We want to find a general solution of this equation The general solution will

We want to find a general solution of this equation

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We want to find a general solution of this equation. The general solution will need to have two arbitrary constants so that a specific solution can be found once y 0 and y 1 are given (just as the general solution of a first-order equation contains one arbitrary constant). As the equation is linear and homogeneous, two very useful properties apply (compare this with homogeneous systems of linear equations, A x = 0 ): a constant multiple of a solution is a solution, the sum of two solutions is a solution. 164
11 11.6. Homogeneous second-order difference equations Activity 11.5 Show this. Given two sequences x t and z t which satisfy the homogeneous difference equation y t +2 + a 1 y t +1 + a 2 y t = 0, show that x t + z t and cx t , where c is a constant, also satisfy this equation Therefore, it follows that if we know two solutions x t and z t of the difference equation, then y t = Ax t + Bz t is also a solution for any constants A, B R . We next set about finding solutions. Knowing what we do about the general solution of the homogeneous first-order equation (a geometric progression), let’s try a solution of the form y t = m t where m is some constant to be determined. Substituting y t = m t into the difference equation we obtain y t +2 + a 1 y t +1 + a 2 y t = m t +2 + a 1 m t +1 + a 2 m t = m t ( m 2 + a 1 m + a 2 ) = 0 . If m = 0, we get y t = 0, so we ignore this possibility. Then y t = m t is a solution of the difference equation if and only if m satisfies m 2 + a 1 m + a 2 = 0. This equation, z 2 + a 1 z + a 2 = 0. is known as the auxiliary equation . Let’s look at an example. Example 11.6 We find the general solution of the difference equation y t +2 - 5 y t +1 + 6 y t = 0 . The auxiliary equation is z 2 - 5 z + 6 = 0 . This equation factors, z 2 - 5 z + 6 = ( z - 2)( z - 3) = 0, with solutions z = 2 and z = 3. Therefore both x t = 2 t and z t = 3 t are solutions and the general solution is y t = A (2 t ) + B (3 t ) , A, B R . Now suppose we are given initial conditions y 0 = 1 and y 1 = 4. Then the difference equation determines all remaining numbers in the sequence. We have y 2 = 5 y 1 - 6 y 0 = 5(4) - 6(1) = 14, y 3 = 5 y 2 - 6 y 1 , and so on. The specific solution of the difference equation with these initial conditions is found by substituting t = 0 and t = 1 into the general solution. We have y 0 = A + B = 1 and y 1 = 2 A + 3 B = 4 . Solving these equations for A and B , we find B = 2 and A = - 1. Therefore the solution of y t +2 - 5 y t +1 = 6 y t = 0 with initial conditions y 0 = 1 and y 1 = 4 is y t = - (2 t ) + 2(3 t ) . We can immediately check that y 0 = - 2 0 + 2(3 0 ) = - 1 + 2(1) = 1 and y 1 = - 2 + 2(3) = 4 as required. A further check is to find that y 2 = - (2 2 ) + 2(3 2 ) = - 4 + 18 = 14 as we found earlier. 165
11 11. Difference equations The auxiliary equation z 2 + a 1 z + a 2 = 0 of a second-order difference equation is a quadratic equation and the general solution to the difference equation depends on whether the auxiliary equation has two distinct solutions, or just one solution, or no (real) solutions. Thus, the form of general solution depends on the value of the discriminant, a 2 1 - 4 a 2 . We consider each case in turn.

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