$y'=\frac{x-1}{y^2\:},\:y\left(0\right)=2$
$y\cdot ln\left(x\right)\frac{dy}{dx}=\left(\frac{\left(y+1\right)^2}{x^2}\right)$
$x^6y^3-1$
$x+11=-26$
$10\cdot45$
$x^3+y^3=x$
$\left(d^2-4d+4\right)y=e^x$
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