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Difference Equations

Part 5: A Third-Order Equation

In the preceding Part we outlined a program by which one might hope to solve a linear difference equation

yk+n + a1 yk+n-1 + .... + an-1 yk+1 + an yk = zk.

We start with the homogeneous equation

yk+n + a1 yk+n-1 + .... + an-1 yk+1 + an yk = 0.

To find its solutions, we construct the auxiliary equation

rn + a1 rn-1 + ... + an-1 r + an = 0.

If this equation has n distinct roots r1, r2, ..., rn, then the exponential sequences {r1k}, {r2k}, ..., {rnk} are a basis for the solution space of the homogeneous equation, so the general solution can be found by taking all linear combinations of this basis.

Now, if we can find any one solution of the nonhomogeneous equation -- by any means whatsoever -- we get the general solution of the nonhomogeneous equation by adding this one particular solution to the general solution of the homogeneous equation.

We will carry out this program for a single third-order equation to illustrate the steps of the general process.

  1. Find the general solution of the difference equation

    yk+3 - 4 yk+2 - 7 yk+1 + 10 yk = 0.

  2. Show that {k2 - 2k} is a solution of the nonhomogeneous difference equation

    yk+3 - 4 yk+2 - 7 yk+1 + 10 yk = - 24 k + 10.

    Then find the general solution of this equation.

  3. Find the unique solution of the equation in Step 2 that satisfies the initial conditions y0 = 1, y1 = 0, and y2 = 1. Check your work by showing that your proposed solution actually satisfies all the conditions of the problem.

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