${d^2y}\over {dt^2}$ rather than ${dy}\over {dt}$ - what does it change$?$ I have the problem "Find a positive value of $k$ for which $y = \sin(kt)$ satisfies $\frac{d^2y}{dt^2} + 9y = 0.$
Any example I can find for solving this equation either just uses $y'$ or has a typical $\frac{dy}{dt}$ 
So what exactly would be changed by doing this$?$
 A: You're not being asked to solve this differential equation from scratch.
You've been told that a solution looks like $y=\sin kt$ for some number $k$. 
Your task is to find $k$.
Find $\frac{\mathrm dy}{\mathrm dt}$ and then find $\frac{\mathrm d^2y}{\mathrm dt^2}$. Substitute $y=\sin kt$ and your expression for $\frac{\mathrm d^2y}{\mathrm dt^2}$ into the equation $\frac{\mathrm d^2y}{\mathrm dt^2} + 9y = 0$, and see if you can tell what $k$ needs to be for the equation to work for all $t$.
You'll get two possible solutions for $k$, but you're told that $k>0$, so you can ignore one of them.
Follow this link to know how to solve differential equations of the form
$$a\frac{\mathrm d^2y}{\mathrm dx^2} + b\frac{\mathrm dy}{\mathrm dx} + cy = f(x)$$
A: Notice:


*

*$$\frac{\text{d}y(t)}{\text{d}t}=y'(t)$$

*$$\frac{\text{d}^2y(t)}{\text{d}t^2}=y''(t)$$


So, find the solution to $y''(t)+9y(t)=0$ which will look like $y(t)=\sin(kt)$
You can do two things, fill the given solution in, or find the general solution first:


*

*Hint for method 1:


$$\frac{\text{d}^2}{\text{d}t^2}\left[\sin(kt)\right]+9\sin(kt)=0\Longleftrightarrow$$
$$-k^2\sin(kt)+9\sin(kt)=0\Longleftrightarrow$$
$$-(k^2-9)\sin(kt)=0$$


*

*Hint for method 2, using Laplace Transform:


$$y''(t)+9y(t)=0\Longleftrightarrow$$
$$\mathcal{L}_{t}\left[y''(t)+9y(t)\right]_{(s)}=\mathcal{L}_{t}\left[0\right]_{(s)}\Longleftrightarrow$$
$$\mathcal{L}_{t}\left[y''(t)\right]_{(s)}+9\mathcal{L}_{t}\left[y(t)\right]_{(s)}=0\Longleftrightarrow$$
$$\mathcal{L}_{t}\left[y''(t)\right]_{(s)}+9y(s)=0\Longleftrightarrow$$
$$s^2y(s)-sy(0)-y'(0)+9y(s)=0\Longleftrightarrow$$
$$(s^2+9)y(s)-sy(0)-y'(0)=0\Longleftrightarrow$$
$$y(s)=\frac{y(0)s+y'(0)}{s^2+9}\Longleftrightarrow$$
$$y(s)=\frac{sy(0)}{s^2+9}+\frac{y'(0)}{s^2+9}\Longleftrightarrow$$
$$\mathcal{L}_{s}^{-1}\left[y(s)\right]_{(t)}=\mathcal{L}_{s}^{-1}\left[\frac{sy(0)}{s^2+9}+\frac{y'(0)}{s^2+9}\right]_{(t)}\Longleftrightarrow$$
$$y(t)=\mathcal{L}_{s}^{-1}\left[\frac{sy(0)}{s^2+9}\right]_{(t)}+\mathcal{L}_{s}^{-1}\left[\frac{y'(0)}{s^2+9}\right]_{(t)}\Longleftrightarrow$$
$$y(t)=y(0)\mathcal{L}_{s}^{-1}\left[\frac{s}{s^2+9}\right]_{(t)}+y'(0)\mathcal{L}_{s}^{-1}\left[\frac{1}{s^2+9}\right]_{(t)}\Longleftrightarrow$$
$$y(t)=y(0)\cos(3t)+\frac{y'(0)\sin(3t)}{3}\Longleftrightarrow$$
$$y(t)=\text{C}_1\cos(3t)+\frac{\text{C}_2\sin(3t)}{3}$$
