3y′+4y=23y'+4y=23y′+4y=2
∫1y3y2−9dy\int\frac{1}{y^3\sqrt{y^2-9}}dy∫y3y2−91dy
y′+y2+y=0y'+y^2+y=0y′+y2+y=0
−7⋅x<7-7\cdot x<7−7⋅x<7
dydx=2xy2+4y\frac{dy}{dx}=2xy^2+4ydxdy=2xy2+4y
4x2−16+644x^2-16+644x2−16+64
(2m2+3)5\left(2m^2+3\right)^5(2m2+3)5
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