The differential equation below has an equilibrium at $x=1$:
$\frac{dx}{dt} = (e^{x}-1)(x-1)$
The problem asks to find the linearized equation but in terms of $h$, where $h=x-a$.
$\frac{dh}{dt}=f'(a)h$
The correct answer is $(e-1)h$, but I don't see how to get there.
I've tried first replacing $x$ in the original equation with $h+a$, simplifying, and then working with series. But I don't arrive at the above answer.
What are some approaches to this problem?
You find the derivative of the right hand side at $x=1$: $$ f'(x)=e^x(x-1)+e^x-1 $$ So, $$ f'(1)=e-1 $$ and your linearized equation is the one you have.