problem in solving recurrence relation

Question

I'm not able to find particular solution of
$a_n-2a_{n-1}$=$3*2^n$
What i've tried

1. Given RR is $a_n-2a_{n-1}$=$3*2^n$
2. For the particular solution observe the r.h.s of the equation(1)
3. It is $3*2^n$=(a constant)*$2^n$
4. Consider the P.S =(a constant)*$2^n$
5. $a_n^{(p)}$=A*$2^n$
6. $a_{n-1}$=A*$2^{n-1}$
7. Substituting this value in Eq..(1)
8. A*$2^n-2$A*$2^{n-1}=$$3*2^n$
9. $A*2^n(1-1)=3*2^n$
10. $A*2^n0=3*2^n$
11. This were i'm stuck i'm not getting $A$ value Please help..:(

Answer 1

If you are looking for a particular solution, you could try $a_0=0$ which would give $$a_1=2\times 0 + 3\times 2^1 = 6$$ $$a_2=2\times 6 + 3\times 2^2 = 24$$ $$a_3=2\times 24 + 3\times 2^2 = 72$$ $$\cdots$$ and see by inspection and prove by induction that $a_n=3n \, 2^n$.

A similar process of inspection and induction would give the general solution $a_n=(3n+a_0)2^n$.

Answer 2

In these "resonant" type cases, try solutions of the form $$a_n \propto n\lambda^n$$

Note that solutions to $a_n - 2a_{n-1} = 0$ can be added to the solution to give further solutions.

Answer 3

Let $a_m=b_m+(cm+d)2^m$

$$3\cdot2^n=b_n+(cn+d)2^n-2\{b_{n-1}+(cn-c+d)2^{n-1}\}=b_n-2b_{n-1}+c2^n$$

Choose $c=3$