Find particular solution of $y''+4y=12$ if the point $(0,5)$ has horizontal tangent line

Question

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

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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!

Answer 1

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 $y'(0) = 0$.

Answer 2

Your general solution is correct.

At $(0,5)$ you have horizontal tangent line that is $y'(0)=0$

Thus your initial conditions are $y(0)=5$ and $y'(0)=0$

Answer 3

Yes, your initial conditions are correct, except for the $k \in \mathbb{R}$ thing. You know that you have a horizontal tangent, i.e. a slope at $x = 0$, which is equal to zero: $$y'(0) = 0$$

Answer 4

hint

$$y_p=3$$ is a partcular solution.

the general solution is

$$y=C_1\cos(2x)+C_2\sin(2x)+3$$

a horizontal tangent at $(0,5)$ mean that

$$y(0)=5$$ and $$y'(0)=0$$

these conditions give $$C_1=2$$ and $$C_2=0.$$