00:01
We are given recurrence relations and initial conditions.
00:04
We are asked to solve these recurrence relations with the initial conditions.
00:12
The recurrence relation is a .n equals a .n minus 1 plus 6, a .n minus 2, with initial conditions a0 equals 3 and a1 equals 6.
00:24
This is a linear homogenous occurrence relation.
00:30
And so the characteristic equation is r squared minus r minus 6, equals 0, and we can factor this as r plus 2 times r minus 3 equals 0.
00:51
So we have the two roots are r equals negative 2 and r equals 3, which are distinct.
00:58
And so the general form for the solution is an equals alpha 1 times negative 2 to the n plus alpha 2 times 3 to the n, plus alpha 2 times 3 to the n.
01:10
Using our initial conditions, we have that 3, which is a0, is equal to alpha 1 plus alpha 2.
01:24
And we have that 6, which is a1, is equal to negative 2 alpha 1 plus 3 alpha 2.
01:34
And so if we add twice equation 1 to equation 2, we get 12 is equal to 4 alpha 2.
01:45
2, so that alpha 2 is equal to 3, and by back substitution, alpha 1 is equal to 0.
02:01
This isn't quite right.
02:05
This should be, this is twice, and this is going to be 5 alpha 2.
02:15
So we have the 5 alpha 2 is equal to 12, so that alpha 2 is equal to 12.
02:22
And therefore the back substitution alpha 1 is equal to negative 12 or 3 minus 12ths which is 15 minus 12 or 5 or 3 5ths and so we have a particular solution is 3 5ths times negative 2 to the n plus 12 5ths times 3 to the end and in part b we're given the recurrence relation a n equals 7 a n minus 1 minus 10 a n minus 2 in the initial conditions a 0 equals 2 and a 1 equals 1 this is a linear homogenous recurrence relation and so the characteristic equation is r squared minus 7r plus 10 equals 0 this can be factored as r minus 2 times r minus 5 equals 0 so we have the roots r of the characteristic equation are 2 and 5 these roots are distinct so so it follows that the general solution is alpha 1, 2 to the n plus alpha 2, 5 to the n.
03:48
And using our initial conditions, we have that 2, which is a0, is the same as alpha 1 plus alpha 2.
03:55
And we have that 1, which is a1, is the same as 2 alpha 1 plus 5 alpha 2.
04:03
And so if we subtract twice equation one from equation two, we get 1 minus 4 as negative 3 is equal to 5 minus 2 is 3 alpha 2.
04:20
And so we have that alpha 2 is equal to negative 1, and therefore the alpha 1 of x substitution is equal to 3.
04:29
And so we have the particular solution is 3 times 2 to the n minus 5 to the n.
04:45
In part c, we're given the recurrence relation, a .n equals 6 a .n minus 1, minus 8, an minus 2, with initial conditions a0 equals 4 and a1 equals 10.
04:58
This is a linear homogenous recurrence relation, and it follows that the characteristic equation is r squared minus 6r.
05:10
Plus 8 equals 0.
05:13
This can be factored as r minus 2 times r minus 4 equals 0, so that the roots are r equals 2 and 4, which are distinct, and so the general solution is alpha 1, 2 to the n, plus alpha 2, 4 to the n.
05:37
And using our initial conditions, we have that 4, which is a 0, is alpha 1.
05:42
Plus alpha 2.
05:44
And we have that 10, which is a 1, is 2 alpha 1 plus 4 alpha 2.
05:52
And if we subtract twice equation 1 from equation 2, we get 10 minus 8 is 2, and 4 minus 2 is 2 alpha 2.
06:05
So we get that alpha 2 is equal to 1.
06:08
And so by back substitution, alpha 1 is equal to 3.
06:13
And therefore we have the particular solution is 3 times 2 to the n plus 4 to the n.
06:29
In part d, we're given the relation an equals 2an minus 1 minus a .n minus 2, an minus 2.
06:37
And initial conditions, a0 equals 4 and a1 equals 1.
06:43
This is a linear homogenous recurrence relation, and so it follows that the characteristic equation is r squared minus 2r plus 1 equals 0.
06:55
And this can be factored as r minus 1 squared equals 0.
07:00
So the characteristic roots are r equals 1, the multiplicity of 2...