$y'\:-\:2y\:=\:-2$
$\frac{x^3+0x^2-2x+4}{3x^2-1}$
$x^2y\:\:-\:5x^2y\:\:$
$y\left(x^2+1\right)^3\frac{dy}{dx}=x\left(1-y^2\right)^3$
$\frac{dy}{dx}=\frac{xy}{y+1}$
$-2\left(7\right)\:-\:3\left(-3\right)$
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