Finding $C_n=2C_{n-1}+4$

Question

Find a closed form for

$C_n=2C_{n-1}+4$

$C_0=3$

Let set

$1.C_n=D_n+d$

$ 2.D_n=\lambda D_{n-1}$

Using equation 1 we get

$D_n+d=2C_{n-1}+4\Rightarrow D_n+d=2(D_{n-1}+d)+4 \Rightarrow D_n=2D_{n-1}+d+4$

Using equation 2 we get

$ \lambda D_n=2D_{n-1}+d+4$ setting $\lambda=2$ we get

$d=-4$

now $C_0=D_0-4\rightarrow 3=D_0-4\Rightarrow D_0=7$

So $C_n=D_0+d=7*2^n-4\in O(2^n)$

How this process is called, was just thought the way with no explanations

Answer 1

I'd call it something like "solving the homogeneous part of a linear difference equation," in exact analogy with solving the homogeneous part of a linear differential equation.