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Consider the differential equation $dy/dt=y^8+3y^6-y^2-1$. Sketch the phase line and classify the equilibrium points.

Since when $y=0$, the derivative is negative and when $y>1$ the derivative is positive and when $y<1$ the derivative is positive. I know due to the intermediate value theorem, there lies an equilibrium point in between $-1$ and $0$, and, $0$ and $1$. How would I go about the algebra to solve $0=y^8+3y^6-y^2-1$? or is this more qualitative solution enough?

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  • $\begingroup$ Hint: Take $z=y^2$. $\endgroup$ – Michael Galuza Jun 27 '15 at 19:33

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