Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

4 Answers

up vote 1 down vote accepted

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

share|improve this answer
add comment

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.

share|improve this answer
add comment

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

share|improve this answer
add comment

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

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.