Homogeneous second-order difference equations

An equation of the form

y_t+a₁yt−1+a₂yt−2 = 0, t≥2

in which a₁ and a₂ are constants, is a homogeneous linear second-order difference equation with constant coefficients (or recurrence equation of the same

description). It is called second-order because the equation involves two previous terms of the sequence (or equivalently, because the difference between the highest index and lowest index is two). If you are given y₀ and y₁, then the equation determinesy₂ and all remaining numbers in the sequence. The equation may also be written as

y_t+2+a₁y_t+1+a₂y_t = 0, t ≥0.

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 y₀ and y₁ 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, Ax=0):

a constant multiple of a solution is a solution, the sum of two solutions is a solution.

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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+a₁y_t+1+a₂y_t= 0, show that x_t+z_t and cx_t, where cis 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 wherem is some constant to be determined.

Substituting yt=m^t into the difference equation we obtain

y_t+2+a₁y_t+1+a₂y_t=m^t+2+a₁m^t+1+a₂m^t=m^t(m²+a₁m+a₂) = 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 m²+a₁m+a₂ = 0. This equation,

z²+a₁z+a₂ = 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+2−5y_t+1+ 6y_t= 0.

The auxiliary equation is

z²−5z+ 6 = 0.

This equation factors, z²−5z+ 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 y₀ = 1 andy₁ = 4. Then the difference equation determines all remaining numbers in the sequence. We have

y₂ = 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 found by substituting t = 0 andt= 1 into the general solution. We have

y₀ =A+B = 1 and y₁ = 2A+ 3B = 4.

Solving these equations for A and B, we find B = 2 andA=−1. Therefore the solution of y_t+2−5y_t+1 = 6y_t= 0 with initial conditions y₀ = 1 and y₁ = 4 is

y_t =−(2^t) + 2(3^t).

We can immediately check that y0 =−2⁰+ 2(3⁰) = −1 + 2(1) = 1 and y₁ =−2 + 2(3) = 4 as required. A further check is to find that

y₂ =−(2²) + 2(3²) = −4 + 18 = 14 as we found earlier.

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The auxiliary equation z²+a₁z+a₂ = 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, a²₁−4a₂. We consider each case in turn.

When the auxiliary equation has two distinct solutions, α and β, the general solution is

yt =Aα^t+Bβ^t (A, B constants).

In any specific case, A and B are determined by the initial valuesy₀ and y₁, as in Example 11.6.

When the auxiliary equation has just one solution, α, the general solution is y_t=Ctα^t+Dα^t= (Ct+D)α^t.

As in the previous case, the values of the constants C and D can be determined by using the initial values y0 and y1.

The auxiliary equation has no real solutions when the quantity a²₁−4a₂ is negative.

In that case, 4a₂−a²₁ is positive, and hence so is a₂. Thus there is a positive square root r of a₂; that is, we can define r =√

a₂. In order to write down the general solution in this case we define the angle θ by

cosθ=−a₁

2r =− a₁ 2√

a2

. Then the general solution in this case is

y_t =Er^tcosθt+F r^tsinθt, where E and F are constants.

That these general solutions are as stated can be verified by substitution into the difference equations, but you are not expected to do so for this subject. If you study complex numbers in the future, you will be able to see where the last type of general solution comes from.

Instead, we will look at some examples.

Example 11.7 We find the general solution of the difference equation yt−6yt−1+ 5yt−2 = 0.

The auxiliary equation is z²−6z+ 5 = 0, that is (z−5)(z−1) = 0, with solutions 1 and 5. The general solution is therefore

yt=A(1^t) +B(5^t) = A+B5^t, for arbitrary constants A and B.

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11.6. Homogeneous second-order difference equations

Example 11.8 We find the general solution of the difference equation y_t+ 6yt−1+ 9yt−2 = 0.

The auxiliary equation is z²+ 6z+ 9 = 0, that is (z+ 3)² = 0, with solution z =−3.

The general solution is therefore

y_t=A(−3)^t+Bt(−3)^t = (A+Bt)(−3)^t, for arbitrary constants A and B.

Example 11.9 Let us find y_t if

y_t−2y_t−1+ 4y_t−2 = 0, and y₀ = 1, y₁ = 1−√

3. Here, the auxiliary equation, z²−2z+ 4 = 0, has no real solutions, so we are in the third case. In the notation used above, we have

r =√

4 = 2. It follows that

cosθ=−(−2)/(2r) = 2/4 = 1/2, so θ=π/3. The general solution is therefore

y_t= 2^t(Ecos (πt/3) +F sin (πt/3)).

Putting t= 0, and using the given initial condition y₀ = 1, we haveE = 1. Similarly y₁ = 1−√

3 implies that

2 (Ecos(π/3) +F sin(π/3)) = 2 1 2 +F

√3 2

!

= 1−√ 3, so that 1 +√

3F = 1−√

3. Therefore F =−1 and the required solution is y_t= 2^t(cos (πt/3)−sin (πt/3)).

Let’s check this solution. It should satisfy the initial conditions y₀ = 1, y₁ = 1−√ 3.

In addition we can obtain y₂: y₂ = 2y₁−4y₀ = 2(1−√

3)−4(1) =−2−2√ 3.

Using the solution,

y₀ = 2⁰(cos 0−sin 0) = 1 and

y₁ = 2(cos(π/3)−sin(π/3)) = 2((1/2)−(√

3/2)) = 1−√ 3.

So far, so good. Now for y₂. We have

y₂ = 2²(cos(2π/3)−sin(2π/3)) = 4(−(1/2)−(√

3/2)) =−2−2√ 3, which is as it should be.

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