# second-order difference equation

I have a second-order difference equation question.

yn + 2 - 78yn = 23n^2

What is the value of root in auxiliary equation?

I have tried searching for videos online but I don't really quite understand on how to solve it. Also, what does Y" and Y' represent?

Could someone explain to me in steps so I could understand this topic better.

Thanks!

• the auxiliary equation is $x^2-78=0$. – André Nicolas Mar 18 '14 at 2:43
• Thank you. Is it possible to show me on how to derive to this equation? – user136154 Mar 18 '14 at 2:57
• We are looking for solutions of $y_{n+2}-78y_n=0$. We look for solutions of the shape $y_n=r^n$, where $r\ne 0$. Substituting we get $r^{n+2}-78r^n=0$. Since $r\ne 0$, this simplifies to $r^2-78=0$. The same idea, applied for example to the Fibonacci recurrence $a_{n+2}=a_{n+1}+a_n$, gives $r^2-r-1=0$. – André Nicolas Mar 18 '14 at 3:01

We are looking for solutions of the recurrence $y_{n+2}-78y_n=0$.
We look for solutions of the shape $y_n=r^n$, where $r\ne 0$. Substituting in our recurrence, we get $r^{n+2}-78r^n=0$. Since $r\ne 0$, this simplifies to $r^2-78=0$. That is the auxiliary equation. the variable we use does not matter, we could say that the auxiliary equation, often called the characteristic equation, is $x^2-78=0$. The roots are $\pm\sqrt{78}$.
The same idea, applied for example to the Fibonacci recurrence $a_{n+2}=a_{n+1}+a_n$, gives $r^2-r-1=0$.
The general solution of $y_{n+2}-78y_n=0$ is therefore $A(\sqrt{78})^n +B(-\sqrt{78})^n$, where $A$ and $B$ are arbitrary constants.
To finish solving the recurrence we started with, we need to find a particular solution of $y_{n+2}-78y_n=23n^2$.