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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. – Did Feb 8 '12 at 16:00
up vote 2 down vote accepted



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

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


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

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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. – Inquest Feb 8 '12 at 17:07

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

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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 monotone – Norbert Feb 8 '12 at 16:55

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