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In the preceding Part we outlined a program by which one might hope to solve a linear difference equation
We start with the homogeneous equation
To find its solutions, we construct the auxiliary equation
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.
Then find the general solution of this equation.
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