Javatpoint Logo
Javatpoint Logo

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.


Example1: Solve the difference equation 2ar-5ar-1+2ar-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 = Particular Solution and 2.

Therefore, the homogeneous solution of the equation is given by

ar(h)= C1Particular Solution+C2 .2r..........equation (i)

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

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

Hence, the particular solution is
Particular Solution


Example2: Solve the difference equation ar-4ar-1+4ar-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
            ar(n)=(C1+C2 r).2r.............. 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
            ar(P)=(1+2r).2r.


Example3: Solve the difference equation 9ar-6ar-1+ar-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 = Particular Solution

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

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

Hence, the particular solution is
            ar(P)=6r.Particular Solution.


(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 ar+2-3ar+1+2ar=Zr ........equation (i)

Where Z is some constant.

Solution: The general form of solution is = A. Zr

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

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

Therefore, the particular solution isParticular Solution


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

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

Now to find the value of A, put this solution on L.H.S of the equation (i), then this becomes
            = A. 5r+2-5.A5r+1+6.A5r
            = 25A. 5r-25A.5r+6A.5r
            = 6A.5r ............equation (ii)

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

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


Example3: Find the particular solution of the difference equation ar+ 2+ar+1+ar=r.2r..........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
            = 2r+2 [A0+A1 (r+2)]+2r+1 [A0+A1 (r+1)]+2r (A0+A1 r)
            = 4. 2r (A0+A1 r+2A1 )+2.2r (A0+A1 r+A1 )+2r (A0+A1 r)
            = r. 2r (7A1 )+2r (7A0+10A1)............equation (ii)

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

Therefore, the particular solution is Particular Solution


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)
        = En-1 [E f(x)] = En-1 f(x+h)
        = En-2 [E f(x+h)] = En-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 yn+r+C1 yn+r-1+C2 yn+r-2+⋯+Cn yn=R (n)..............equation (i)

Equation (i) can be written as
        C0 Er yn+C1 Er-1 yn+C2 Er-2 yn+⋯+Cn yn=R (n)
        (C0 Er+C1 Er-1+C2 Er-2+⋯+Cn) yn=R (n)
Putting C0 Er+C1 Er-1+C2 Er-2+⋯+Cn=P(E)

So     P (E) yn=R (n)
        ∴     yn=Particular Solution................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 Er+C1 Er-1+C2 Er-2+⋯+Cn)A
            = (C0+C1+C2+⋯+Cn)A
            = P (1) A
            Particular Solution
Therefore, using equation (ii), the particular solution of (i) is
            yn=Particular Solution,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 Er+C1 Er-1+⋯+ Cn} (A.Zn)
                                          =A{C0 Zr+n+C1 Zr+n-1+⋯+Cn Zn}
                                          = A{C0 Zr+C1 Zr-1+⋯+Cn }. Zn
                                          =AP(Z).Zn

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

Therefore, Particular Solution, provided P (Z) ≠ 0

Thus,       yn=Particular Solution, P (Z) ≠ 0

If A = 1, then    yn=Particular Solution

When P (Z) = 0 then for equation

          (i) (E-Z) yn= A. Zn

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

          (ii) (E-Z)2 yn= A. Zn

For this, the particular solution becomes Particular Solution

          (iii) (E-Z)3 yn= A. Zn

For this, the particular solution becomesParticular Solution 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+∆)

Particular Solution

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

⇒       Particular Solution =(b0+b1 ∆+b2 ∆+⋯.+bmm+⋯)
Particular Solution.R(n)=( b0+b1 ∆+b2 ∆+⋯.+bmm+⋯).R(n)
            = b0 R(n)+b1 ∆ R(n)+⋯+bmm 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

            yn=b0 R(n)+b1 ∆ R(n)+⋯.+bmm 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         Er[Zn R(n)]=Zr+n R (n+r)=Zr.Zn.Er.R(n)=Zn (ZE)rR(n)

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

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

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

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

The particular solution is given by
            ar=Particular Solution.12

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

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

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

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

Thus, the particular solution is given by

Particular Solution
Next TopicTotal Solution




Please Share

facebook twitter google plus pinterest

Learn Latest Tutorials


B.Tech / MCA