Lesson 7

Here is an example: \[F(n)=2F(n-1)+n.\] The solution to the corresponding homogeneous equation is \(c2^{n}\). To find the non-homogeneous addition, try \(an+b\). For the function \(c2^{n}+an+b\) to satisfy the recurrence relation, the following condition must hold: \[an+b=2(a(n-1)+b)+n.\] This is equivalent to: \[0=(a+1)n+(b-2a).\] This equation holds for all \(n\) when \(a=-1\) and \(b=-2\). Thus, the solution is: \[F(n)=c2^{n}-n-2.\]