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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!

user136154
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    the auxiliary equation is $x^2-78=0$. – André Nicolas Mar 18 '14 at 02:43
  • Thank you. Is it possible to show me on how to derive to this equation? – user136154 Mar 18 '14 at 02:57
  • 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 03:01

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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 $y_{n+2}-78y_n=23n^2$.

André Nicolas
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