Particular Solution

(a) Homogeneous Linear Difference Equations and Particular Solution:

We can find the particular solution of the difference equation when the equation is of homogeneous linear type by putting the values of the initial conditions in the homogeneous solutions.

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Example1: Solve the difference equation 2a_r-5a_r-1+2a_r-2=0 and find particular solutions such that a₀=0 and a₁=1.

Solution: The characteristics equation is 2s²-5s+2=0
(2s-1)(s-2)=0
s =  and 2.

Therefore, the homogeneous solution of the equation is given by

a_r(h)= C₁+C₂ .2^r..........equation (i)

Putting r=0 and r=1 in equation (i), we get
      a₀=C₁+C₂=0...........equation (a)
      a₁= C₁+2C₂=1...........equation.(b)

Solving eq (a) and (b), we have
C₁=-and C₂=

Hence, the particular solution is

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Example2: Solve the difference equation a_r-4a_r-1+4a_r-2=0 and find particular solutions such that a₀=0 and a₁=6.

Solution: The characteristics equation is
            s²-4s+4=0 or (s-2)²=0             s = 2, 2

Therefore, the homogeneous solution of the equation is given by
            a_r(n)=(C₁+C₂ r).2^r.............. equation (i)

Putting r = 0 and r = 1 in equation (i), we get
            a₀=(C₁+0).2⁰ = 1          ∴C₁=1
            a₁=(C₁+C₂).2=6          ∴C₁+C₂=3⇒C₂=2

Hence, the particular solution is
            a_r(P)=(1+2r).2^r.

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Example3: Solve the difference equation 9a_r-6a_r-1+a_r-2=0 satisfying the conditions a₀=0 and a₁=2.

Solution: The characteristics equation is

            9s²-6s+1=0 or (3s-1)²=0
            s = 

Therefore, the homogeneous solution of the equation is given by
            a_r(h)=(C₁+C₂ r). ..........equation (i)

Putting r = 0 and r = 1 in equation (i), we get
            a₀=C_1=0
            a₁= (C₁+C₂).=2.           ∴C₁+C₂=6⇒C₂=6

Hence, the particular solution is
            a_r(P)=6r..

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(b) Non-Homogeneous Linear Difference Equations and Particular Solution:

There are two methods to find the particular solution of a non-homogeneous linear difference equation. These are as follows:

1. Undetermined coefficients method
2. E and ∆ operator method.

1. Undetermined Coefficients Method: This method is used to find a particular solution of non-homogeneous linear difference equations, whose R.H.S term R (n) consist of terms of special forms.

In this method, firstly we assume the general form of the particular solutions according to the type of R (n) containing some unknown constant coefficients, which have to be determined. Then according to the difference equation, we will determine the exact solution.

The general form of a particular solution to be assumed for the special forms of R (n), to find the exact solution is shown in the table.

Form of R (n)	General form to be assumed
Z, here z is constant	A
Zr, here z is constant	Zr
P (r), a polynomial of degree n	A0 rn+A1 rn-1+⋯..An
Zr. P (r), here P(r) is a polynomial of the nth degree in r. Z is a constant.	[A0 rn+A1 rn-1+⋯..An].Zr

Example1: Find the particular solution of the difference equation a_r+2-3a_r+1+2a_r=Z^r ........equation (i)

Where Z is some constant.

Solution: The general form of solution is = A. Z^r

Now putting this solution on L.H.S of equation (i), we get
            = A Z^r+2-3AZ^r+1+2AZ^r=(Z²-3Z+2) A Z^r.........equation (ii)

Equating equation (ii) with R.H.S of equation (i), we get
            (Z²-3Z+2)A=1
            A =(Z≠1, Z≠2)

Therefore, the particular solution is

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Example2: Find the particular solution of the difference equation a_r+2-5a_r+1+6a_r=5^r .............equation (i)

Solution: Let us assume the general form of the solution= A. 5^r.

Now to find the value of A, put this solution on L.H.S of the equation (i), then this becomes
            = A. 5^r+2-5.A5^r+1+6.A5^r
            = 25A. 5^r-25A.5^r+6A.5^r
            = 6A.5^r ............equation (ii)

Equating equation (ii) to R.H.S of equation (i), we get
            A =

Therefore, the particular solution of the difference equation is =.5^r.

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Example3: Find the particular solution of the difference equation a_r+ 2+a_r+1+a_r=r.2^r..........equation (i)

Solution: Let us assume the general form of the solution = (A₀+A₁r). 2^r

Now, put these solutions in the L.H.S of the equation (i), we get
            = 2^r+2 [A₀+A₁ (r+2)]+2^r+1 [A₀+A₁ (r+1)]+2^r (A₀+A₁ r)
            = 4. 2^r (A₀+A₁ r+2A₁ )+2.2^r (A₀+A₁ r+A₁ )+2^r (A₀+A₁ r)
            = r. 2^r (7A₁ )+2^r (7A₀+10A₁)............equation (ii)

Equating equation (ii) with R.H.S of equation (i), we get
            7A₁=1             ∴ A₁=
            7A₀+10A₁=0         ∴ A₀=

Therefore, the particular solution is 

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2. E and ∆ operator Method:

Definition of Operator E: The operator of E on f(x) means that give an increment to the value of x in the function. The operation of E is, put (x+h) in the function wherever there is x. Here h is increment quantity. So Ef(x) = f(x+h)

Here, E is operated on f(x), therefore, E is a symbol known as shift operator.

Definition of Operator∆: The operation ∆ is an operation of two steps.

Firstly, x in the function is incremented by a constant and then former is subtracted from the later i.e.,
            ∆f(x)=f(x+h)-f(x)

Theorem1: Prove that E ≅1+∆.

Proof: The operation of ∆ on f(x) is of two steps. First, increment the value of x in the function. So, whenever, there is x in f(x) put x+h (here h is constant increment), which means operation of E on f(x) i.e.,
            f (x+h)=Ef(x).

Second, subtract the original function from the value obtained in the first step, hence
            ∆f(x)=Ef(x)-1f(x)=(E-1)f(x)

So, the operation of ∆ on f(x) is equivalent to the operation of (E-1) on f(x).

Therefore, we have
            E ≅1+∆.

Theorem2: Show that Eⁿ f(x)=f(x+nh).

Proof: We know that E f(x) =f (x+h)

Now       Eⁿ f(x)=E.E.E.E.........n times f(x)
        = E^n-1 [E f(x)] = E^n-1 f(x+h)
        = E^n-2 [E f(x+h)] = E^n-2 f(x+2h)
        ......................
        ......................
      = E f[x+ (n-1) h] = f(x+nh).

Theorem3: Show that E Cf(x) = CE f(x)

Proof: We know that E C f(x) = C f(x+h) = CE f(x+h). Hence Proved.

There is no effect of the operation of E on any constant. Therefore, the operation of E on any constant will be equal to constant itself.

By E and ∆ operator method, we will find the solution of
            C₀ y_n+r+C₁ y_n+r-1+C₂ y_n+r-2+⋯+C_n y_n=R (n)..............equation (i)

Equation (i) can be written as
        C₀ E^r y_n+C₁ E^r-1 y_n+C₂ E^r-2 y_n+⋯+C_n y_n=R (n)
        (C₀ E^r+C₁ E^r-1+C₂ E^r-2+⋯+C_n) y_n=R (n)
Putting C₀ E^r+C₁ E^r-1+C₂ E^r-2+⋯+C_n=P(E)

So     P (E) y_n=R (n)
        ∴     y_n=................equation (ii)

To find the particular solution of (ii) for different forms of R (n), we have the following cases.

Case1: When R (n) is some constant A.

We know that, the operation of E on any constant will be equal to the constant itself i.e.,
            EA=A
Therefore,     P (E) A = (C₀ E^r+C₁ E^r-1+C₂ E^r-2+⋯+C_n)A
            = (C₀+C₁+C₂+⋯+C_n)A
            = P (1) A
            
Therefore, using equation (ii), the particular solution of (i) is
            y_n=,P(1)≠0

P (1) is obtained by putting E = 1 in P (E).

Case2: When R (n) is of the form A. Zⁿ, where A and Z are constants

We have,     P (E) (A. Zⁿ)={C₀ E^r+C₁ E^r-1+⋯+ C_n} (A.Zⁿ)
                                          =A{C₀ Z^r+n+C₁ Z^r+n-1+⋯+C_n Zⁿ}
                                          = A{C₀ Z^r+C₁ Z^r-1+⋯+C_n }. Zⁿ
                                          =AP(Z).Zⁿ

To get, P (Z) put E=Z in P (E)

Therefore, , provided P (Z) ≠ 0

Thus,       y_n=, P (Z) ≠ 0

If A = 1, then    y_n=

When P (Z) = 0 then for equation

          (i) (E-Z) y_n= A. Zⁿ

For this, the particular solution becomes A.  Zⁿ=A. n Zⁿ-1

          (ii) (E-Z)² y_n= A. Zⁿ

For this, the particular solution becomes 

          (iii) (E-Z)³ y_n= A. Zⁿ

For this, the particular solution becomes and so on.

Case3: When R (n) be a polynomial of degree m is n.

We know that E≅1+∆
So,       P (E) =P (1+∆)

Which can be expanded in ascending power of ∆ as far as upto ∆^m

⇒        =(b₀+b₁ ∆+b₂ ∆+⋯.+b_m ∆^m+⋯)
⇒ .R(n)=( b₀+b₁ ∆+b₂ ∆+⋯.+b_m ∆^m+⋯).R(n)
            = b_{0 R(n)+b1 ∆ R(n)+⋯+bm ∆^m R(n)}

All other higher terms will be zero because R (n) is a polynomial of degree m.

Thus, the particular solution of equation (i), in this case will be

            y_n=b₀ R(n)+b₁ ∆ R(n)+⋯.+b_m ∆^m R(n).

Case4: When R (n) is of the form R(n).Zⁿ,where R(n) is a polynomial of degree m and Z is some constant

We have         E^r[Zⁿ R(n)]=Z^r+n R (n+r)=Z^r.Zⁿ.E^r.R(n)=Zⁿ (ZE)^rR(n)

Similarly, we have
 [Zⁿ R(n)]=Zⁿ  .(R(n))= Zⁿ [P(Z+Z∆)]^-1.R(n)

Thus, the particular solution of equation (i), in this case will be
            y_n=Zⁿ [P(Z+Z∆)]^-1.R(n)

Example1: Find the particular solution of the difference equation
            2a_r+1-a_r=12.

Solution: The above equation can be written as
            (2E-1) a_r=12

The particular solution is given by
            a_r=.12

Put E=1, in the equation. The particular solution is a_r=12

Example2: Find the particular solution of the difference equation a_r-4a_r-1+4a_r-2=2^r.

Solution: The above equation can be written as
           (E²-4E+4) a_r=2^r

Therefore,       P (E) = E²-4E+4 = (E-2)²

Thus, the particular solution is given by