Respuesta :
All the solutions to the recurrence relation are derived below. And the general solution will be given below.
[tex]\rm a_n = \alpha _1 \cdot 2^n + \alpha _2 \cdot 3^n - n \cdot 2^{n+ 1} + \dfrac{3}{2}n + \dfrac{21}{4}[/tex]
What is an auxiliary solution?
A Volterra integral expression of the second sort for the thickness of a piezoelectric coating on the surface is obtained by solving an auxiliary question in the abdomen and pelvis.
The equation is given below.
[tex]\rm a _n = 5a_{n-1} - 6a _{a-2} + 2^n + 3n[/tex]
Then the equation associated with a homogeneous linear recurrence relation, then we have
[tex]\rm a _n - 5a_{n-1} + 6a _{a-2} = 0[/tex]
The relative equation will be
r² - 5r + 6 = 0
On solving, we have
r = 2, 3
Then we have
[tex]\rm a_n^h = \alpha _1 2^n + \alpha _2 3^n[/tex]
Let a suitable coefficient C such that [tex]C\cdot 2^n = a_n^{p_1}[/tex]
Then replace this in the recurrence relation, then we have
[tex]\rm C\cdot 2^n = 5C \cdot 2^{n-1} - 6C \cdot 2^{n -2}[/tex]
Let
[tex]a_n^{p_1} = b \cdot n \cdot 2^n\\\\a_n^{p_1} = -2n \cdot 2^n[/tex]
Then we have
dn + e = {d(n - 1) + e} - 6 {d(n - 2) + e} + 3n
dn + e = (-d + 3)n + (7d - e)
In comparing we have
d = -d + 3
d = 3/2
and
7d - e = e
e = 21/4
Then the recurrence relation can be written as
[tex]\rm a_n^{p_1} = \dfrac{3}{2}n + \dfrac{21}{4}[/tex]
Then the general solution will be
[tex]\rm a_n = a_n ^{h} + a_n ^{p_1} +a_n ^{p_2} \\\\a_n = \alpha _1 \cdot 2^n + \alpha _2 \cdot 3^n - n \cdot 2^{n+ 1} + \dfrac{3}{2}n + \dfrac{21}{4}[/tex]
More about the auxiliary solution link is given below.
https://brainly.com/question/18521479
#SPJ4
