Recent content by jjosephh

if f(n) = A*2^n + (B+Cn+Dn^2)*3^n A = 3 18D + 9Cn = 7 so what can i do ... ??

I could'nt calculate the particular solution I said F(n) = B1*2n + B2*3n and after the equation B1 = 3 But there is no B2.. what should be f(n) ? could you help me please ..

r1,r2 = 3 so the solution must be an+2 = rn an+2 = A1*r1n + A2*n*r2n an+2 = A1*31n + A2*n*32n I think i need help :)

by the way a(0) = 1and a(1)=4 n>=2

B1 = 3 B2 = -7/9 and the solution is ; (I got before hom. solution A1= -11/9 A2 = 4/3) an = -11/9*2n + 4/3*3n +3*2n -7/93n

you mean ; f(n) = B1(2n + 3n) ??

Hello again, an+2 - 6an+1 + 9an = 3*2n + 7*3n what about the particular solution ? should be f(n) = B12n + B23n

Thank you for all helps I have another question ; Solve the recurrence relation an+2 - 6an+1 + 9an = 3*2n + 7*3n where n>=0 and a0 = 1 a1 = 4 I think there are two path to solve this problem. First path is find homogenous solution and the second is particular solution So ; In homog...

I found r1 = -1 -i r2 = -1 +i an = K1 r1n + K2 r2n I'm trying to find K1 and K2 a0 = 1 => K1 + K2 = 1 a1 = 3 => K1(-1-i) + K2(-1+i) = 3 Can you help me to find K1 and K2

The problem is ; Solve the recurrence relation a n + 2a(n-1) + 2a(n-2) = 0 where n ≥ 2 and a0 = 1 a 1 = 3 I said; a n = rn and found this equation ; r2 + 2r +2 = 0 I have to find r1 and r2 to write ; K1r12 + K2r22 = a n I know a0 and a 1 so i can find K1 and K2 But i have problem with...

How to solve recurrence relation ? The Example is ; Solve the recurrence relation a n + 2a n-1 + 2a n-2 = 0 where n ≥ 2 and a 0 = 1 a 1 = 3 a n = nth order a n-1 = (n-1)th order. a n-2 = (n-2)th order. I've started the solving ; a n = r^n so the equation will be ; r^n + 2r^(n-1)...