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Consider the following differential equation, $$\frac{du}{dt}=w+u-u^3.$$

Suppose that the parameter changes slowly in time depending on the value of $u$. That is, consider the system of equations $$\frac{du}{dt}=w+u-u^3.$$ $$\frac{dw}{dt}=-\epsilon u,$$ where $\epsilon>0$ is very small. Using your bifurcation diagram from part a, sketch what a solution looks like for small $\epsilon$.

So I have a graph of my bifurcation diagram of a time-independent parameter right here enter image description here

and now I am tasked with considering this time-dependent parameter. I'm very new to this. Previously I've learned how to work with bifurcations through the MATCONT program in MATLAB, but I don't think there's a way I can set my parameter as a function of $t$. I need help on how to work with this problem.

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After a bit of thinking, I came up with a solution.

when we suppose that the parameter changes slowly in time we look back to our equations $$-w=u-u^3$$ and $$\frac{d}{du}(-w)=\frac{d}{du}(u-u^3)$$ . In the second equation the derivative of $w$ with respect to time is no longer 0, but defined as $-\epsilon u$ so our second equation becomes $\epsilon u=1-3u^2$. Setting this equation equal to zero gives $3u^2+\epsilon u-1=0$, using the quadratic formula to solve for $u$ gives $$u=\frac{-\epsilon\pm\sqrt{\epsilon^2-4(3)(-1)}}{2(3)}=\frac{-\epsilon\pm\sqrt{\epsilon^2+12}}{6}$$ Because we consider a small $\epsilon$ we approximate $u$ as $u\approx\pm\frac{\sqrt{12}}{6}$. Plugging these back into the first equation gives the bifurcation points $(u^*,w^*)=(\frac{\sqrt{12}}{6},-\frac{2}{3\sqrt{3}}),(-\frac{\sqrt{12}}{6},\frac{2}{3\sqrt{3}})$. Using the plot from a) this can be adapted for our new bifurcation points as followed: bifurplot2

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