How can I show that this family of curves may be described by this differential equation? I have a homework problem in which I wish to show that the family of curves given by
$x^2 + y^2 = c x$
where $c$ is an abitrary constant may be described by the differential equation
$\frac{dy}{dx} = \frac{y^2-x^2}{2xy}$
I thought that I could use implicit differentiation to differentiate the original equation to get the second equation, but instead I get the equation
$\frac{c-2x}{2y}=\frac{dy}{dx}$
As you can see the derivative I get is not in the form of the equation that I am supposed to get. I do not see a way for my solution to even become similar to the proposed solution as one contains constants whereas the other does not.
What is the correct procedure I should use to solve the problem?
 A: You should do as follows:
Differentiate the original equation, getting
$${x^2} + {y^2} = cx$$
$$2x + 2yy' = c$$
Now substitue from the last equation onto the first one:
$${x^2} + {y^2} = \left( {2x + 2y\frac{{dy}}{{dx}}} \right)x$$
$$\frac{{{x^2} + {y^2}}}{{2x}} = x + y\frac{{dy}}{{dx}}$$
$$\frac{{{x^2} + {y^2} - 2{x^2}}}{{2x}} = y\frac{{dy}}{{dx}}$$
$$\frac{{{y^2} - {x^2}}}{{2x}} = y\frac{{dy}}{{dx}}$$
$$\frac{{{y^2} - {x^2}}}{{2xy}} = \frac{{dy}}{{dx}}$$
as required.
Is this from Spiegel's Applied Equations? I'm asking since when I edited I wanted to add some info from the book and I found a problem that stated:
"Find the orthogonal trayectories of $x^2+y^2 = cx$"
And suggested the followging: solve for $c$ to get
$$\frac{{{x^2} + {y^2}}}{x} = c$$
$$\frac{{2{x^2} + 2xyy' - {x^2} - {y^2}}}{{{x^2}}} = 0$$
$${x^2} - {y^2} + 2xyy' = 0$$
$$\frac{{{y^2} - {x^2}}}{{2xy}} = \frac{{dy}}{{dx}}$$
The idea is to get rid of $c$, as you can see in any of the solutions.
A: This is what I got too: differentiate $x^2 + y^2 = c x$ with respect to $x$, we get
$$2x + 2y\frac{dy}{dx} = c, $$
which implies that 
$$\frac{dy}{dx}=\frac{c-2x}{2y},$$
as you got. But you can multiplying $x$ to get:
$$\frac{dy}{dx}=\frac{c-2x}{2y}\cdot\frac{x}{x}=\frac{cx-2x^2}{2xy}=\frac{(x^2 + y^2 )-2x^2}{2xy}= \frac{y^2-x^2}{2xy},$$
where we have used $cx=x^2 + y^2$ in the third equality. 
On the other hand, you can also solve the differential equation: $\displaystyle\frac{dy}{dx} = \frac{y^2-x^2}{2xy}$. Let $y=tx$. Then $dy=tdx+xdt$. Therefore the differetial equation can be written as
$$tdx+xdt=\frac{t^2x^2-x^2}{2tx^2}dx=\frac{t^2-1}{2t}dx,$$
which implies that 
$$xdt=-\frac{t^2+1}{2t}dx.$$
Integrating it, we get
$$\ln(t^2+1)=\int\frac{2t}{t^2+1}dt=-\int\frac{dx}{x}=-\ln x+C.$$
Hence, we have
$\ln[x(t^2+1)]=C$. Recall $t=y/x$, we get 
$\displaystyle\frac{y^2}{x}+x=e^C.$ If we write $e^C=c$, we get
$$y^2+x^2=cx.$$
A: This is a homogeneous differential equation as
$$\frac{dy}{dx}=F(x,y)=\frac{y^2-x^2}{2xy}$$
with $F(tx,ty)=F(x,y)$. One can solve it by putting $y=ux$ and so it becomes
$$x\frac{du}{dx}+u=F(1,u)=\frac{u^2-1}{2u}$$
so,
$$x\frac{du}{dx}=-\frac{u}{2}-\frac{1}{2u}$$
that can be promptly integrated to give
$$y^2+x^2=Cx$$
A: First multiply by $\frac{x}{x}$ $$\frac{c-2x}{2y}\cdot\frac{x}{x}=\frac{x(c-2x)}{2xy}=\frac{cx-2x^2}{2xy}$$ From the first equation we know that $$cx=x^2+y^2$$ so $$\frac{cx-2x^2}{2xy}=\frac{(x^2+y^2)-2x^2}{2xy}=\frac{y^2-x^2}{2xy}$$ Therefore $$\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$$
