Slope Field and "satisfying" a differential equation? Disclaimer: I am not a student trying to get free internet homework help. I am an adult who is learning Calculus from a textbook. I am deeply grateful to the members of this community for their time.
$\frac{dy}{dx}=(y-1)^2cos(\pi x)$
Q: There is a horizontal line with equation y=c that satisfies this differential equation.  Find the value of c.    A: The line y=1 satisfies this differential equation, so c=1.
I don't even know what this question is asking.  Can someone clarify what is being asked?  When you draw the slope fields, the y=1 dots are horizontal (dy/dx = 0)  I'm missing the connection, if there is one.

A: Suppose $y = c$ satisfies this equation for some value of $c$. You know that the derivative of a constant is $0$, i.e. $\dfrac{dy}{dx} = 0$. Then, to satisfy the differential equation:
$$
0 = (c - 1)^2 \cos(\pi x)
$$
for all $x$. Since $\cos(\pi x)$ is usually not equal to zero, you're going to have to ensure that $(c-1)^2 = 0$. The only way to do this is to have $c = 1$; that is, $y = 1$ is the solution you're looking for.
Further, to see how you might "guess" this solution from the slopefield, think of the lines on the slope field as indicating the direction of a current. Note that if you start at any of the points on $y=1$, the line is horizontal there, so the current would drag you to another point on $y=1$.
EDITS:
(1) "solving" a differential equation can mean various things. Often times, we use some prescribed method to go from the equation to the solution (examples are direct integration (as you mention) or the method of separation of variables (if you haven't learned about this yet, don't worry)). 
But another valid method of solving differential equations is guessing (and then possibly checking and refining guesses)! If we guess a solution, the way that we check whether or not it's correct is to see if it satisfies the differential equation. If you can plug the solution into the left- and right-hand sides of the differential equation, and they're equal, then you've guessed correctly.
Alternatively, if you have '"solved" the DiffEq by finding the integral", you can still (double-) check that it is a solution (i.e. that it "satisfies the diff. eqn.") in the same manner: plug it into the LHS and RHS, check that they're equal.
(2): it is important to realize that differential equations don't have a unique solution (until you've specified initial values). At best, you can identify a family of similar functions that all solve the diff. eqn. For instance, for the diff. eqn.
$$
\dfrac{dy}{dx} = 5y
$$
the function $y = ce^{5x}$ is a solution for any $c$. That is, $2e^{5x}, -3.7 e^{5x}, $etc. are all solutions. Finally, $y = 0$ is also a solution, and though it doesn't look like the other solutions at first glance, you can realize this is the same as $y = 0 e^{5x}$.
Something similar is happening in this diff. eqn.: in general, the other solutions are going to be more complicated functions of $x$, but this problem is asking for a special solution.
(3) the lines in the slopefield will be tangent to the various solutions. In some sense, the tangent line of a function tells you "where the function will go next". So the slopefield gives us a way to roughly estimate/sketch solutions. 
For instance, if I had the initial condition $y(0) = 0$, then I know the solution goes through $(0,0)$. I note that the tangent line there has a slope of $1$ (and is confirmed by looking the diff. eqn., and plugging in $(0,0)$ : $\dfrac{dy}{dx} = (0 - 1)^2 \cos(\pi 0) = 1$), so if I want to guess where the solution goes next, I would move over $1$ and up $1$, and then repeat the process.
Maybe it would help your intuition to see a better picture of this slopefield. Go to http://www.math.psu.edu/cao/DFD/Dir.html
and use the following settings:
y' = (y-1)^2 * cos(3.14 * t)
T steps : 40
Y steps : 20
Arrow length : 0.1
Tmin : -2
Tmax : 2
Ymin : -2
Ymax : 2

can you better visualize how you might "follow" these arrows, one to the next, to sketch a solution?
(and if you can't get the applet to work, the output looks like this:  )
