What is the explanation of taking $dx$ to right hand side if $\frac{d}{dx}$ is an operator?

Question

Say we have $\frac{dy}{dx}=e^x$, then it is solved by taking $dx$ to right hand side of the equation and integrated.

My question is if $\frac{d}{dx}$ is an operator, then how do can take $dx$ to right? What is the explanation and reason behind this mathematically and intuitively?

Answer 1

In general, for a function of $n$ variables $y(x_1,\dots,x_n)$, its differential is

$$dy = \frac{\partial y}{\partial x_1}dx_1+\dots+\frac{\partial y}{\partial x_n}dx_n$$

In case $n=1$, we obtain (saying just $x$ instead of $x_1$):

$$dy = \frac{\partial y}{\partial x}dx$$

Which is commonly abused like this:

$$\frac{dy}{dx} = \frac{\partial y}{\partial x}$$

Note that in general dividing differentials makes no sense, much like dividing vectors in $n$-dimensional vector space. But your somewhat can divide them in 1-dimensional space, and that's exactly what is happening here with the differentials.

Answer 2

It is easier to understand if you look at taking $dx$ to the right side and integrating as one operation. In fact, integration does not make any sense if no differential is present.

This way you are just applying an operator to both sides.

$$\int \frac{dy}{dx}dx=y+C$$

$$\int e^xdx=e^x+C$$

Answer 3

Don't think too far. This is just a brainless, but easy to memorize step in a solution recipe. Of course it can be proven (and most probably you have seen a proof) that the end result obtained in this way is correct.

Check the answers to the question quoted in Hans Lundmark's comment.