What type of differential equation is it and how can it be solved? The equation is:
$$\frac{dA(t)}{dt}=aA(t)\cdot k^{2}(t)+b\cdot k(t)$$
If not the last part, I would use the separation of variables, but how to solve it in this form? 
 A: Reform your given equation to the known ODE form :
$$\frac{dA(t)}{dt}=aA(t)\cdot k^{2}(t)+b\cdot k(t) \Leftrightarrow \frac{dA(t)}{dt}-aA(t)k^2(t)=bk(t)$$
Let $$μ(t)= e^{\int-atk^2(t)dt}=e^{-a\int tk^2(t)dt}$$
and multiply both sides of the reformed ODE by that $μ(t)$.
Substitute : 
$$-atk^2(t)e^{-a\int tk^2(t)dt} = \frac{d}{dt}\big(e^{-a\int tk^2(t)dt}\big)$$
It will then be : 
$$\frac{\frac{dA(t)}{dt}}{e^{a\int tk^2(t)dt}} + \frac{d}{dt}\big(e^{-a\int tk^2(t)dt}\big)A(t)= \frac{bk(t)}{e^{a\int tk^2(t)dt}}$$
Integrate both sides with respect to $t$ :
$$\int \frac{d}{dt}\bigg(\frac{A(t)}{e^{a\int tk^2(t)dt}}\bigg)dt =\int \frac{bk(t)}{e^{a\int tk^2(t)dt}}dt$$
It will then be : 
$$\frac{A(t)}{e^{a\int tk^2(t)dt}}=b\int e^{-a\int tk^2(t)dt}k(t)dt + c$$
Dividing both sides with the initial $μ(t)$ will yield the solution : 
$$A(t) = e^{a\int tk^2(t)dt}\bigg(b\int e^{-a\int tk^2(t)dt}k(t)dt + c\bigg)$$
A: This is first order linear. You solve by multiplying by the integrating factor $$\exp\left(\int ak^2(t)\,dt\right).$$ 
A: It is a linear equation. Indeed, the first step is to solve the equation without the last part. Next is you are assuming that C=C(x) and put it into full equation. Some reduction will appear. An example of the theory in Polish (as I guess from your name) you have here:
http://smurf.mimuw.edu.pl/node/271
