Computing the solution of a recurrence relation Let $y_k=\dfrac{16}{3}y_{k-1}-\dfrac{5}{3}y_{k-2}$ 

with $y_0=1$ and $y_1=\dfrac{1}{2}$ for $k \geq 2, k \in \mathbb{N}$.

To find the solution $\lbrace y_k \rbrace_{k \in \mathbb{N_0}}$ I used:

Firstly, I computed it for $k=2$:

$y_2=\dfrac{16}{3}\dfrac{1}{2}-\dfrac{5}{3}1=1$

What has to be done next? Induction isn't helpful here.

I don't see how can it be found for $\lbrace y_k \rbrace_{k \in \mathbb{N_0}}$.
$$y_k=\frac{{{5}^{k}}}{28}+\frac{{{3}^{3-k}}}{28}$$

$$[1,1/2,1,9/2,67/3,2009/18,15067/27,452009/162,3390067/243,101702009/1458,762765067/2187]$$

Solution details:

Char. equation is
$${{t}^{2}}-\frac{16 t}{3}+\frac{5}{3}=0$$
$$t_1=5,\quad t_2=\frac13$$
Then solution is
$$y_k=A\,5^k+B\,\left(\frac13\right)^k$$
From initial conditions
$$A\,5^0+B\,\left(\frac13\right)^0=1,\\
A\,5^1+B\,\left(\frac13\right)^1=\frac12$$
$\Rightarrow$
$$A=\frac{1}{28},\quad B=\frac{27}{28}$$
$\Rightarrow$
$$y_k=\frac{{{5}^{k}}}{28}+\frac{{{3}^{3-k}}}{28}$$

see for example https://www.youtube.com/watch?v=aHw7hAAjbD0
$$y_k=\frac{{{5}^{k}}}{28}+\frac{{{3}^{3-k}}}{28}$$

$$[1,1/2,1,9/2,67/3,2009/18,15067/27,452009/162,3390067/243,101702009/1458,762765067/2187]$$

Solution details:

Char. equation is
$${{t}^{2}}-\frac{16 t}{3}+\frac{5}{3}=0$$
$$t_1=5,\quad t_2=\frac13$$
Then solution is
$$y_k=A\,5^k+B\,\left(\frac13\right)^k$$
From initial conditions
$$A\,5^0+B\,\left(\frac13\right)^0=1,\\
A\,5^1+B\,\left(\frac13\right)^1=\frac12$$
$\Rightarrow$
$$A=\frac{1}{28},\quad B=\frac{27}{28}$$
$\Rightarrow$
$$y_k=\frac{{{5}^{k}}}{28}+\frac{{{3}^{3-k}}}{28}$$

see for example https://www.youtube.com/watch?v=aHw7hAAjbD0