$\frac{dy}{dx}\left(3x^2+4y=\log\left(xy\right)\right)$
$1-12.x>13.x+18$
$\frac{du}{dt}+\frac{u}{t}=-\frac{1}{t^2}$
$3x^2\:\frac{dy}{dx}=2x^2+y^2$
$a^2-121=0$
$12.15-24\:\frac{1}{2}$
$2st^4-8st^2-90s$
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