Basic question about derivative and integrals

Question

I'm struggling with the very basic meaning of derivative.

Let's say that I have the following derivative: $$ \frac{df(t)}{dt} = -kf(t) $$ If I get the idea, this means that at each time step t the value of $f(t)$ will decrease the constant $k$ times the value of $f(t)$. If $f(0)$ is possitive, and for all positive values of $t$, given that every time-step $f(t)$ will be smaller, the decreasing rate of it will fall less every time-step, so at the limit, there is no change when $f(t)$ cross the 0.

If I try to find the original function that maps from $t$ to $f(t)$ I should integrate that? right? My problem is that I know how to integrate $\frac{df(t)}{dt} = -kt$ but not $\frac{df(t)}{dt} = -kf(t)$, which is an absolut different equation (that quickly goes below 0).

Answer 1

A trivial solution to $f'(t) = -kf(t)$ is $f \equiv 0$. Other solutions can be found as follows, choosing an inital data $t=t_0$: $$f'(t) = -kf(t) \Rightarrow \frac{f'(t)}{f(t)} = -k \Rightarrow \int_{t_0}^t \frac{f'(s)}{f(s)}\,ds = -k(t-t_0),$$ and this gives $$\log \left\lvert \frac{f(t)}{f(t_0)}\right\rvert = -k(t-t_0) \Rightarrow f(t)=\pm \lvert f(t_0)\rvert e^{-k(t-t_0)}.$$

Answer 2

$$f'(t)=-k f(t)$$ $$\frac{f'(t)}{f(t)}=-k$$ $$\int \frac{f'(t)}{f(t)}dt=-k\int dt$$ $$\int \frac{f'(t)}{f(t)}dt=-k(t+c_1)$$ For the left, let $u=f(t)$. Therefore $du=f'(t)dt$. Then substitute: $$\int \frac{1}{u}du=-k(t+c_1)$$ $$\ln |u|=-k(t+c_1)$$ $$\ln |f(t)|=-k(t+c_1)$$ $$f(t)=e^{-k(t+c_1)}$$ In general, the same process can be applied to any equation of the form $$f'(t)=p(t)\cdot f(t)$$ To yield the solution $$f(t)=e^{\int p(t)dt}$$ These are called separable differential equations.