The question is linearize the equation $\frac{du}{dt}=ru\left(1-\frac{u}{C}\right)$ about the solution $u=v^*=C$. So I let $f(u)= ru\left(1-\frac{u}{C}\right)$. I tried linearizing this. $$f'(u)= r-\frac{2u}{C}$$
$$\begin{align} L &=& f(u) + f'(u)(u-u_0)\\ &=&f(C) + f'(C)(u-C)\\ &=&0 + (r-2)(u-c) \end{align}$$
But the answer is $-rCv$. What did I do wrong?
I don't know what the answer means. Actually linearize means, for stationary solution $C$ we know that
$$\frac{dC}{dt}=f(C)=0$$ We suppose there exist a solution $u=C+\epsilon\phi,\ \ \ \phi=u-C,\epsilon<<1$ then $$\frac{d(C+\epsilon \phi)}{dt}=f(C+\epsilon\phi)=f(C)+f'(C)\epsilon\phi+O(\epsilon^2)$$ then $$\epsilon\frac{d\phi}{dt}\sim \epsilon f'(C)\phi=\epsilon(-r)\phi$$ From this we get linearized ODE: $$\frac{d\phi}{dt}+r\phi=0,(\phi=u-C)$$
and the formal way is:
Let $u=C+\epsilon\phi$ ,then $$\epsilon\frac{d\phi}{dt}=r(C+\epsilon\phi)(1-\frac{C+\epsilon\phi}{C})= -r\phi+O(\epsilon^2)$$ So we can get same linearized equation $$\frac{d\phi}{dt}+r\phi=0$$