Too basic Differential equation question

Question

Assume that $$\frac{d s_t}{d t} =\frac{1}{q_0}$$ where $q_0>0$ is constant.

I would like to find $s_t$

What I found is $s_t=s_0+ \frac{1}{q_0}t$ for any constant $s_0>0$.

Is this true?

Answer 1

Starting from

$$\frac{d s_t}{d t} =\frac{1}{q_0}$$

you correctly deduced that

$$d s_t=\frac{1}{q_0}dt \implies s_t=\frac{1}{q_0}t+ s_0$$

where $s_0$ is an arbitrary constant (you don't need the additional restriction that $s_0 > 0$ as it is an arbitrary constant of integration). To check your work, all you need to do differentiate both sides with respect to $t$

$$\frac{d}{dt}(s_t)=\frac{d}{dt}\left(\frac{1}{q_0}t+ s_0\right)=\frac{1}{q_0}$$