particular solution of non homogenous difference equation

Question

I have a difference equation and I am struggling to find the particular solution for it. $$ y_n-y_{n-1}=n^2 $$ I have already found the yc and I need to find yp to find out the total solution of this difference equation

Answer 1

HINT

Try $y_n = An^3+Bn^2+Cn+D$ and compute the value of $A,B,C,D$ that would fit by plugging in...

UPDATE

Note that $$ \begin{split} y_{n-1} &= A(n-1)^3+B(n-1)^2+C(n-1)+D \\ &= A \left[n^3-3n^2+3n-1\right] + B \left[n^2-2n+1\right] + Cn-C+D \\ &= An^3 + n^2 [B-3A] + n[3A-2B+C] + 1[D-C] \end{split} $$

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UPDATE 2

From your comments, you are suggesting that the correct solution has $A=1/3,B=1/2,C=1/6,D=0$, which yields $$y_n = n^3/3 + n^2/2 + n/6.$$

Let's check that is indeed the case. Note that $$ \begin{split} y_{n-1} &= (n-1)^3/3 + (n-1)^2/2+(n-1)/6 \\ &= \frac{n^3-3n^2+3n-1}{3} + \frac{n^2-2n+1}{2} + \frac{n-1}{6} \\ &= \left[\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} \right] + n^2\left[\frac{-3}{3}\right] + n \left[ \frac{3}{3} + \frac{-2}{2}\right] + \frac{-1}{3}+\frac{1}{2} - \frac16\\ &= y_n -n^2, \end{split} $$ and therefore $y_n - y_{n-1} = n^2$, as desired. Hence, this is a good solution.

Answer 2

gt6989b's answer isn't bad, though it is somewhat messy. It's easier to compute it in terms of rising factorials:

\begin{align}y_n&=An^{(3)}+Bn^{(2)}+Cn^{(1)}+Dn^{(0)}\\&=An(n+1)(n+2)+Bn(n+1)+Cn+D\end{align}

We then have

\begin{align}y_n-y_{n-1}&=3An^{(2)}+2Bn^{(1)}+Cn^{(0)}\\&=3An(n+1)+2Bn+C\end{align}

We also have

$$n^2=n^{(2)}-n^{(1)}$$

Thus, $A=\frac13,B=-\frac12,$ and $C=0,$ with $D=y_0$.

Answer 3

Note that it is a telescoping relation, thus $\displaystyle y_n-y_0=\sum\limits_{k=1}^n k^2=\dfrac{n(n+1)(2n+1)}6$

At the same time since: why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$

Assuming we don't know $\sum k^2$, we are reduced to find a solution of the form $An^3+Bn^2+Cn+D$ as in the other answers...