OK, I understand the circumstances now.
The blackboard shot shows a spring without any weight attached at left. The line y=0 is indeed chosen at the point where force from spring and force from gravity cancel each other: the point where the acceleration is zero. Not the forces, but the sum of the forces (the vector sum: one is down and equal in magnitude to the other, which is up). Note that the height of this y=0 depends on m (which isn't the case when the spring is horizontal on a table).
Middle drawing shows a weight attached, and we are led to assume this is a snapshot at the moment immediately after the hand (or other support) under the weight was taken away: speed is zero, no equilibrium at all, spring not stretched yet, so acceleration g downwards. Professor leaves it to us to guess if y(0) = y
0 or not. A second look tells us not (y(0) = y
initial so to his credit: he works out the more general case).
The one on the right is then a generic drawing of an intermediate point, for which the differential equation will probably be set up in the following (sorry, don't have time to watch the whole thing).
When I now look at your notes in post #1, I see more or less the same, only you start at the lower end.
In your left picture you write y0 - equilibrium and I am pretty sure that you don't mean a minus sign but a ##\equiv##, i.e. ##y = 0 \Leftrightarrow## equilibrium.
Your middle picture leads me to look at it as the point of maximum stretching at time 0, at the moment the weight is let go: y(0) = A, spring is stretched by A+y
0 and ##\dot y(0) = 0##.
You work it out and worry about the constant. But on the way, you step from ## \vec F_S = k\left ( y_0 - y(t) \right)\hat\imath = -k \; y(t) \hat\imath##.
a) It should be ##-y_0##
b) If you leave the ##-y_0## in, your problems are over, since you already know that ##-k\; y_0 + mg = 0 ##.
(So I could have skipped the first seven paragraphs altogether: you understand things quite well but go a little too fast. Just like almost all of us
)
Unstuck now ?
