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We want to find a general solution of this equation. The general solution will need tohave two arbitrary constants so that a specific solution can be found oncey0andy1aregiven (just as the general solution of a first-order equation contains one arbitraryconstant).As the equation is linear and homogeneous, two very useful properties apply (comparethis with homogeneous systems of linear equations,Ax=0):a constant multiple of a solution is a solution,the sum of two solutions is a solution.164
1111.6. Homogeneous second-order difference equationsActivity 11.5Show this. Given two sequencesxtandztwhich satisfy thehomogeneous difference equationyt+2+a1yt+1+a2yt= 0, show thatxt+ztandcxt,wherecis a constant, also satisfy this equationTherefore, it follows that if we know two solutionsxtandztof the difference equation,thenyt=Axt+Bztis also a solution for any constantsA, B∈R.We next set about finding solutions. Knowing what we do about the general solution ofthe homogeneous first-order equation (a geometric progression), let’s try a solution ofthe formyt=mtwheremis some constant to be determined.Substitutingyt=mtinto the difference equation we obtainyt+2+a1yt+1+a2yt=mt+2+a1mt+1+a2mt=mt(m2+a1m+a2) = 0.Ifm= 0, we getyt= 0, so we ignore this possibility. Thenyt=mtis a solution of thedifference equation if and only ifmsatisfiesm2+a1m+a2= 0. This equation,z2+a1z+a2= 0. is known as theauxiliary equation.Let’s look at an example.Example 11.6We find the general solution of the difference equationyt+2-5yt+1+ 6yt= 0.The auxiliary equation isz2-5z+ 6 = 0.This equation factors,z2-5z+ 6 = (z-2)(z-3) = 0, with solutionsz= 2 andz= 3. Therefore bothxt= 2tandzt= 3tare solutions and the general solution isyt=A(2t) +B(3t),A, B∈R.Now suppose we are given initial conditionsy0= 1 andy1= 4. Then the differenceequation determines all remaining numbers in the sequence. We havey2= 5y1-6y0= 5(4)-6(1) = 14,y3= 5y2-6y1, and so on.The specific solution of the difference equation with these initial conditions is foundby substitutingt= 0 andt= 1 into the general solution. We havey0=A+B= 1andy1= 2A+ 3B= 4.Solving these equations forAandB, we findB= 2 andA=-1. Therefore thesolution ofyt+2-5yt+1= 6yt= 0 with initial conditionsy0= 1 andy1= 4 isyt=-(2t) + 2(3t).We can immediately check thaty0=-20+ 2(30) =-1 + 2(1) = 1 andy1=-2 + 2(3) = 4 as required. A further check is to find thaty2=-(22) + 2(32) =-4 + 18 = 14 as we found earlier.165
1111. Difference equationsThe auxiliary equationz2+a1z+a2= 0 of a second-order difference equation is aquadratic equation and the general solution to the difference equation depends onwhether 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 thediscriminant,a21-4a2. We consider each case in turn.