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

Solve: $\frac{dy}{dx}-\frac{dx}{dy}=\frac{y}{x}-\frac{x}{y}$
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raefx88y
If u and v are real numbers $\ne 0$ then
$v-\frac{1}{v}=u-\frac{1}{u}$
holds if either .

Elaine Verrett
$xy{\left(\frac{dy}{dx}\right)}^{2}-\frac{dy}{dx}{y}^{2}+\frac{dy}{dx}{x}^{2}-xy=0$
$y\left(\frac{dy}{dx}\right)\left(x\left(\frac{dy}{dx}\right)-y\right)+x\left(x\left(\frac{dy}{dx}\right)-y\right)=0$
$\left(x\left(\frac{dy}{dx}\right)-y\right)\left(y\left(\frac{dy}{dx}\right)+x\right)=0⇒$
$\frac{dy}{dx}=\frac{-x}{y}⇒\int ydy=-\int xdx$
and $\frac{dy}{dx}=\frac{y}{x}⇒\int \frac{dy}{y}=\int \frac{dx}{x}$

RizerMix
$v-\frac{1}{v}=u-\frac{1}{u}$ ) Now are reciprocals, so $\frac{dy}{dx}-\frac{dx}{dy}=\frac{y}{x}-\frac{x}{y}$