00:01
Here we are given a .n is equal to 2 times a .n.
00:10
Minus 1 plus 3 to the power of n.
00:16
So to find the solutions, we're first going to find the root characteristic equation.
00:22
So we're going to set a .n.
00:26
A being equal to r.
00:27
And then a n minus 1 will then be equal to 1.
00:33
All right, and then all other functions of n will be 0.
00:38
So all other functions of n, we will set to 0.
00:50
So when we do that for the above equation, we end up with r is equal to 2.
01:00
All right, so that tells us that the solution to the homogeneous linear recurrence relation.
01:11
So, anr, our homogeneous part is going to be alpha times 2 to the power of n.
01:27
And that's because r is equal to 2.
01:31
And so now we're interested in finding the particular solution.
01:37
Okay, so here we have a function of n being equal to 3 to the power of n.
01:53
So the particular solution, an and we'll put a little p, so most of the particular portion, is going to be some constant p -not times 3 to the power of n.
02:14
And we know that this is the case because 3 is not a root of the characteristic.
02:22
Equation.
02:24
So our characteristic equation here we just have r is equal to two.
02:32
Had r been equal to three, we would have had to do something else, but it isn't, so we don't.
02:38
Great.
02:39
And so then we just want to find the particular solution that will satisfy our recurrent relation.
02:46
All right, so what we know is that an is equal to two times a n minus one plus 3n.
03:00
Okay, so then we'll say p .0 times 3n is going to be equal to 2 times p .0 times 3 to the power of n minus 1 plus 3n...