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.
 A: I can think of two ways to solve this. One would be to draw the slope field for this ODE. That would essentially tell you how to draw any solution curve. The other method is exactly what you described: for each initial condition, find the arbitrary constant and plot up the solution. The equation is linear, and there are existence and uniqueness theorems for linear equations, so no, you can't have infinitely many solutions going through an initial value.
So, for one example. You say you have the general solution - that's great. I'll just re-derive it here for completeness.
\begin{align*}
\frac{dy}{dx}&=y\cos^2(x)\\
\int\frac{dy}{y}&=\int\cos^2(x)\,dx \\
\ln|y|&=\frac{x}{2}+\frac{1}{4}\,\sin (2 x)+C \\
y&=C\exp\left(\frac{x}{2}+\frac{1}{4}\,\sin (2 x)\right).
\end{align*}
If we take the $y(0)=1/2$ initial condition, we find that
\begin{align*}
\frac12&=C\\
y&=\frac12\,\exp\left(\frac{x}{2}+\frac{1}{4}\,\sin (2 x)\right).
\end{align*}
Then you can plot that on an interval containing $x=0.$
A: Another approach is to draw a direction field plot and then superimpose the solution curves for the three initial conditions. 
The solution is given by $$y(x) = c_1 e^{\frac{x}{2}+\frac{1}{4} \sin (2 x)}$$
You should get

A: 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 $A = \pm \frac{1}{2}$
