$\frac{dy}{dx}=y^2-9$
$\frac{dy}{dx}=y^2-4$
$\frac{dy}{dx}=y-y^2$
$\frac{dy}{dx}=y\left(y+2\right)$
$\frac{dy}{dx}=\left(1+e^{-x}\right)\left(y^2-1\right)$
$\frac{dy}{dx}=\frac{x}{2x-y}$
$\frac{dy}{dx}=\frac{x+3y}{x-y}$
Given a function f(x) and the interval [a,b], the definite integral is equal to the area that is bounded by the graph of f(x), the x-axis and the vertical lines x=a and x=b
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