Find particular solution of $y''+4y=12$ if the point $(0,5)$ has horizontal tangent line Find the particular solution of $$y''+4y=12$$ if the point $(0,5)$ has horizontal tangent line (parallel to the $x$-axis).

I know the general solution of $y''+4y=12$ is $$y=C_1\cos{(2x)}+C_2\sin{(2x)}+3\quad\text{for some}~C_1,C_2\in\mathbb R.$$ Now we have to find the particular solution but I don't know if the initial conditions are correct: $$\begin{cases}y''+4y=12\\\color{red}{y(0)=5}\\\color{red}{y'(0)=k,\quad\text{for some}~k\in\mathbb R}.\end{cases}$$ Are the $\color{red}{\text{initial conditions}}$ correct?
Thanks!

The equation of the tangent line at $(0,5)$ is $$y(x) = y'(0)(x-0) + y(0) = y'(0)x + 5$$
This is parallel to the $x$-axis if and only if $y'(0) = 0$.
