$xy'=-4y+xe^{\frac{x^5}{5}}$
$81b^2-36a^2$
$20y^2-y-8$
$y'+y=e^{3x}$
$\frac{1\:-\:2.1\:z^{-1}}{\left(1\:+\:0.5\:z^{-1}\right)\left(1\:-\:0.8\:z^{-1}\right)}$
$\left(x^3+y^3\right)dx+3xy^2\:dy=0\:\:\:\:$
$2^7\cdot2^{-2}$
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