Unclear with slope field and initial values. So I am working with a first order ODE: $$\frac{dy}{dx}= y\cos^2(x)$$
We are told to draw by hand three separate solution curves for each initial value points: $y(0)=0, y(0)=\frac{1}{2}, y(0)=-\frac{1}{2}.$
I already have the general solution. Am I supposed to solve for $C$ with each of the initial values? I am confused by the idea of the solution curve. I mean the line $y=0$ passes through the singular point $y(0)=0,$ but is that a solution curve? Couldn't you construct an infinite number of lines to pass through any initial value? I am very lost. $x=0$ passes through $y(0)=\frac{1}{2}$. Please help.

You want to find function $y$, such that $\frac{dy}{dx} = y(x) \cos^2(x)$. Assume for a while that $y$ does not attain $0$ value, then we have: $\frac{dy}{y(x)} = \cos^2(x)dx$, and by integrating side by side, we get:
$\ln|y(x)| = \int \cos^2(x)dx = \frac{1}{2}\int (\cos(2x) + 1)dx = \frac{1}{4}\sin(2x) + \frac{1}{2}x + C = \frac{1}{2}(\sin(x)\cos(x) + x) + C $
By that we get:
$|y(x)| = \exp(\frac{1}{2}(\sin(x)cos(x) + x) + C) = A\exp(\frac{1}{2}(\sin(x)\cos(x) + x)) $, where $A = e^C > 0 $.
Taking rid of absolute value ( and realise that function $y \equiv 0 $ is a solution, too)$, we have:
$y(x) = Ae^{\frac{1}{2}(\sin(x)\cos(x)+x)}$ - general solution.
Now taking your three cases:
$y(0) = 0$, then $0 = Ae^{\frac{1}{2}(0 + 0)} = A $, so your solution is $y \equiv 0$
$y(0) = \pm\frac{1}{2} $, then $A = \pm \frac{1}{2}$
