$\frac{dx}{dy}\left(y^4=\left(x+y\right)^3\right)$
$\left(4t^2u^6-4r^3s\right)^2$
$2x^2+13x-24=0$
$12\cdot5\frac{1}{2}$
$y'-y=1+te^t;y\left(0\right)=0$
$m^9-27$
$\frac{dx}{dy}\:+\:5x\:=10$
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