$$ \frac{dy}{dt} = Ky$$
$$ \frac{1}{y}\frac{dy}{dt} = K$$
...then my teacher did something illogical he decomposed the Differential operator $\frac{dy}{dt} $ into $\frac{1}{dt} *dy$ and used this "form" with the laws of arithmetic to "cancel" the $\frac{1}{dt}$ by multiplying both sides by $dt$
$$ dt*\frac{dy}{dt}\frac{1}{y} = K* dt$$
$$ dy =K *y *dt$$
Why is this possible?
Is it related to the chain rule?
Answer
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 $\text{1D}$).
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