Answer to Question #219496 in Discrete Mathematics for Srilekha

Question #219496

Solve the Recurrence relation an-3an-1-4an-2=4.3n where a0=1,a1=2


1
Expert's answer
2021-07-22T06:05:27-0400
"a_n-3a_{n-1}-4a_{n-2}=4\\cdot3^n"


The associated linear homogeneous equation is 


"a_n-3a_{n-1}-4a_{n-2}=0"

The characteristic equation is


"r^2-3r-4=0"

"(r+1)(r-4)=0"

"r_1=-1, r_2=4"

The solution is


"a_n^{(h)}=\\alpha_1(-1)^n+\\alpha_2(4)^n"



Because "F(n)=4\\cdot3^n," a reasonable trial solution is


"a_n^{(p)}=C\\cdot3^n,"

where "C" is a constant.

Substitute


"C\\cdot3^n-3C\\cdot3^{n-1}-4C\\cdot3^{n-2}=4\\cdot3^n"

"9C-9C-4C=36"

"C=-9"

Hence the particulr solution is


"a_n^{(p)}=-9\\cdot3^n"

All solutions are of the form


"a_n=\\alpha_1(-1)^n+\\alpha_2(4)^n-9\\cdot3^n"

where "\\alpha_1" and "\\alpha_2" are a constants.


"a_0=1, a_1=2"


"a_0=\\alpha_1(-1)^0+\\alpha_2(4)^0-9\\cdot3^0=1"

"a_1=\\alpha_1(-1)^1+\\alpha_2(4)^1-9\\cdot3^1=2"


"\\alpha_1+\\alpha_2=10"

"-\\alpha_1+4\\alpha_2=29"



"\\alpha_1=2.2"

"\\alpha_2=7.8"

"a_n=2.2(-1)^n+7.8(4)^n-9\\cdot3^n"


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