(D 2,
where
cos (ax
+
b),
+ b) is evaluated by the general method.
We now consider a few examples.
Example 8 : Find particular integral of
(D4
10D2 + 9) y
+
= CQS
+ 3)
(2x
Solution : Particular Integral is
1
-
(-22) (-22)
+
lo(-22)
+
9
1
-
(-4) (-4)
+
10 (-4)
1
-
16-40
+
9
+9
cos (2X
cos (2x
cos (2x
+
+
3)
3)
+ 3)
1
cos (2x + 3)
15
Let us look at another example.
=
-
---
Example 9 :Find particular integral of
(D4 - 1) y = sinx
75
Second md Hlgber Order
Ordinary WlmntiPl Eqaationm
Solution : Particular integral of Eqn. (40) is
YPa '
1
sinx
(D4 - 1)
--
1
(D2 - 1) (D"
-
1
sinx ('.' D2
(-1 -1) (D2 + 1)
1)
sinx
+
1 = 0 for D2 = - 12)
- - -1
1
sinx
2 D2+1
-
=--[
1
2
1
+ i) (D - i)
(D
1
1
D-i
--
1
]
sinx
-1
1
D + i
sinx
sinx - -sinx]
D + i
1
4i
= - -
LeiX S
1
= - [2xcosx
8
S
e-'" sinx dx -e-IX
I
eiXsinx dx
- sinx]
We can alternatively deal with sine and cosine functions in 11 above by considering
f(D) y = ei(ax+b)= cos(ax
+ b) +
isin (ax
By superposition Theorem 3, we then have
are read as 'real part of'
and 'imaginary part of',
respectively.
and
In particular, suppose we want to solve
Then let us consider
-
1
(D
+
ai) (D - ai)
iax
+
b)
lei-
-
Method of Symbolk Operators
D + ai
2ai
D - ai
-.[I--
I
1
-
eim - - (here D
2ai
ei-
- - ix (cos ax
-2a
NOW,V
Im U
=
D~
+
= 0
for D = ai)
1
+ i sin ax) + (cos ax + i sin ax)
4a2
-x
=
COS
ax
2a
1
Similarly,
- ai
a2
+ - sin4a2ax
cos ax = Re U =
x sin ax
2a
+
cos ax
--4a2
Remark : You must have noticed that by using the above alternative approach the
1
term
~2
+
1
sinx in Example 9 could have been evaluated very easily, thus
avoiding long manipulations. Here we would like to remark that choosing an
appropriate method for the evaluation of a particular integral for a given equation
is a skill, which comes through practice only.
You may now try this exercise.
-
-
E6) Solve the following equations:
+ 2 n ' ~+ ~n4) y = cos mx
a) ( D ~
2
b) (D + m2) (D2
+
n2) y
=
cos
In many problems you may find that the polynomial f(D) is not an even function.
It may have an even polynomial factor. That is, suppose we come across the
situation when
f(D) = g(D) h(D), where g(D) is an .even polynomial factor.
Then
1
sin (ax
f(D)
+
1
b) =
-
sin(=
g(D) h(D)
+ b)
h(-D)
sin (ax
f(D) h(D) h(-D)
+
b)
Now h(D) is an odd function, therefore h(D) h(-D) is even.
Also g(D) h(D) h(-D), being the product of two even functions, is an even
polynomial in D and therefore can be written as $(D'). Thus,
1
f (D)
sin(=
+ b)
1
=
--h(-D) sin (&
9(D2)
Similarly, --- cos (ax
f(D)
+ b)
+ b)
= h(-D)
A function f is an even
function if f(-x) = f(x)
and odd function if
f(-X) = - f(x).
second and HIgber Order
Ordl-r~ D i f f e m Eqmuom
~~
Now, P.I. in Eqns. (41) and (42) can be evaluated as in Case I or Case I1 above
according as $(-aZ) # 0 or $(-aZ) = 0.
Let us now consider an example.
Example 10 : Solve ( D ~+ D Z- D -1) y =
COS~X
Solution : The A.E is
Hence the complete solution is,
y = yc
+ yp = cleX + (c, +
c3x) e-'
-
2sin2x
25
cos2x
25
-- -
You may now try the following exercise.
E7) Solve the following equations:
a) (D'
b) (D'
+ D + 1) y = s i n b
+ 2n cosaD + n2) y
= a cosnx
111. b(x) is a polynomial
Let f(D) be a polynomial in D of degree n and b(x) be a polynomial in x of
degree, say k. Then consider the equation
f(D) y = b(x).
P.I. is given by y, =
1
b(x).
f(D)
Let us first assume that D is not a factor of f(D). We then arrange f(D) in
ascehding powers of D. Now divide 1 by f(D) until a quotient of the form
co + cl D + cz D~ + ....+ ckDkfor constants q,, cl ,...., ck and a remainder
involving terms of power higher than k in D is obtained. We have written the
quotient only upto terms involving kth power in D, because b(x) is a polynomial in
1
x of degree k and terms of the expansion -beyond kth power of D when
f(D)
operated on b(x) would yield zero.
Method of Symbolic ~ p m t o m
If we write the remainder in the form Dk+lg(D) for some polynomial g in D,
then
and
+
cID+c2DZ+....+ c ~ D b(x),
~ ) (the second term on the
r.h.s. of Eqn. (44) is zero),
which is the required P.I.
=
(co
...(45)
Let us now assume that D is a factor of f(D) of the order h, where h < n.
Then f(D) = Dh g(D), g(D) being a polynomial in D of degree n-h.
1
and -b(x)
f (Dl
1
. ,1 b(x)
=
D
We then proceed with
1
b(x) exactly as above and get the particular integral.
g(D)
In either of the cases, the method is to express l/f(D) as 11 + polynomial of I)]-'
and then expand by using the binomial expansion. We now illustrate the method
with the help of a few examples.
-
Example 11 : Find the particular integral of
+
y"
+
y = x4
+ 2x +
1.
1
Solution : Here P.I. =
(x4 + 2x + 1)
1 + DZ + D3
= [l + (DZ + D3)]-' (x4 +2x + 1)
= (1 - D2 - D3 + D4 + 2D5 + ....) (x4 + 2x + 1)
= (x4 + 2x +1) - 12x2-24x + 24
= x4 - 12 x2 - 22x + 25
y"'
Let us consider another example.
Example 12 : Find the particular integral of
y " ' + y " + y ' + y = x 4 + 2 x + 1 .
Solution : Here P.1 =
1
(x4
1 + D + D Z + D 3
= (1
=
(1
- D4-I) (x4 - 4x3
+
D4
+
D8
+
2x
+
1)
+ 2~ - 1)
+... ) (x4 - 4x3 +
+ 2x - 1) + 24
= x 4 - 4 x 3 + 2x + 23
= (x4 - 4x3
Let us consider an example in which D is a factor of f(D).
2x - 1)
Second and Higber Order
Example 13 : Solve (D3
Ordlnary Differential E.q~l)tjo~ts
+ 3D2 + 2D) y
=
x2
Solution : A.E is
m3
+
3m2
+
+ 2m = 0
a m(m
2)(m + 1) = 0
a m = 0,-1,-2.
:.
C.F = yc
= C,
+
c2e-'
+~
~
e
-
~
~
Hence the complete solution is
y
=
yc
+
y, = cl
+ cze-" +
1
c3e-2x+ - (2x3 - 9x
12
+ 21)
And now an exercise for you.
E8) Solve the following differential equations:
+ D2) y = x
(D3 - 3D2 - 6D + 8) y = x
(D2 + D - 2) y = 2 (1 + x - x2)
(D4 + 2D3 - 3DZ)y = x2 + 3eZX+ 4sinx
a) (D4 - 2D3
b)
c)
d)
IV.
b(x) = eaXV(x), a constant
Consider an equation of the form
f(D) y = eaXV(x),
where a is a constant and V is a function of x.
We wish to find the particular integral
By the shift formula for operator polynomial (Theorem 4), we know that
f(D) eaXV(x) = eax f(D
+ a ) V(X)
Now put
f(D
+ a) V = V1;then V =
1
f(D
+ a)
v1
Since V is a function of x, V1 will also be a function of x.
On substituting the value of V in terms of V1 in Eqn. (47). we get
1
V1 = eax V1
f(D + cr)
Operating on both sides of Eqn. (48) with I
f(D)
-we get
f(D) eaX
.
where Vl is any function of x.
Let us consider the following examples.
Example 14 : Find the particular integral of
(D2
+
1) y = xe2'
Solution : P.1 =
I
xeZX
D2 + 1
= e2X
(D
- -eZY
25
+
1
2)2
+
1
x (using relation (49))
(Sx - 4)
dS
Example 15 : Solve - + 2y = x2e3'
dx2
+
ex cos2x
Solution : A.E is
I
m 2 + 2 = 0
* m = t i f i
:. C.F = yc = c1 C
- 121
elx ( l l x
+ c2 sinfix
O S ~ X
- 6) +
ex (2D
+
1).
1
4D2 - 1
cos2x
Method of Symbolic Operators
1
Second and Higher Order
Ordinary Differentlnl Equations
1
=-
121
e3' ( l l x
- 6) +
ex (2D
+
1)
1
4(-22 ) - 1
ex
-17
e3' ( l l x - 6)
+
- [2(-2sinx)
1
- e3' ( l l x - 6)
+
( f:
- -
121
121
---
sin2x -
COS~X
+ COS~X)]
17
Hence the complete solution is
1
c2sindZx + -e3' ( l l x - 6)
121
Let us look at another example.
y = clcosfix
+
+
1
17
- ex (4sin2x
Example 16 : Find the particular integral of
(D2 - 2D
+
1) y = xeXsinx
Solution : P.I. is
-
I~
(-e(l + O X
(x
= Im [-ex (cosx
=
-ex (2cosx
+ 2i)j
+ i sinx) (x + 2i) ]
+ x sinx)
You may now try the following exercises:
E9) Solve the following equations:
+ 2) y = e2'sinx
(D2 - 2D + 1) y = x2e3'
(D3 - 2D2 - 19D + 20) y = xeX + 2e-4Xsinx
(D3 - 3D2 + 3~ -1) y = xeX + ex
a) (D2
b)
c)
d)
+
3D
E10) Solve the following equations:
a) (D2 - I) y = x2cosx
b) (D2 - 4D
+ 4) y
= 8x2 eZXsin2x
+ (1 + x2) ex
c) ( D -~ 1) y
=
xsinx
d) (D4 - 1) y
=
x2sinx
- cos2x)
Note that methods of symbolic operators for finding particular solutions are
applicable only to equations with constant coefficients. They do not work for
equations with variable coefficients. Quite often operators with variable coefficients
are not factorisable. Even if so, factors do not commute as you have already seen
in Sec. 8.2. However, these methods can be applied t o certain particular types of
equations with variable coefficients namely, Euler's equations and equations
reducible to Euler's form. Euler's equations, which you have already studied in
Unit 7, can be reduced to equations with constant coefficients by using certain
transformation of the variables. In the next section we shall apply operator
methods of such equations.
8.5 EULER'S EQUATIONS
Consider the nth order Euler's equation, namely,
On substituting x
eZor z
=
lnx, we can reduce Eqn. (50) to the form
--
.... (D'- n - 1) + a l Dr(D'- 1) ... (D'- n - 2) + ...
.... + a,-lD' + a,] y = b(e7
+ A,D'"-' + ... + A,-,D' + A,] y = b(ez)
[a,,D1(D' - 1)
or [A0D'"
=
-
...(51)
d
where D ' = - and A,,, Al ,...., A, are constants.
dz
Eqn.(51) is a linear differential'equation with constant coefficients and can be
treated by the operator methods. If its solution is y = g(z), then the solution of
Eqn. (50) is y = g(1nx).
Let us now consider the following example.
d 2~ - dy
Example 17 : Solve x2 ---x-dx
dx
+
y
=
21nx
d
Solution : Putting z = lnx, and denoting - by D ' , Eqn. (52) can be written as
dz
[Df(D'- 1) - D ' + 11 y = 22
( D ' ~- 2D + 1) y = 22
A.E. is
+
1= 0
m2-2m
3 m = 1, 1.
:. C.F = (cl + cZz)eZ
=
2 [l - 2D'
=
2 [l
= 2(2
+
+
+
2D'
~ ' ~ 1z- l
+....I
2) = 22
z
+
4
.'. Complete solution is
y = (c,
= (cl
+ c2z)eZ+
22
+
+ 2111~+ 4
C2
111~)X
+
4
You may now try the following exercise.
E l 1) Solve the following equations:
- 3xD
~
a) ( x ~ D
+ 4) y
= 2x2
Method of Symbolic Operators
Second and Higher Order
Ordiisry Differentid Egoations
4
c)
xZ
x3
d) (x2D2 - XD + 4) y = cos (lnx) + xsin (Inx)
There are some differential equations that are.easily reducible to the Euler's
equations and hence to the equations with constant coefficients. We shall now take
up such equations.
Equations reducible to the Euler's Form.
Consider an equation of the form
d "Y
dn-'y
dy
(ax+ b)" - + a, (ax + b)"-' - +..-.+ a,,-' (ax+b) dx "
dxn-I
dx
where the coefficients a,, a,
,...., a,
+
any
=
f(x),
...(53)
are constants.
We can transform equations of the form (53) to Euler's equation when the
independent variable x is changed to z by means of the substitution z = ax+ b.
Eqn. (53), under this substitution, reduces t o
-
Eqn. (54) can then be reduced t o an equation with constant coefficients by the
substitution t = Inz.
In practice, instead of making two substitutions, we make only one substitution,
viz., et = ax + b, and then from Eqn. (53) we can directly derive an equation
with constant coefficients.
We now illustrate through an example how this is achieved.
dy - 36y = 3x2 + 4x + 1
3(3x + 2) dx
. dx
d
2 = e Zand denoting - = Dl ' , we have
- dz
Example 18 : Solve (3x +2)'
Solution : Putting- 3x
dy
dy
I=----
-d 2~- dx'
-
+
dz -
-3.3
D'y
( 3 ~ + 2 ) ~
-d2y +
...(55)
nfv
...(56)
3
+ --(Dr4)
3x + 2
,
,
3
9(D" - D ' ) y
(3x
+ 2)2
Substituting from Eqns. (56) and (57) into Eqn. (59, we get
...(57)
Method of Symbolic Operators
:.
Complete solution of Eqn. (55) is
And now an exercise for you.
p~
E12) Solve the following differential equations:
b) (1
d2y
dy
+x )~
+ (1 + x) +y
dx2
dx
= 4cos [In (1
+ x)]
We now end this unit by giving a summary of what we have covered in it.
8.6 SUMMARY
In this unit we have covered the following :
d
1. The symbol D is used for --- and an nth degree polynomial f(D) in Q is
dx
called symbolic operator.
I
1
2. -b(x) is that function of x which when operated upon by f(D) gives
f(D)
1
b(x) and is a particular integral of f(D) y = b(x). Here -2.- is called
f(D)
inverse operator.
3.
If y,, y2 ,...., ym are the solutions of equations
f(D) y = b,(x), f(D) y = b2(x) ,..., f(D) y = bm(x) respectively, then
Y = Y1 + Y2 + ....+ Ym is a solution of f(D) y = b,(x) + ...+ bm(x)
4.
f(D) [eay] = e" f(D + a) y
1
Y
(D - m,) (D - m2) ... (D - m,)
(1
e-'lXb(x) dx) dx)
1
6.
--- b(x) = e"
D-a
7.
-e"
1
f(D)
1
1
S
= -- e",
f(a)
.... dx)
dx
e-"b(x)dx
if f(a) # 0.
1
eax
e " = - _ xP
_
8. -e" =
, if fk(a) = 0
f(D)
(D - a)P 4 0 )
P! +(a)
for k = I
I
,...., (p - I),
but 4(a) # 0.
1
1
9. --- C O S (+~ ~
b) = -cos (ax
4(D2)
4(- a2)
+
b) , if 4(-a2) # 0.
Second and Higher Order
Ordinary Differential Equations
1
1
10. --- sin(=
4(D2)
+
1
11. -cos(ax
4P2)
+ b)
=
1
12. -sin(ax
+ b)
=
4032,
+
b) = -sin (ax
4(- a2)
1
+
13. -bk(x) = (c,,
f(D)
1
1
1
1
b), if 4( - a2) # 0.
cos (ax + b) if 4k(-a2) = 0
(D' + a21p $(D')
for k = 1, 2 ,...., p-1 but $(-a2) # 0
'
+ b) if 4k(-a2) = 0
(D' + a2)p $(D')
for k = 1 ,..., p - 1 but $(-a2) # 0.
sin (ax
c1D + ... + c ~ D bk(x)
~ ) ; bk(x) a polynomial of degreek.
15. Euler's equation (xnDn + alxn-I D"-I + .... + an-] xD + an) y = b(x) can be
reduced to an equation with constant coefficients with the help of the
substitution x = eZ.
16. Differential equation
[(ax + b)" Dn + al (ax+ b)"-' D"-I +...+ an-] (ax + b ) D + a,] y = b(x)
can be either reduced to Euler's equation by using the substitution
ax + b = z or it can be reduced to an equation with constant coefficients by
means of the transformation ax + b = e'.
El) a) ( D + 2 ) ( 2 D - 1 )
b) (D - 1) (D
C) (D
+
+
.
2) (D - 3)
2)3 (2D - 1)
d) (D - 4) (D'
+ 4D +
5)
E2) We have
1
1
Thus, - -is a particular integral of g(D) h(D) y = b(x)
h(D)
g(D)
E3) yp = ex - 2.
1
secnx
E4) a) P.1 =
D~ + n2
1
- 2n1
-
1
-(1
2ni
-
2ni
D
[.nix
+
J ,-i
FeniX S
(1
1
in
nx
+
KZ)
secnx
secnx dx - e-'
nX
S
- i tan nx) dx -
(1
I
1
x sin nx - - cosnx In (secnx)
n
n
e-'
1
b) - - (3sinx + cosx) + - e-" (2sinx + cosx)
10
5
-'[
-
I
ei nx secnx dx
+
i tan nx) dx
I
Method of Symbolic Operators
E5) a) y = (c,
+ cZx)ex +
4
3
- e(5/2)X
c) Given equation can be written as
Then solving, we get
E6) a) y
=
b) y
=
E7) a) C.F
-1
=
-
:.
1
+ c2x) cosnx + (c3 + c4) sinnx + (n2 - m2)2 cosmx
(cl cosmx + c2 sinmx) + (c3.cosnx + c4 sinnx)
sinmx
sinnx
+ 4(n2 x- m2) [m
n
(cl
(P.)+
[
e-("2)X cl cos
1
(2cos2x
13
+
(P
c2 sin
X)]
3sin2x)
Complete solution is
y = e(-1/2)x
:(
cos
x)
b) y = e-(ncOsa)x
[cl cos (nsinu) x
+
:(
c2 sin
+ c2 sin (nsinn) x] +
a
sin nx
2n2 cosn
Second and Higher Order
Ordinary Differential Equations
E8) a) y = cl
:.
+
c2x
+
(c3
+
c4x) ex
+
x2
+
X
-
6
Complete solution is
4
5
+
- sinx
2
cosx
5
+
-
+ c2e-" +
E9) a) y =
P.1 =
1
D3 - 2D2 - 19D
+
2e-4X
= ex
D3
=
1
eZX(1 1 sinx - 7 cosx)
170
-
- ex
+
+
1
(D - 4)3 - 2(D - 4)'-
1
D2 - 20D
-
-
ex - 1
l [
20
D
+
+
e4'
20
D
+ 2e4X
19(D - 4)
+ 20
D
20
-
sinx
1
sinx
D3 - 14D2 + 45D
ioD2]-lX + ~ ~ - 4 x
-D
20D
-
(xeX + 2e-" sinx)
20
I
+ order (D2)
+
1
14
+ 45D
x
7 -22D
sinx
(7 - 22D) (7 + 22D)
+
1
e-4X(7 - 22D)
49
+ 484
sinx
sinx
ex
:.
The complete solution is
y = cleX
+ ~~e~~+ ~
~ e - ~ ~ - 20
c) y
- 22 cosx)
+ xsinx +
cosx
= (cl + c2x) e2" + e2' (3sin2x - 4xcos2x - 2x2 sin2x)
1
1
= cleX+ c2e-' - - (cosx + xsinx) + - xeX(9 - 3x + 2x2)
2
12
d) y = cle-x
-
=
+
c2eX+ c3 cosx
+ c., sinx +
4
3
- x2 sinx
8
E l l ) a) y = (c,
b) y
(7 sinx
( )
EIO) a) y = cleX + c2e-'
b) y
Method of Symbolic Operators
e
(7 sinx - 22 cosx)
533
(cl
+ c2 lnx) x2 + ( ~ n xx2
)~
1
1
X
+ c2 lnx) +In X
x
1-x
c) The given equation can be written as
(x3D3- 4x2D2 + 5xD - 2) y = x3
d
Let z = lnx and denoting - by D ' , the given equation reduces to
dz
[D' ( D ' - 1) (D' - 2) - 4D' (D' - 1) + 5D' - 21 y = e3'
( D ' ~- 7 D f 2 + 11D' - 2) y = 'e3'
C.F
. . = c1e2z + e(3/2)z (C2e(fi1/2)z+ C3e(-fil/2)z
- C,X2 + x5/2 (c2xfil/2 + C3X-fil/2
1
-
1
(D' - 2) ( D -~ 5D
P.1 =
1
5
-
i
e3z =
+
1)
e 3z
x
-5
Hence solution is
d) The given equation is
( x ~ D
- XD
~ + 4) y = cos (111~)+ xsin (lnx)
d
Let z = lnx and denoting - by D', the given equation reduces t o
dz
[Df(D' - 1) - D ' + 41 y = cosz + eZsinz
( D f 2- 2D'
4) y = cosz + eZsinz
+
A.E. is
rn2-2m
+
4 = 0
m =
C.F = eZ(cl cos f i z
P.1 =
-
- 2 2-6
+ c2 sin f i z )
1
cosz
D ' ~ - ~ D+' 4
+
1
-1-2D'
+ ez
+4
cosz
-
= l ~ i f i
L
= x [c, cos ( f i l m )
1
D f 2- 2D'
(D'
+
+4
+
c2 sin (film)]
ez sinz
1
112-2(Df
+
1)
+4
sinz
89
Second and Higher Order
Ordinary Differential Equations
1
4
1
sinz
cosz
+
ez
~ ' ~
D ' ~
+3
= (3
+ 2D')
9
= (3
+ 2D')
-cosz
= (3
+ 2D')
-+
3
13
= - cosz
-
1
9 + 4
cosz
13
-
2
13
+
- - sinz
1
+ ez sinz
-1 + 3
e
sinz
2
1
2
- eZsinz
Hence solution is
Y =' clx cos(fi1nx)
+
+
c2x sin ( f i lnx)
3
2
+cos(1nx) - - singnx)
13
13
X
- sin (lm)
2
E12) a) The given equation is
(2x - 1)' D2 + (2x - 1) D - 2y = 0
Let (2x
-d2y
dx2
- 1) =
-
d
eZand denoting - = D', we have
dz
4
(2x - 1)'
( D f 2- D') y
Substituting them in the given equation, we have
[4(Dt2 - D') + 2Df - 21 y = 0
a ( 4 ~ -' 2D'
~ - 2) y = 0
a (2Dt2 - D' - 1) y = 0
A.E is
2m2 - m - 1. = 0
:. The solution is
b) The given equation is
[(I
+ x2) D2 +
Let 1
+x
-d2y
- dx2
(1
+ x) D +
I ] y = 4cos (ln(1
d
dz
+ x))
= eZand denoting - by D', we have
1
D' (D' - 1) y
il+x)2
Then the given equation reduces to
[D' (D' - 1) + D ' + l] y = 4 ~ 0 s ~
A.E is
m2-m + m + 1 = 0
a m = *i.
:.
+
C.F = c, cosz
P.1 = 4
1
D ' ~
+1
c2sinz
cosz
=
clcos (In (1
Z
=
4 - sinz
2
=
+
x))
2 In (1
+
c,sin(In (1
+ x))
+ x) sin (In (1 + x))
Hence the required solution is
y = cl cos (In (1
+ x)) +
czsin (In (1
+
x))
+
2 In (I
+
x) sin (In (1
+ x)).