$$y_p = A\sin{kx} + B\cos{kx}$$
where k is the coefficient for the trig function radian measure.
$$y'p = -2A\sin{2x} + 2B\cos{2x}$$
$$y''p = -4A\cos{2x} - 4B\sin{2x}$$
A problem results here. When taking $k=2$, you should see
$$y_p = A \sin 2x + B \cos 2x$$
and then you get the derivatives
$$y_p' = 2A \cos 2x - 2B \sin 2x \;\;\;\;\; y_p'' = -4A \sin 2x - 4B \cos 2x$$
You seem to have forgotten to taken the derivative of the trigonometric functions themselves when getting $y_p'$.
This is mostly just an arithmetic error, though, and the overarching idea (and the complimentary solution) are correct.
Edit:
After an edit made to the OP, that error was fixed. The other error that results is finding $A,B$, the latter in particular.
$A=3/26$ is correct. However, an arithmetic error seems to have resulted in finding $B$. Using substitution into $-4A-6B=0$,
$$-4\left( \frac 3 {26} \right) - 6B = 0 \implies B = \frac{-1}{6} \cdot 4\left( \frac 3 {26} \right) = \frac{-12}{156} = \frac{-1}{13} \ne \frac{-12}{216} = \frac{-1}{18}$$
We must consider the equation
$y’’+a_1y’+a_2y=0$
with initial conditions $y’(t_0)=m_1$ and $y(t_0)=m_2$.
You can observe that by Cauchy theorem there exists a unique solution of your equation in a neighborhoods of $t_0$. By some theorems of extension of solutions you have that you can extend your unique solution to all $\mathbb{R}$. In this way you get that it is possibile define a bijective map
$\Psi: \mathbb{R}^2\to X$
such that for each couple $(m_1,m_2)$ we consider the solution of the differential equation $\Psi(m_1,m_2)$ of that initial data.
You can observe that the map $\Psi$ is linear because the equation is linear. So the map $\Psi$ is an isomorphism and $X$ has dimension 2.
If you want, you can generalize the result to a generic omogeneus equation of degree $n$ using the same argument.
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