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Solve: $\frac{dy}{dx} - \frac{dx}{dy} = \frac{y}{x} - \frac{x}{y}$

What I have done till now: $\left(\frac{dy}{dx}\right)^2 -1= \frac{dy}{dx}\left(\frac{y}{x} - \frac{x}{y}\right)$ $xy\left(\frac{dy}{dx}\right)^2 - xy= \frac{dy}{dx}\left({y^2} - {x^2}\right)$ Now I am stuck.

EDIT: I have been through a course on ODEs (More like a course which added tools to our toolbox to solve ODEs) but it has been a long time since then. Everytime someone helps me with an answer it is an "Oh Yes! I forgot about that" moment. Is there any resource (Succinct PDF preferably) which can help me revise the tools used for solving ODEs? This looked good but I want more examples and methods.

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    This is a quadratic polynomial in $dy/dx$ hence one can solve it for $dy/dx$. The roots are $dy/dx=y/x$ and $dy/dx=-x/y$. Then one can solve these two first order differential equations.2012-02-08

4 Answers 4

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The differential equation reads

$f\left(\frac{\partial y}{\partial x}\right)=f\left(\frac{y}{x}\right)$ and $\frac{\partial y}{\partial x}=\frac{y}{x}$ has solution $y(x)=c\ x.$

The kernel of the right hand side of the equation is a factor of $\pm 1$, which gets sucked up by $c$ already.

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    f is not monotone2012-02-08
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A hint: If $u$ and $v$ are real numbers $\ne0$ then $v-{1\over v}=u-{1\over u}$ holds iff either $v=u$ or $v=-{\displaystyle{1\over u}}$.

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    Awesomely Clever hint.2012-02-08
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$xy\left(\frac{dy}{dx}\right)^2-\frac{dy}{dx}y^2+\frac{dy}{dx}x^2-xy=0$

$y\left(\frac{dy}{dx}\right)\left(x\left(\frac{dy}{dx}\right)-y\right)+x\left(x\left(\frac{dy}{dx}\right)-y\right)=0$

$\left(x\left(\frac{dy}{dx}\right)-y\right)\left(y\left(\frac{dy}{dx}\right)+x\right)=0 \Rightarrow$

$\frac{dy}{dx}=\frac{-x}{y} \Rightarrow \int y \,dy=-\int x\,dx $

$\text{and}$

$\frac{dy}{dx}=\frac{y}{x} \Rightarrow \int \frac{dy}{y}=\int \frac{dx}{x}$

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With the hint from @christian-blatter: $ v-\frac{1}{v}= u-\frac{1}{u} $ $ \iff u,v\neq0 \quad\text{and}\quad v-u=\frac{1}{v}-\frac{1}{u} =-\frac{v-u}{uv} $ $ \iff u=v\neq0 \quad\text{or}\quad uv=-1 $ (i.e. $v=\delta u^\delta\neq0$ for $\delta=\pm1$).

Now $\frac{dy}{dx}$ and $\frac{dx}{dy}$ are reciprocals, so $ \frac{dy}{dx} - \frac{dx}{dy} = \frac{y}{x} - \frac{x}{y} $ $ \implies\qquad \frac{dy}{dx}=\frac{y}{x} \qquad\text{or}\qquad \frac{dy}{dx}=-\frac{x}{y} $ $ \implies\qquad \int\frac{dy}{y}=\int\frac{dx}{x} \qquad\text{or}\qquad \int\;y\;dy=-\int\;x\;dx $ $ \implies\qquad \ln|y|=\ln|x|+c_1 \qquad\text{or}\qquad \frac{y^2}{2}=\frac{x^2}{2}+c_2 $ $ \implies\qquad y=\left(\pm e^{c_1}\right)x \qquad\text{or}\qquad y^2=x^2+2c_2 $ $ \implies\qquad y=ax \qquad\text{or}\qquad y^2=x^2+b $