♻️Independent Variable Substitution

Quick Notes

We can redefine the differential expression by assigning the dependent variable x to a new variable, u = u(x)

$$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = \frac{dy}{du} / \frac{dx}{du}$$

This last expression is a handy trick, which allows us to find two differential equations -- the first differentiating a function y(u) and the second differentiating a function x(u) -- and equate them with the original differential expression.

$$\int_x \frac{dy}{dx} dx = \int_{u(x)} \frac{dy}{du}du$$

Sometimes, this makes the original equation easier to solve.

Detailed Notes

It's common in simplifying differential equations to replace the differential marking dx with

$\frac{dx}{d\theta} d \theta$

I call dx a "marking" instead of a "variable" or "term" because, when looking at a differential equation, dx does not denote a well-defined number. dx simply marks which variable is independent during the differential operation.

The expression of the derivative

appears to divide two numbers, contradicting our insistence that markings like dy and dx are not well-defined. Before the formalization of calculus, and in some modern extensions of classical real number theory into hyperreal domains, we called these markings "infintessimals".

Redefinition using Limits

Redefine the expression

$$\frac{dy}{dx} = \frac{d}{dx}[y] = \lim_{ \Delta x \rarr 0} \frac{\Delta y}{\Delta x}$$

as the limit of every change in y divided by the corresponding change in x, taken via successively smaller changes in x. Since changes are determined by pairs of real numbers, the division on the right hand side is a normal operation on real numbers and well-defined.

Notably, the division occurs just once with each pair of changes; the limit gives us a boundary which we equate to the result of the differential operator.

The division occurring inside the limit operation allows us to perform a pairwise replacement of xs

$$\frac{\Delta y}{\Delta x} \frac{ \Delta z }{\Delta z} = \frac{ \Delta y } {\Delta z }\frac{ \Delta z } {\Delta x} = \frac {\Delta y }{\Delta z} / \frac{\Delta x}{\Delta z}$$

We can do this, because the delta variables are real numbers and the division is done with real number division, so all the normal algebraic rules apply cleanly. And we do this before taking the limit, now with regards to delta z sized steps increasingly close to zero.

$$\lim_{\Delta z \rightarrow 0} \frac{\Delta y}{\Delta z} / \frac{\Delta x}{\Delta z} = \frac{dy}{dz} / \frac{dx}{dz}$$

and equivalently

$$\lim_{\Delta x \rightarrow 0, \Delta z \rightarrow 0}\frac{\Delta y}{\Delta x}\frac{\Delta z}{\Delta z} = \lim_{\Delta x \rightarrow 0}\frac{\Delta y}{\Delta x} \lim_{\Delta z \rightarrow 0}\frac{\Delta z}{\Delta z}$$

As the second limit on the right evaluates to one, we see that the resulting differentials are equivalent, by definition.

In this last expression, the division performed is not a division of real numbers. It's an artifact of this notation. At most, we perform division of specific outputs of the differential evaluated at a point; but we don't even need to say that much about what happens here -- it's "only" notation.

Redefinition using the Chain Rule

The Fundamental Theorem of Calculus (F.O.C.) is a powerful tool that correlates integration operations with differentiation operations. We can use F.O.C. to write

$$\int y’(x) dx = y(x) + C$$

Once again, the use of dx is notational. The dx tells us we have a function y differentiated with respect to x.

On the right hand side, we can substitute for x directly, allowing us to write

$$y(u) + C\text{, where } u = u(x)$$

If we took just this expression and differentiated it with respect to u, we'd need to apply the chain rule with respect to x

$$\frac{d}{dx}[ y(u(x)) + C] = y’(u(x))u’(x)$$

This is precisely the expression we'll need to replace x with u in the left hand side of the expression provided by the F.O.C.

$$\int y’(u(x))u’(x) dx$$

and we can continue to appeal to F.O.C. and equate the two sides

$$\int y’(u(x))u’(x) dx = y(u(x)) + C$$

Now, understanding the notation that y' marks the same thing that dy/dx marks, we can write the above as

$$\int \frac{dy}{du}\frac{du}{dx}dx = y(u(x)) + C$$

Lastly, observe that when we integrate, we're still implicitly integrating with respect to boundaries on x. If we change these boundaries to boundaries on u, resulting from a transformation of each x to some u, then we no longer need to include that term. In other words

$$\int_x \frac{dy}{dx} dx = \int_{u = u(x)} \frac{dy}{du}du$$

and we see right away that we've "canceled" the dx markings. Yet we've haven't actually done any division here; we've merely rewritten each occurrence of x with an occurrence of u, shifting our integration bounds according to the new mapping.