1

$a_{n+1} - a_n = 3n^2 - n$ ;$a_0=3$

I need help for solving the particular solution. Based on a chart in my textbook if you get $n^2$ the particular solution would be $A_2n^2 + A_1n + A_0$ and $n$ has the particular solution of $A_1n+A_0$.
So given $3n^2 - n$, my first thought was that if the equation was $n^2-n$ you can have something like $An^2 + Bn+C - (Bn + C) = An^2$.

Is this process correct if I simply had $n^2-n$ ? If so how would the $3$ in $3n^2$ affect this step?

Extreme112
  • 253

2 Answers2

2

take a particular solution in the form $$a_n = An^3 + Bn^2+Cn,\\ 3n^2 - n = a_{n+1} - a_n = A(n+1)^3 - An^3 + B(n+1)^2 - Bn^2+ C\\= 3An^2 + n(3A+2B) + A+B+C$$ equating the coefficients gives $$A = 1, B = -2, C=1. $$

the homogeneous solution is $ a_n = D $ so the general solution is $$a_n = n^3 -2n^2 +n + D, a_0 = 3 \to D = 3. $$

abel
  • 29,170
  • Whats the reason for $An^3$? I thought you could just equate $a_n$ to the quadratic equation since the right hand side of the equation seems to be in that form already. – Extreme112 Mar 17 '15 at 01:33
  • 1
    you are going to be adding $1^2 + 2^2 +\cdots + n^2$ to get the $a_n$ among authors. so you will need a cubic term $n^3$ and quadratic term is not sufficient. – abel Mar 17 '15 at 01:36
1

hint: $a_n = (a_n-a_{n-1})+(a_{n-1}-a_{n-2})+\cdots +(a_2-a_1)+(a_1-a_0)+a_0$

DeepSea
  • 77,651