$0=e^x\left(1+y^2\right)dx+\frac{1}{x}dy\:$
$\left(2xy+3y-2x-3\right)dx=\left(3xy-3y-4x+4\right)dx$
$4x-2y^2+2y^2+2-3x^3$
$\frac{dy}{dx}=-\frac{y}{x^2-4}$
$9x^2+x=0$
$( 5 x ^ { \frac { 1 } { 2 } } ) ^ { 2 }$
$4\frac{x^3-3x^2+2x+4}{x-7}$
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