If $u_{n+2}-4u_{n+1}+4u_n=0$ then why $u_n=A2^n+Bn2^n$? Let $V=\{u\in\mathbb R^{\mathbb N}\mid\ u_{n+2}-4u_{n+1}+4u_n=0\}$, a vector space.

1) Show that $V$ has dimension $2$.

2) Show that $V=\text{Span}((2^n)_{n\in\Bbb{N}},(n2^n)_{n\in\Bbb{N}}).$



Let $u_n=r^n$. Then $$r^{n+2}-4r^{n+1}+4r^n=0\iff 0=r^2-4r+4=(r-2)^2\iff r=2.$$
Since I have a unique root, $u_n=A2^n+Bn2^n=D2^n$, so I can't get a sum of two linear independent solution. Is there a way as for ODE ? (i.e. for example if $f''-4f'+4=0$, then by setting $D(f)=f'$, then $D^2f-4Df+4=(D-2)^2f=0$ and thus $f$ is is a generalized eigenvector, and we can easily find a basis of $\{f\mid f''-4f+4=0\}$. Is there a similar way for sequences ?
Let $D:\mathbb R^{\mathbb N}\to \mathbb R^{\mathbb N}$ defined as $$D\big((u_n)_{n\in\mathbb N}\big)=\big(u_{n+1}\big)_{n\in\mathbb N}.$$

Then $V=\ker (D-2)^2$. Let $(u_n)_{n\in\mathbb N}\in \ker(D-2)^2\setminus \ker(D-2)$. 

Set $v_n=u_{n+1}-2u_n$. Then $$v_{n+1}-2v_n=0\implies v_{n+1}=2^{n+1}v_0,$$
and thus $$u_{n+1}=2^{n}v_0+2u_n=2^{n}v_0+2^{n}v_0+2^2u_{n-1}=...=(n+1)2^{n+1}v_0+2^{n+1}u_0$$
$$=(n+1)2^{n+1}(u_1-2u_0)+2^{n+1}u_0$$
as wished.
First let $v_n=u_{n+1}-2u_n$ so that $v_{n+1}-2v_n=0$

Then we clearly have that $v_n=2^nv_0$ and $u_{n+1}-2u_n=2^nv_0$ or $$u_n=2u_{n-1}+2^{n-1}v_0$$ and $$2u_{n-1}=4u_{n-2}+2^{n-1}v_0$$ so that $u_n=4u_{n-2}+2\cdot2^{n-1}v_0$

and in general $$2^ru_{n-r}=2^{r+1}u_{n-r-1}+2^{n-1}v_0$$ with the consequence that $u_n=2^{r+1}u_{n-r-1}+(r+1)\cdot 2^{n_1}v_0$ and with $r=n-1$ $$u_n=2^nu_0+n\cdot 2^{n-1}v_0$$