Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given the first-order difference equation

$y(k) = u(k) + y(k-1)$ for $k = 1, 2, 3, \dots$

with the input signal u(k) = k, and the initial condition y(–1) = 0. I am trying to verify that its solution also satisfies the second-order difference equation

$y(k) = 2y(k-1) - y(k-2) + 1$

with the initial conditions $y(0) = 0$ and $y(–1) = 0$.

Using the method of undetermined coefficients I attained

$y(k) = -(-1)^k + k^2$ (solution for the 1st order difference equation)

I'm having trouble finding a solution to the 2nd order difference equation. How do I solve for its particular solution?

share|cite|improve this question
Your solution for the first-order equation seems wrong. $-(-1)^k + k^2 \neq -(-1)^{k-1} + (k-1)^2 = k^2 - 2k + 1$. Hint: A solution for the first equation is $\frac{k(k+1)}{2}$, since $y(k) = \sum_{i=0}^k k$ – fgp Oct 21 '12 at 20:20
up vote 1 down vote accepted

Hint $\ $ Note $\rm\ y(k) - y(k-1) = k\:\Rightarrow\:y(k-1) - y(k-2) = k-1,\:$ hence, subtracting

$$\rm\:y(k)-2y(k-1)+y(k-2) = 1$$

With $\rm\:\nabla f(k) := f(k)-f(k-1)\:$ it's $\rm\ \nabla y = k\:\Rightarrow\: \nabla^2 y = \nabla k = k-(k-1) = 1.$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.