# 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?

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Ignoring zeros (so for example you can divide by $x$), $c=\dfrac{x^2 + y^2}{x}$ implies $\dfrac{c-2x}{2y} = \dfrac{y^2-x^2}{2xy}$ by substitution – Henry Apr 19 '15 at 21:57

## 4 Answers

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 $$y^2+x^2=cx.$$

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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

$$y^2+x^2=Cx$$

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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 $$\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$$

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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.

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