What is the logic behind decomposing a derivative operator symbol. In the population growth equation? [duplicate]

Solution 1:

The proper way to think about this is as follows:

$$\frac{dy}{dt} = Ky \Longleftrightarrow \frac{1}{y}\frac{dy}{dt} = K.$$

Thus

$$\frac{1}{y}\frac{dy}{dt} - K = 0.$$

However

$$\frac{1}{y}\frac{dy}{dt} = \frac{d}{dt}\log|y(t)|.$$

Thus if we integrate our equation with respect to $t$, what we find is that

$$ \int \left(\frac{d}{dt}\log|y(t)|-K\right)\,dt = C.$$

Using the fundamental theorem of calculus we get

$$ \log|y(t)| - Kt = C$$

which can be easily solved. As you see, we used the relation $\frac{1}{y}\frac{dy}{dt} = \frac{d}{dt}\log|y(t)|$ which relies on chain rule (since $y$ is a function of $t$). The symbolic manipulation your professor did (which many people do) is really just a repackaging of the chain rule. It's not rigorous since the notation $\frac{dy}{dt}$ is not meant to represent a fraction - it is merely notation adapted from $\frac{\Delta y}{\Delta x}$ representing slopes of secant lines. It is convenient notation as such calculations show (you can kind of think of it as a fraction without too many issues in $\text{1D}$).

Solution 2:

This kind of manipulation can be made rigorous, but even without rigor the two equations $$ \frac{dy}{dt} = Ky $$ and $$ dy = K y \, dt $$ can be seen to be consistent.

The first says the rate of change of $y$ (with respect to $t$) is proportional to the value of $y$. The second says that a small change $dt$ in $t$ causes the small change $Ky\, dt$ in $y$.

(I think there's an error in your version of the second equation. If you fix it I'll edit this answer.)