I could use a process check. In general since the solution is $$y = y_c + y_p$$ where $y_c$ is the complementary solution and $y_p$ is the particular, I can break it down.
The equation in question is this:
$$y" - 3y' = sin2x$$
Let's find the complementary solution first:
$$y_c = c_1e^{r_1x} + c_2e^{r_2x}$$
so let's solve the auxiliary equation:
$$r^2 - 3r = 0$$ $$r(r - 3) = 0$$ so the roots are $r = {0, 3}$
so the complementary equation is:
$$y = c_1 + c_2e^{3x}$$
Let's find the particular solution second:
so the form of the particular since it's a trig function on the right side is: $$y_p = A\cos{kx} + B\sin{kx}$$
where k is the coefficient for the trig function radian measure.
$$y'p = -2A\sin{2x} + 2B\cos{2x}$$ $$y''p = -4A\cos{2x} - 4B\sin{2x}$$
so substituting:
$$(-4A\cos2x - 4B\sin2x) - 3(-2A\sin2x + 2B\cos2x) = \sin{2x}$$
$$(-4B + 6A)sin2x + (-4A - 6B)cos2x = sin2x$$
so matching coefficients $(-4B + 6A) = 1$ and $-6B - 4A = 0$ and $-6B = 4A$ so $B = \frac{-2}{3}A$
solving further:
$\frac{8}{3}A + 6A = 1$ so $\frac{26}{3}A = 1$ so $A = \frac{3}{26}$ and so $B = \frac{-12}{216}$
so then the general solution is:
$$y = c_1 + c_2e^{3x} + \frac{3}{26}cos2x - \frac{12}{216}sin2x$$
Is my process right? Did I make an error?