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I have and want to solve the following difference equation:

$ x_{n+2} - 4X_n = 27n^2, x_0 = 1, x_1 = 3 $.

First, I find a solution to the homogeneous case (by solving the auxiliary equation $ (\lambda -2)(\lambda + 2) = 0 $), which is:

$ x^c_{n} = A2^n+B(-2)^n $

Then I have to "guess", what's called the particular integral on the ODE world. And as $27n^2$ is a polynomial, I guess with the generic same degree polynomial $ x^p_{n} = Cn^2+Dn+E $.

Now, I substitute $ x^p_{n} $ in the initial equation to find $ C=-9, D=-12 and E=-20 $.

Therefore, as the solution $ x_n = x^c_{n} + x^p_{n} $, we have:

$ x^n = A2^n+B(-2)^n-9n^2-12n-20 $.

We substitute in the initial conditions and get:

$ x_0 = A + B - 20 = 1 $

$ x_1 = 2A - 2B - 9 - 12 - 20 = 3 $

Which means $ A = \frac{43}{2} $ and $ B = -\frac{1}{2} $.

Therefore, the solution to this difference equation should be:

$ x_n = \frac{43*2^n}{2} - \frac{(-2)^n}{2} - 9n^2 - 12n - 20 $

But, when substituting into the initial equation it's clearly not a solution.

Where did I go wrong?

Jena Rayner
  • 81

1 Answers1

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Here's the theory to justify your guess:

Denote $\mathbb{E}$ the forward shift operator such that $\mathbb{E} X_n=X_{n+1}, \mathbb{E}^2 X_n=X_{n+2}$, etc.

Note that $(\mathbb{E}-1)^3 n^2=0$ (the third difference of a second order polynomial is zero, just like third derivative of a second order polynomial is zero), then

$$(\mathbb{E}^2-4)X_n=27n^2 \implies (\mathbb{E}-1)^3 (\mathbb{E}^2-4) X_n = 0\\ $$

Therefore $$X_n = c_1 2^n + c_2 (-2)^n + c_3 + c_4 n + c_5 n^2.$$

Also see this example.

Neat Math
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