67. A material point moves in such a way that its speed is proportional to the distance traveled. At the initial moment, the point was 1 m away from the origin, and after 2 s - at a distance of $e$ m. Find the law of motion of the material point (see problem 21).
Solution. Let the speed of the material point be denoted by $v$. As is known, speed is the derivative of the path with respect to time, i.e., $v=\frac{d s}{d t}$. According to the problem, the speed is proportional to the distance traveled, i.e., $v=\frac{d s}{d t}=k s$, where $k$ is the proportionality constant.

Thus, the differential equation of this problem is $\frac{d s}{d t}=k s$. By separating the variables, we get $\frac{d s}{s}=k d t$. The general solution of this equation is $\ln s=k t+C$.

We need to find a particular solution, i.e., from all possible motions according to this law, we need to find the one where the point is initially 1 meter away from the origin. Generally speaking, the distance of the material point from the origin at the initial moment could have been any value, not necessarily 1 meter. This choice is analogous to selecting one curve from a family of curves in geometric problems.

Using the initial conditions $t=0, s=1$ to determine $C$, we have $\ln 1=k \cdot 0+C$, from which $C=0$. Therefore, the law of motion can be written as $\ln s=k t$.

The value of $k$ remains unknown. It can be determined from the condition that after 2 seconds, the point is at a distance of $e$ (where $e$ is the base of the natural logarithm). Thus, $\ln e=k \cdot 2$ or $2 k=1$, from which $k=1 / 2$.

Therefore, the law of motion of the material point is $\ln s=t / 2$, i.e., $s=e^{t / 2}$.