$$\frac{dc}{dt}=a-bc$$ where $a,b$ are constants
a. Find the solution if at $t=0$, $c=k$
b.Prove that for all k $\lim_{t \to \infty}c(t)=\frac{a}{b}$
Finding the homogeneous solution
$$c_{h}:\frac{dc}{dt}+bc=0$$
$$c_{h}:c=\frac{e^t}{k}$$ where $k=\frac{1}{e^b}$
Variation of Parameters:
$$c_{p}=\frac{e^t}{k(t)}$$
$$c'_{p}=\frac{e^tk(t)-k'(x)e^t}{[k(t)]^2}$$
substitution in the original ODE:
$$\frac{e^tk(t)-k'(x)e^t}{[k(t)]^2}=a-\frac{be^t}{k(t)}$$
$$\frac{e^tk(t)-k'(x)e^t+be^tk(t)}{[k(t)]^2}=a$$
How should I continue from here?
Hint. You already established that the homogeneous solution is $c_h(t)=Ce^{-bt}$ (not $e^{t}\cdot e^b$) with $C\in\mathbb{R}$. What you need now is a particular solution $c_p(t)$.
Note that this linear ODE has a solution which is identically constant. Are you able to find it?
P.S. If you prefer to use the method of Variation of Parameters then let $c_p(t)=k(t)e^{-bt}$. Then $$a=c_p'(t)+bc_p(t)=k'(t)e^{-bt}-bk(t)e^{-bt}+bk(t)e^{-bt}=k'(t)e^{-bt}\Rightarrow k'(t)=ae^{bt}$$ Hence $k(t)=\frac{a}{b}e^{bt}$ which implies that $c_p(t)=\frac{a}{b}$.