0
$\begingroup$

I have been given this difference equation and asked to solve it:

$$y_{n+2} + 2y_{n+1} -3y_{n} = 5 \cdot 2^n + 12$$

where $y_{0} = 7$ and $y_{1} = -9$

I know it sounds weird but we have never actually been taught difference equations. I've looked on the internet and from what I've read i think it's a second order equation but apart from that I have no idea where to even start. Can anyone help me?

$\endgroup$
2
$\begingroup$

Follow Wilf's "generatingfunctionology". Define the ordinary generating function: $$ Y(z) = \sum_{n \ge 0} y_n z^n $$ Using the properties of the generating function: $$ \begin{align*} \frac{Y(z) - y_0 - y_1 z}{z^2} + 2 \frac{Y(z) - y_0}{z} - 3 Y(z) &= 5 \cdot \frac{1}{1 - 2 z} + 12 \cdot \frac{1}{1 - z} \\ Y(z) &= \frac{1}{1 - 2 z} - 2 \cdot \frac{1}{1 - z} + 3 \cdot \frac{1}{(1 - z)^2} + 5 \cdot \frac{1}{1 + 3 z} \end{align*} $$ Thoses are mostly geometric series: $$ \begin{align*} y_n &= 2^n - 2 + 3 \binom{-2}{n} + 5 \cdot (-3)^n \\ &= 2^n + 5 \cdot (-3)^n + 3 n + 1 \end{align*} $$

$\endgroup$
1
$\begingroup$

Start by solving homogenous equation $y_{n+2}+2y_{n+1}-3y_n=0$. To do so, write the characteristic polynomial (it will also be of order two), and its roots $\lambda_1$ and $\lambda_2$ will give you the idea of the answer, $y_n= c_1(\lambda_1)^n+c_2(\lambda_2)^n +$ something, which can be guessed from the expression on the RHS. The initial conditions are used to identify the coefficients.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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