SOLUTIONS OF PROBLEMS IN SECOND-ORDER DIFFERNCE EQUATIONS WITH FORCING FUNCTIONS EQUAL TO ZERO

EDITOR: B. SOMANATHAN NAIR

Example 1:  Obtain the solution of the equation

                                                y(n) + 4y(n‒1) + 3y(n‒2) = 0  (1)

                                                                                                             

Solution: The transient part of the solution can be obtained from the characteristic equation

                                     (C² + 4C + 3)y = 0    (2)                                            

The roots of Eq. (2) are ‒3, and ‒1, which are real and unequal. Hence, we use Solution I:

                                    y(n) = A(‒3)ⁿ + B(‒1)ⁿ  (3)                                          

            The constants A and B are evaluated with initial conditions. Let the initial conditions be given as y(0) = 1, and y(1) = 0. With n = 0, we get

                                    y(0) = A + B = 1  (4)                                                   

With n = 1, from Eq. (3), we get

                                y(1) = ‒3A ‒ B = 0 (5)                                                    

Solving, we find A = -1/2, B = 3/2. So the solution is

                        y(n) = (‒1/2)(‒3)ⁿ + (3/2)(‒1)ⁿ  (6)

                                      

The solution can be checked by substituting the value of y(n) from Eq. (6) into Eq. (1). For this, we determine y(0),  y(1), and y(2) from Eq. (6). Thus at n = 0, we get

                                   

                        y(0) = ‒1/2 + (3/2) = 1  (7)

                       

With n = 1, Eq. (6) gives

                                                y(1) = (‒1/2)( ‒3) + (3/2)( ‒1) = 0 (8)

                                                                         

When n = 2, Eq. (6) gives

                                            y(1) = (‒1/2)( ‒3)² + (3/2)( ‒1)² = 3 (9)

                                                                       

Now consider Eq. (1) again, which states that

                            y(n) + 4y(n‒1) + 3y(n‒2) = 0      

Now, substituting for y(0),  y(1), and y(2) from Eqs. (7), (8), and (9), respectively, we get

                                                y(2) + 4y(1) + 3y(0) = ‒3+0+3 = 0              

The obtained value of y(n) thus agrees with the given equation. Hence it is a valid solution.

Example 2:  Obtain the solution of the equation

                                                y(n) + 2y(n‒1) + y(n‒2) = 0    (1)

Solution: The characteristic equation is

                               (C² + 2C + 1)y = 0            (2)                                                                

and the roots of the characteristic equation are

           

                           C₁, C₂ = ‒1 and ‒1            (3)                                             

Therefore, the solution is

                                                            y(n) = (‒1)ⁿ (A + Bn)    (4)

                                                                                               

As before, to evaluate A and B, let y(0) = 1, and y(1) = 0. Then with n = 0, we get

                                                                        1 = A                    (5)

and with y(1) = 0, we find

                                                            0 = (‒)(A + B)

                                   

which gives

           

                                    B = -A = -1  (6)                                                                     

Thus the complete solution is therefore

                                    y(n) = (‒1)ⁿ (1‒n)    (7)                                                      

Example 3:  Obtain the solution of the equation

                                                 y(n) + 4y(n‒1) + 8y(n‒2) = 0    (1)

                                                      

Solution: The characteristic equation is

                                    (C² + 4C + 8)y = 0          (2)                                   

           

and the roots of the characteristic equation are

                                                C₁, C₂ = ‒2‒2j, ‒2+2j             (3)

We get, with α ± jβ = ‒ 2 ± j2, we choose Solution III. For this, we find

­                        rⁿ = √(α² + β²) =√(2² + 2²) = 2√2   (4)

                   θ = tan^‒1 (α/β) = tan^‒1 (2/2) = 3π/4  (5)

                                                                        

Using these, the solution can now be written as

            y(n) = (2√2)ⁿ [A cos (3nπ/4) + B sin (3nπ/4)]  (6)

                       

Example 4: Obtain the solution of the equation

                                                    y(n) +y(n‒2) = 0    (1)

                                                      

                                                              

Solution: The characteristic equation is

                                    (C²  + 1)y = 0  (2)                                                      

and the roots of the characteristic equation are

                                    C₁, C₂ = ±j    (3)                                                                     

Since j corresponds to 90^º or π/2, we use Solution IV. Thus

                         y(n) = A cos (nπ/2) + B sin (nπ/2) (4)                              

CHECK: The solution given by Eq. (4) may now be checked as follows. First, as before, n = 0, we have

                                       y(0) = A       (5)                                 

When n = 2

                                            y(2) = A cos (2π/2) + B sin 2nπ/2)

                                                 = A cos (π) = ‒A (6)                      

Substituting for y(0) and y(2) in (4), we get        

                                    y(2) + y(0) =‒A+ A = 0         (7)                 

The result we find in Eq. (7) shows the accuracy of the solution for y(n).