$\sqrt{4x}$
$4y^{'\:}+5y=3x+4e^x$
$\left(y^2+ty^2\right)y'+t^2-yt^2=0$
$7-2-\left(6-5\right)$
$624\cdot\frac{1}{4}$
$x\:+\:2y\:-\:3x\:+\:3y\:-\:y\:+\:x\:+\:u\:-\:v\:\:$
$1-\cos\left(0.45\right)^2$
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