MAT 241-Variation of Parameters, Operator and Difference Equations

Variation of Parameters, Operator Techniques and
Difference Equation
Lawal G.S
University of Ibadan
gslawal@hotmail.com
Thursday 27th April, 2017
Lawal G.S (Department of Mathematics) MAT 241 Thursday 27th April, 2017 1 / 54
Overview
1 Variation of Parameters
2 Operator Techniques
3 Difference Equation
Variation of Parameters
Variation of Parameter: Procedure

Let
φ(D)y = a0 (x)y n + a1 (x)y n−1 + a2 (x)y n−2 + ... + an−1 (x)y + an (x)y
and suppose that the complementary solution of
φ(D)y = R(x)
is
y = C1 y1 (x) + C2 y2 (x) + C3 y3 (x) + ... + Cn yn (x)
Replace the arbitrary constants C1 , C2 , C3 , ...Cn by the functions
K1 , K2 , K3 , ..., Kn and seek to determine these functions so that
y = K1 y1 (x) + K2 y2 (x) + K3 y3 (x) + ... + Kn yn (x)
is a solution of
φ(D)y = R(x)
To determine these n-functions, we must impose n restrictions on them

and since one of these is that the differential equation be satisfied, if
follows that the remaining (n − 1) may be taken at will. The conditions
which lead to the greatest simplicity are given by the equations;
K10 y1 + K20 y2 + K30 y3 + ... + Kn0 yn = 0

K10 y10 + K20 y20 + K30 y30 + ... + Kn0 yn0 = 0
.........................................................
K1 y1 + K20 y2n−2 + K30 y3n−2 + ... + Kn0 ynn−2 = 0
0 n−2
K10 y1n−1 + K20 y2n−1 + K30 y3n−1 + ... + Kn0 ynn−1 = R(x)
where the last equation represents the condition that the given differential
equation be satisfied. From these equations, K10 , K20 , K30 , ..., Kn0 .
K1 , K2 , K3 , ..., Kn can then be found by integration leading to the required
equation. This method is applicable whenever the complementary solution
can be found including cases where a0 , a1 , a2 , ..., an are not constants.
Example I:
Solve the differential equation using method of variation of parameters
(D 2 − 3D + 2)y = e 5x
Solution:
The differential equation can also be written as
y 00 − 3y 0 + 2y = e 5x
The complementary solution
yc = c1 e x + c2 e 2x
To find the solution, we assume the general solution
y = K1 e x + K2 e 2x (i)
where K and K are suitable functions

MAT 241 of x to be determined.
Lawal G.S (Department of Mathematics) Thursday 27th April, 2017 5 / 54
Continuation...
Differentiating (i) yields
y 0 = K10 e x + K20 e 2x + K1 e x + 2K2 e 2x (ii)
Since there are two functions, we must arrive at two conditions for
determining them. One of these conditions is that
K10 e x + K20 e 2x = 0 (iii)
such that (ii) becomes
y 0 = K1 e x + 2K2 e 2x (iv )
Differentiating (iv) again, we obtain
y 00 = K10 e x + 2K20 e 2x + K1 e x + 4K2 e 2x (v )
substituting (i), (iv) and (v) into the given differential equation, we
obtain Thus,
Continuation...
y 00 − 3y 0 + 2y = K10 e x + 2K20 e 2x + K1 e x + 4K2 e 2x − 3(K1 e x + 2K2 e 2x )

+2(K1 e x + K2 e 2x ) = e 5x
(1)
such that
K10 e x + 2K20 e 2x = e 5x (vi)
solving (iii) and (vi) simultaneously to determine K10 and K20 , we have
K10 e x + K20 e 2x = 0
K10 e x + 2K20 e 2x = e 5x
from(iii)
K10 = −K20 e x (vii)
substitute (vii) into (vi), then
(−K20 e x )e x + 2K20 e 2x = e 5x
Continuation...
−K20 e 2 x + 2K20 e 2x = e 5x
K20 e 2x = e 5x
1
K20 = e 3x =⇒ K2 = e 3x + c2
3
using K20 = e 3x , then from (vii),

K10 = −(e 3x )e x = −e 4x
integrating implies that
1
K10 = − e 4x + c1
4
By assumption,
y = K1 e x + K2 e 2x
Continuation...
hence
1 1
y = (− e 4x + c1 )e x + ( e 3x + c2 )e 2x
4 3
rearranging, we have
1 5x
y = c1 e x + c2 e 2x + e
12
Example II:
Solve the differential equation
(D 2 + 4)y = 8 sin 2x
Solution:
The differential equation can also be written as
y 00 + 4y = 8 sin 2x
The complementary solution is
yc = c1 cos 2x + c2 sin 2x
To find the solution, we assume the general solution
y = K1 cos 2x + K2 sin 2x (∗)
where K and K are suitable functions

MAT 241 of x to be determined.
Lawal G.S (Department of Mathematics) Thursday 27th April, 2017 10 / 54
Continuation...
Differentiating (*) yields
y 0 = −2K1 sin 2x + K10 cos 2x + 2K2 cos 2x + K20 sin 2x (∗∗)
Since there are two functions, we must arrive at two conditions for
determining them. One of these conditions is that
K10 cos 2x + K20 sin 2x = 0
such that (**) becomes
y 0 = −2K1 sin 2x + 2K2 cos 2x
Differentiating again
y 00 = −4K1 cos 2x − 2K10 sin 2x − 4K2 sin 2x + 2K20 cos 2x
Thus,
y 00 + 4y = −2K10 sin 2x + 2K20 cos 2x = 8 sin 2x (∗ ∗ ∗)
Continuation...
From the two equations,
K10 cos 2x + K20 sin 2x = 0
0
−2K1 sin 2x + 2K20 cos 2x = 8 sin 2x
From here,
−K20 sin 2x
K10 =
cos 2x
substituting into the second equation, we obtain
−K20 sin 2x

−2 sin 2x + 2K20 cos 2x = 8 sin 2x
cos 2x
Multiply all by cos 2x, then
2K20 sin2 2x + 2K20 cos2 2x = 8 sin 2x cos 2x
2K20 = 8 sin 2x cos 2x
Continuation...
K20 = 4 sin 2x cos 2x =⇒ K2 = sin2 2x + c2

Also
−K20 sin 2x sin 2x
K10 = =⇒ K10 = (−4 sin 2x cos 2x)
cos 2x cos 2x
and
K10 = −4 sin2 2x
integrating, we obtain
1
K1 = −2x + sin 4x + c1
2
Using the assumption that
y = K1 cos 2x + K2 sin 2x
we have

1
−2x + sin 4x + c1 cos 2x + sin2 2x + c2 sin 2x

y=
2 th
Lawal G.S (Department of Mathematics) MAT 241 Thursday 27 April, 2017 13 / 54
Continuation...
Hence
1
y = c1 cos 2x + c2 sin 2x − 2x cos 2x + sin 4x cos 2x + sin3 2x
2
Tutorial Questions:
Solve using variation of parameters

1 y 00 + 2y 0 − 3y = xe −x
2 y 00 + 4y = csc 2x
3 (D 3 + D)y = 4 tan x
4 (D 2 + 9)y = sec 3x
5 xy 00 − y 0 = x
6 (D 2 + D − 2)y = x 2 e 2x
7 (D 2 − 4)y = 16x 3
Operator Techniques
Operator Techniques
When a0 , a1 , ..., an are constants, the equation
φ(D)y = R(x)
can be written in factored form as
a0 (D − m1 )(D − m2 )...(D − mn )y = R(x) (2.2)
where m1 , m2 , ..., mn are constants and where the other of the factors
(D − m1 )(D − m2 ), ..., (D − mn ) is immaterial. This is not true if
a0 , a1 , ..., an are not constants.The constants, m1 , m2 , ..., mn are the root
of the auxiliary equation and the complementary solution is
y = c1 e m1 x + c2 e m2 x + ... + cn e mn x
To obtain particular solutions,we use the following operator methods:

Operator Techniques
I.Method of Reduction of order:

Let
a0 (D − m2 )(D − m3 )...(D − mn )y = Y1
then (2.2) becomes
(D − m1 )Y1 = R(x)
which can be solved for Y1 . Also let
a0 (D − m3 )(D − m4 )...(D − mn )y = Y2
so that
(D − m2 )Y2 = Y1
which can be solved for Y2 .By continuing in this manner, y can be
obtained. This method yields the general solution if all arbitrary constants
are kept, while it yields a particular solution if the constants are omitted.
Operator Techniques
II.Method of Inverse Operators:

Let
1
R(x)
φ(D)
be defined as a particular solution, yp such that
φ(D)yp = R(x)
1
We call φ(D) an inverse operator.By reference to the entries in the
following table, the labour in finding particular solution of φ(D)y = R(x)
is often reduced.
Operator Techniques
Table of Inverse Operator
1
e mx e −mx R(x)dx
R
(D−m) R(x)
1
e m1 x e −m1 x e m2 x ... e −mn−1 x e m
R R R
(D−m1 )(D−m2 )...(D−mn ) R(x)
1 px e px
φ(D) e φ(p) if φ(p) 6= 0
1 cos(px+q)
φ(D 2 )
cos(px + q) φ(−p 2 )
if φ(−p 2 ) 6= 0
1 sin(px+q)
φ(D 2 )
sin(px + q) if φ(−p 2 ) 6= 0
φ(−p 2 )
1 p k p
φ(D) x = (c0 + c1 D + ... + ck D +)x (c0 + c1 D + ... + cp D p )x p
1 px 1
φ(D) e F (x) e px φ(D+p) F (x)
Operator Techniques
Examples I:
1 4x
Evaluate D−2 (e )
Using the formula
1
(e 4x )
D −2
Z
= e 2x e −2x e 4x dx
1
= e 4x
2
Operator Techniques
Example II:
1
Evaluate (3e −2x )
(D + 1)(D − 2)
Using the formula;
1
(3e −2x )
(D + 1)(D − 2)

1 1
= (3e −2x )
D +1 D −2
Z
1 2x −2x −2x
= e e (3e )dx
D +1

1 3 −2x
= − e
D +1 4
Z
3 3
= e −x e x (− e −2x )dx = e −2x
4 4
Operator Techniques
Alternatively:
Using Partial fraction
1
(3e −2x )
(D + 1)(D − 2)
" #
−1 1
= 3
+ 3
(3e −2x )
D +1 D −2
1 1
=− (e −2x ) + (e −2x )
D +1 D −2
Z Z
−x x −2x
= −e e (e )dx + e 2x
e −2x (e −2x )dx
3
= e −2x
4
Operator Techniques
Example III:
1 px e px
Prove that e = if φ(p) 6= 0.
φ(D) φ(p)
By definition;
φ(D) = a0 (x)D n + a1 (x)D n−1 + a2 (x)D n−2 + ... + an−1 (x)D + an (x)
where a0 , a1 , ..., an are constants. Then,
φ(D)e px = a0 D n + a1 D n−1 + a2 D n−2 + ... + an−1 D + an e px

= a0 p n + a1 p n−1 + a2 p n−2 + ... + an−1 p + an e px

= φ(p)e px
Operator Techniques
Example IV:
Find the general solution of (D 2 − 3D + 2)y = e 5x using the result in
(Example III):
The complementary solution is;
yc = c1 e x + c2 e 2x
Using the result above, the particular solution is;
1 5x 1 5x e 5x
yp = e = e =
D 2 − 3D + 2 (52 ) − 3(5) + 2 12
Hence the general solution is;
e 5x
y = c1 e x + c2 e 2x +
12
Operator Techniques
Example V:
Find the general solution of (D 2 + 1)y = cos 2x

The complementary solution is;
yc = c1 cos x + c2 sin x
U̇sing the result in the table,
1 cos(px + q)
2
cos(px + q) = if φ(−p 2 ) 6= 0
φ(D ) φ(−p 2 )
then we have
1 1 1
yp = cos 2x = cos 2x = − cos 2x
D2 +1 (−4 + 1) 3
Hence the required general solution is y = c1 cos x + c2 sin x − 31 cos 2x
Operator Techniques
Example VI:
1
Evaluate D 3 +D 2 +2D−1
cos 2x
Using the result in example IV,
1 cos(px + q)
cos(px + q) = if φ(−p 2 ) 6= 0
φ(D 2 ) φ(−p 2 )
then we have
1 1
cos 2x = cos 2x
D3 + D2 + 2D − 1 D(−4) − 4 + 2D − 1
1 (2D − 5) (2D − 5)
− cos 2x = − 2 cos 2x = − cos 2x
2D + 5 4D − 25 4(−4) − 25
1 1
= (2D − 5) cos 2x = (−4 sin 2x − 5 cos 2x)
41 41
Operator Techniques
Example VII:
Evaluate (D 3 + D)y = e −2x cos 2x
Using the formula in the table;
1 px 1
e F (x) = e px F (x)
φ(D) φ(D + p)
1 1
yp = e −2x cos 2x = e −2x cos 2x
D3 +D 3
(D − 2) + D − 2
1 1
= e −2x cos 2x = e −2x cos 2x
D 3 − 6D 2 + 13D − 10 −4D + 24 + 13D − 10
1 9D − 14
= e −2x cos 2x = e −2x cos 2x
9D + 14 81D 2 − 196
e −2x e −2x
= (9D − 14) cos 2x = (9 sin 2x + 7 cos 2x)
81(−4) − 196 260 th
Operator Techniques
Tutorial Question II:
Evaluate the following:

1
1 e −2x
D +3
1
2
2
9e 5x − 4e −x
D + D − 12
D −1
3 8 cos x
D + D2 + 1
4
1
4 e −2x cos 2x
(D − 4)(D + 3)(D + 1)
5 Prove that
φ(D)[e px R(x)] = e px φ(D + p)R(x)
Difference Equation
Difference Equation
Definition 3.1.1
An equation which contains independent variable, dependent variable and
successive differences of the dependent variable is called the difference
equation.
Examples 3.1.2
1 yn+2 − 6yn+1 + 9yn = 0
2 (E 2 + 6E + 9)yn = 2n i.e yn+2 + 6yn+1 + 9yn = 2n
3 (∆2 + 3∆ + 2)yn = 1 or ∆2 yn + 3∆yn + 2yn = 1
or (yn+2 − 2yn+1 + yn ) + 3(yn+1 − yn ) + 2yn = 1
or yn+2 + yn+1 = 1
Difference Equation
Order of a difference equation:

Definition 3.1.3
The order of the difference equation is the difference between the largest
and smallest arguments occurring in the difference equation divided by the
unit of argument.
Thus, the order of the difference equation is
largest argument - smallest argument
=
unit of argument
Example 3.1.4
The order of the difference equation yn+2 − 7yn = 5 is
(n + 2) − n
=2
1
Difference Equation
Definition 3.1.5
Degree of a difference equation

The highest degree of yn0 s in the difference equation is called the Degree
of the difference equation.
Example i:
n+2−n
The order and degree of yn+2 + 4yn+1 + 4yn = 2n are = 2 and 1
1
respectively.
Example ii:
2 n+3−n
The order and degree of yn+3 + 3yn+2 + 3yn+1 + yn = 0 are =3
1
and 2 respectively.
Difference Equation
Definition 3.1.6
Solution of Difference Equation

Any function which satisfies the given difference equation is called the
solution of the difference equation.
General solution of the difference equation

A solution in which the number of arbitrary constants is equal to the order
of the difference equation is called the general solution of the difference
equation.
Particular solution
A solution which can be obtained from the general solution by assigning
particular values is called particular solution.
Difference Equation
Formation of difference equation:

Example I: Write the difference equation ∆3 yn + ∆2 yn + ∆yn + yn = 0 in
the subscript notation.
Solution:
∆yn = yn+1 − yn
∆2 yn = yn+2 − 2yn+1 + yn
∆3 yn = yn+3 − 3yn+2 + 3yn+1 − yn

substituting these into the given difference equation, we obtain
(yn+3 − 3yn+2 + 3yn+1 − yn ) + (yn+2 − 2yn+1 + yn ) + (yn+1 − yn ) + yn = 0
yn+3 − 2yn+2 + 2yn+1 = 0

Difference Equation
Example II: Form a difference equation from

yn = A2n + B(−2)n
Solution:
yn = A2n + B(−2)n =⇒ yn − A2n − B(−2)n = 0
yn+1 = A2n+1 + B(−2)n+1 =⇒ yn+1 − 2A2n + 2B(−2)n = 0
yn+2 = A2n+2 + B(−2)n+2 =⇒ yn+2 − 4A2n − 4B(−2)n = 0

Eliminating A and B, we obtain,
yn+2 − 4yn = 0
Difference Equation
Definitions :
Linear Difference Equation

A difference equation in which yn , yn+1 , yn+2 , ..., occur to the first degree
only and not multiplied together is called the Linear difference equation.
A linear difference equation of order k is an equation of the form
a0 yn+k + a1 yn+k−1 + a2 yn+k−2 + ... + ak yn = R(n) 1.1.6
If R(n) = 0, then the equation 1.1.6 is called linear homogeneous

difference equation, otherwise it is called non-homogeneous difference
equation.
Difference Equation
Homogeneous Linear Difference Equation with constant

coefficient
The equation of the form
(a0 E k + a1 E k−1 + a2 E k−2 + ... + an−1 E + an I )yn = 0
or
φ(E )yn = 0
where a0 , a1 , ..., an are all constants is known as homogeneous linear
difference equation with constant coefficient.
Let yn = mn be the solution of the difference equation, then we have
a0 mn+k + a1 mn+k−1 + ... + an mn = 0
=⇒ (a0 mk + a1 mk−1 + ... + an )mn = 0 3.1.7

which shows that mn is the solution of 3.1.7 if m satisfies
a0 mk + a1 mk−1 + ... + an = 0 3.1.8
th
Difference Equation
Continuation...
Equation 3.1.8 is known as the auxiliary equation. This equation has
k-roots m1 , m2 , m3 , ..., mk .Some cases may arise;
Case I: When the roots are real and distinct
If m1 , m2 , m3 , ..., mk are real and distinct roots of auxiliary equation
(3.1.8), then the solution is
yn = c1 (m1 )n + c2 (m2 )n + ... + ck (mk )n
Case II: When some of the roots are equal

If two roots are equal, i.e m1 = m2 , then the solution is
yn = (c1 + c2 n)(m1 )n + c3 (m3 )n + ... + ck (mk )n
Difference Equation
Continuation...
If three roots are equal, i.e m1 = m2 = m3 , then the solution is
yn = (c1 + c2 n + c3 n2 )(m1 )n + c4 (m4 )n + ... + ck (mk )n
If k-roots are equal, i.e m1 = m2 = m3 = ... = mk , then the solution is
yn = [c1 + c2 n + c3 n2 + c4 n3 + ... + ck nk−1 ](m1 )n
Difference Equation
Continuation...
Case III: When the roots are complex number

If complex roots occur then they must be complex conjugate. i.e If
(α + iβ) is the root, then (α − iβ) is also a root where α and β are real.
Then the solution is
yn = c1 (α + iβ)n + c2 (α − iβ)n
which can be written as
yn = γ n (A1 cos nθ + A2 sin nθ)

β
p
where γ = α2 + β 2 , A1 = c1 + c2 , A2 = i(c1 − c2 and θ = arctan α .
Difference Equation
Example I:
Solve the difference equation yn+2 − 2yn+1 − 8yn = 0
Solution:
This equation can also be written as
(E 2 − 2E − 8)yn = 0
Let yn = mn , then
yn+2 − 2yn+1 − 8yn = mn+2 − 2mn+1 − 8mn = 0
m2 − 2m − 8 = 0 =⇒ (m − 4)(m + 2) = 0
m = 4 and m = 2. Hence
yn = c1 (4)n + c2 (−2)n
Difference Equation
Example II:
Solve the difference equation yn+3 + yn+2 − 8yn+1 − 12yn = 0

Solution:
The given difference equation can be written as
(E 3 + E 2 − 8E − 12)yn = 0
Let yn = mn , then the auxiliary equation becomes
m3 + m2 − 8m − 12 = 0
The roots are m = −2, −2, 3

Hence the solution is
yn = (c1 + c2 n)(−2)n + c3 3n
Difference Equation
Example III:
Solve the difference equation yn+2 + 16yn = 0
Solution:
(E 2 + 16)yn = 0
Let yn = mn , then the auxiliary equation becomes
m2 + 16 = 0
The roots are m = ±4i

Hence the solution is
yn = (c1 (4i)n + c2 (−4i)n
yn = 4n {c1 (i)n + c2 (−i)n }

Difference Equation
Continuation...
π n π n

n π π
= 4 c1 cos + i sin + c2 cos − i sin
2 2 2 2

nπ nπ nπ nπ
= 4n c1 cos + i sin + c2 cos − i sin
2 2 2 2

n nπ nπ
= 4 (c1 + c2 ) cos + i(c1 − c2 ) sin
2 2

n nπ nπ
= 4 A cos + B sin
2 2
Difference Equation
Non-homogeneous linear difference equation with constant

coefficient
The equation of the form
(a0 E k + a1 E k−1 + a2 E k−2 + ... + an−1 E + an I )yn = R(n) 3.1.9
or
φ(E )yn = R(n)
where a0 , a1 , ..., an are all constants is known as non-homogeneous linear
difference equation with constant coefficient.
The general solution of (3.1.9) consist of two parts, the complementary
function and the particular integral.
The complementary function is the general solution of the homogeneous
1
equation of (3.1.9) and the Particular Integral = R(n)
φ(E )
Difference Equation
Rules for obtaining the Particular Integral
Case I:R(n) = an
1 n 1 n
P.I = a = a
φ(E ) φ(a)
provided φ(a) 6= 0
If φ(a) = 0, then for the equation
(E − a)k yn = an ; k is a positive integer.
1 n(n − 1)(n − 2)...(n − (k − 1)) n−k n

P.I = k
an = a = Ck an−k
(E − a) k!
When k = 1
1
P.I = an = nan−1
E −a
Difference Equation
Continuation...
When k = 2,
1 n n(n − 1)an−2
P.I = a =
(E − a)2 2!
Case II: When R(n) = sin an or cos an

1 1 e ian − e −ian
P.I = sin an =
φ(E ) φ(E ) 2i

1 1 ia n 1 −ia n
= (e ) − (e )
2i φ(E ) φ(E )

1 1 1
= e ian − e −ian
2i ia
φ(e ) φ(e −ia )
provided φ(e ia ) 6= 0 and φ(e −ia ) 6= 0
Difference Equation
Case III: When R(n) = nk

1 k 1
P.I = n = nk
φ(E ) φ(1 + ∆)
since E = 1 + ∆.
= [φ(1 + ∆)]−1 nk
Expansion of φ(1 + ∆)]−1 in the ascending powers of ∆ by the Binomial
theorem as far as the term in ∆k , and distributing it into each term of
the expansion.
Difference Equation
Example I:
Solve the difference equation yn+2 − 4yn+1 + 3yn = 5.4n
(E 2 − 4E + 3)yn = 5.4n
Solving the homogeneous equation yn+2 − 4yn+1 + 3yn = 0 , we obtain a

complementary solution
yn = c1 + c2 (3)n
To determine the Particular integral P.I, we have
1 1 5.4n
P.I = 5.4n = 5. 2 4n =
φ(E ) 4 − 4(4) + 3 3
Hence the general solution is

5
yn = c1 + c2 (3)n + 4n
Lawal G.S (Department of Mathematics) MAT 241
3 Thursday 27th April, 2017 48 / 54
Difference Equation
Example II:
Solve the difference equation yn+2 − 3yn+1 + 2yn = 6.2n

(E 2 − 3E + 2)yn = 6.2n
Solving the homogeneous equation yn+2 − 3yn+1 + 2yn = 0, we obtain a

complementary solution
c1 + c2 (2)n
To determine the particular integral, P.I, we have
1 1
P.I = 6.2n = 6. 2 2n
φ(E ) E − 3(E ) + 2
but φ(2) = 0, therefore we have
Difference Equation
Continuation...
1 1 1
= 6. .2n = 6. .2n = 6. .2n
(E − 1)(E − 2) (2 − 1)(E − 2) (E − 2)
= 6.n2n−1 = 3n2n
yn = c1 + c2 (2)n + 3n2n
Difference Equation
Example III:
Solve the difference equation yn+2 − 6yn+1 + 8yn = 3n2 + 2
(E 2 − 6E + 8)yn = 3n2 + 2
Ṫhe complementary solution is
yn = c1 2n + c2 4n
The particular solution is

1 1
P.I = (3n2 + 2) = (3n2 + 2)
E2 − 6E + 8 2
(1 + ∆) − 6(1 + ∆) + 8
1
= (3n2 + 2)
1 + 2∆ + ∆2 − 6 − 6∆ + 8
Difference Equation
Continuation...
1 1 1
= (3n2 + 2) = 2
2 (3n + 2)
∆2 − 4∆ + 3 3 (1 − 4∆−∆
)
3
−1
4∆ − ∆2

1
= 1− (3n2 + 2)
3 3
4∆ − ∆2 (4∆ − ∆2 )2

1
= 1+ + + ... (3n2 + 2)
3 3 9
8 44
= n2 + n +
3 9
8 44
y n = c 1 2n + c 2 4n + n 2 + n +
3 9
Difference Equation
Tutorial Questions
Solve the following difference equations

1 (E 2 − 2E − 8)yn = 0
2 2yn+2 − 5yn+1 + 2yn = 0
3 yn+2 − 4yn+1 + 3yn = 4n
4 yn+2 − 4yn = 5.3n
5 2yn+2 − 5yn+1 + 2yn = 0, y0 = 0, y1 = 1
6 yn+2 − 4yn+1 + 4yn = 2n
7 yn+2 − 4yn = n2 + n + 1
Difference Equation
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MAT 241-Variation of Parameters, Operator and Difference Equations

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MAT 241-Variation of Parameters, Operator and Difference Equations

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Variation of Parameters, Operator Techniques and

Thursday 27th April, 2017

Variation of Parameter: Procedure

To determine these n-functions, we must impose n restrictions on them

K10 y1 + K20 y2 + K30 y3 + ... + Kn0 yn = 0

The complementary solution

To find the solution, we assume the general solution

where K and K are suitable functions

y 0 = K10 e x + K20 e 2x + K1 e x + 2K2 e 2x (ii)

K10 e x + K20 e 2x = 0 (iii)

such that (ii) becomes

Differentiating (iv) again, we obtain

y 00 = K10 e x + 2K20 e 2x + K1 e x + 4K2 e 2x (v )

y 00 − 3y 0 + 2y = K10 e x + 2K20 e 2x + K1 e x + 4K2 e 2x − 3(K1 e x + 2K2 e 2x )

using K20 = e 3x , then from (vii),

The complementary solution is

To find the solution, we assume the general solution

y = K1 cos 2x + K2 sin 2x (∗)

where K and K are suitable functions

2K20 = 8 sin 2x cos 2x

K20 = 4 sin 2x cos 2x =⇒ K2 = sin2 2x + c2

Solve using variation of parameters

When a0 , a1 , ..., an are constants, the equation

can be written in factored form as

a0 (D − m1 )(D − m2 )...(D − mn )y = R(x) (2.2)

To obtain particular solutions,we use the following operator methods:

I.Method of Reduction of order:

II.Method of Inverse Operators:

Table of Inverse Operator

where a0 , a1 , ..., an are constants. Then,

φ(D)e px = a0 D n + a1 D n−1 + a2 D n−2 + ... + an−1 D + an e px

= a0 p n + a1 p n−1 + a2 p n−2 + ... + an−1 p + an e px

Using the result above, the particular solution is;

Hence the general solution is;

Find the general solution of (D 2 + 1)y = cos 2x

U̇sing the result in the table,

Hence the required general solution is y = c1 cos x + c2 sin x − 31 cos 2x

Tutorial Question II:

Evaluate the following:

φ(D)[e px R(x)] = e px φ(D + p)R(x)

Order of a difference equation:

Degree of a difference equation

Solution of Difference Equation

General solution of the difference equation

Formation of difference equation:

∆3 yn = yn+3 − 3yn+2 + 3yn+1 − yn

(yn+3 − 3yn+2 + 3yn+1 − yn ) + (yn+2 − 2yn+1 + yn ) + (yn+1 − yn ) + yn = 0

yn+3 − 2yn+2 + 2yn+1 = 0

Example II: Form a difference equation from

yn+1 = A2n+1 + B(−2)n+1 =⇒ yn+1 − 2A2n + 2B(−2)n = 0

yn+2 = A2n+2 + B(−2)n+2 =⇒ yn+2 − 4A2n − 4B(−2)n = 0

Linear Difference Equation

a0 yn+k + a1 yn+k−1 + a2 yn+k−2 + ... + ak yn = R(n) 1.1.6

If R(n) = 0, then the equation 1.1.6 is called linear homogeneous

Homogeneous Linear Difference Equation with constant

=⇒ (a0 mk + a1 mk−1 + ... + an )mn = 0 3.1.7

yn = c1 (m1 )n + c2 (m2 )n + ... + ck (mk )n

Case II: When some of the roots are equal