Plugging an initial value into an implicit solution? I was working on this exact equation,  $x\;dx + (\,y^{-3}-y^{-1}\,)\;dy=0$ and get the implicit solution $\frac{1}{2}x^2 -\ln{|y|}-\frac{1}{3}y^{-2}=\;C$. My next task on this problem is to solve for the initial value $y(2)=1$. Can I just plug in the IV and solve for $C$? In this case is $y(2)=1$ equivalent to  $f(2,1)$? Trying to shake the rust off my math skills.
 A: You have it reversed.  I think the problem means that $y=1$ when $x=2$.
You can plug in the point $(2, 1)$ into the implicit equation to find the particular value of $C$. It isn't that $f(x, y) = f(2, 1)$, it's that the differential equation has an infinite family of solutions, and one of them goes through the point (2,1).
Take a trivial example.  Suppose the equation were $\frac{dy}{dx} = 5$.  The family of solutions would be $y = 5x + C$.  Which solution goes through $(2, 4)$?  $4 = 5 * 2 + C$, so $C = -6$.
Suppose you had $y\,dy = -x\,dx$.  Take a few points on the plane, and compute $\frac{dy}{dx}$ at that point.  For example, at $(3, 4)$, $\frac{dy}{dx} = -\frac{3}{4}$. Draw a small vector from $(3, 4)$ with that slope.  Repeat for a few points. Do you see that you get something like a counterclockwise circulation around the origin? In fact, the solutions are precisely the circles $x^2 + y^2 = C$. Draw the circles, and notice that those arrows are tangent to the appropriate circle.
In fact before the advent of the graphic calculator, people would draw those vectors, and then they would draw the curves free-hand.
