Consider $y'=y(y-50)(y-100)$
(a) Let $y_e$ be the stable equilibrium solution and let $u=y-y_e$, rewrite the equation as a differential equation for $u$.
$y_e=50, u=y-50, u'=(u-50)(u)(u+50)$
(b) What is the linearized equation near $u=0$, and what is its general solution?
Near $u=0$: $u'=-2500u$. General solution: $u=Ce^{-2500t}$
(c) If $y(0)=40$, estimate $y(1)$ and $y(2)$ be linearizing near the equilibrium solution.
I am stuck on part (c) as I don't understand how we can linearize near $0$ if the equilibrium solution is at $50$. Did I make a mistake in part (a) or (b)?
Yes $u(t) = u_0 e^{-2500 t}$ is a good approximation for when $u_0$ small.
So treat $u_0 = y_0 - 50 = -10$. This isn't small so the linearized solution might be rather bad. We can plug it anyway and get
$$ u(1)=-10 e^{-2500}\\ y(1)=50-10 e^{-2500}\\ u(2)=-10 e^{-5000}\\ y(2)=50-10 e^{-5000}\\ $$
40 is closer to 50 than 0 so there is hope for this approximate solution, but you should check.