Mathematics

Mathematics

Problem

For the family of differential equations dydx=y3−py\displaystyle \frac{dy}{dx}=y^{3}-p y, the real parameter pp can vary. An equilibrium value y=y0y=y_0 is called stable if ddy(y3−py)∣y=y0<0\dfrac{d}{dy}\bigl(y^{3}-py\bigr)\big\vert_{y=y_0}<0. Which value of pp listed below results in a slope field with exactly one stable equilibrium solution? a) p=−4p=-4 b) p=3p=3 c) p=0p=0 d) p=−1p=-1

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