well.. no, not exactly. If u = u(x) then du = u'(x) dx holds rigorously, and then you can substitute du/u' = dx in an integral.
tptacek•2mo ago
I'm thinking more along the lines of knocking a '2x' out of an integral from d/dx of like 2x^2.
ajkjk•2mo ago
I don't know what you mean.
krackers•2mo ago
It's funny that most intro calculus courses will make it a point to remind you that "dy/dx" isn't a fraction, then when they get to integration & diffeqs they want you to forget that and start manipulating them as such. I think most intro courses would be better off skipping everything on convergence tests (which feel really arbitrary anyway until you understand more of complex analysis) and instead use that time better explaining differentials (and maybe a peek into differential forms)
eaglefield•2mo ago
The solution of differential equations by separation of variables in physics is also notated in an abusive way. You have some differential equation
dy/dx = g(x)h(y)
You separate the variables by some quick manipulations
dy/h(y) = g(x) dx
And then you have a small step in some coordinate on both sides. So by integrating both sides
\int 1/h(y) dy = \int g(x) dx
you find a solution to your differential equation. Obviously there's a real formal procedure underneath it with also some safeguards. For example you're supposed to check that h(y) doesn't equal 0 at any point. But the happy path in physics is often done without worrying about all that.
tptacek•2mo ago
Yes! Separation of variables the other instance in the back of my mind. I suck at math (I've had basic ODEs for just a couple months now) but are there more examples like this?
I find this whole topic very gratifying because Leibniz notation seems very arbitrary and I'm glad it's not just me. :)
leephillips•2mo ago
More examples? Any undergraduate text in thermodynamics. The entire way the subject is taught depends on treating differentials as numbers. Even in partial derivatives.
d4rkn0d3z•2mo ago
The real formal procedure:
dy/dx = g(x)f(y)
Let h(y) = 1/f(y)
=> dy/dx = g(x)/h(y)
=> h(y) dy/dx = g(x)
Now, we integrate both sides,
int h(y) dy/dx dx = int g(x) dx
But the left hand side is the same as
int h(y) dy by substitution rule of integration.
Therefore,
int h(y) dy = int g(x) dx
Proceed with solving now, no abuse since the substitution rule is provable. QED
tptacek•2mo ago
ajb•2mo ago
ajkjk•2mo ago
tptacek•2mo ago
ajkjk•2mo ago
krackers•2mo ago
eaglefield•2mo ago
dy/dx = g(x)h(y)
You separate the variables by some quick manipulations
dy/h(y) = g(x) dx
And then you have a small step in some coordinate on both sides. So by integrating both sides
\int 1/h(y) dy = \int g(x) dx
you find a solution to your differential equation. Obviously there's a real formal procedure underneath it with also some safeguards. For example you're supposed to check that h(y) doesn't equal 0 at any point. But the happy path in physics is often done without worrying about all that.
tptacek•2mo ago
I find this whole topic very gratifying because Leibniz notation seems very arbitrary and I'm glad it's not just me. :)
leephillips•2mo ago
d4rkn0d3z•2mo ago
dy/dx = g(x)f(y)
Let h(y) = 1/f(y)
=> dy/dx = g(x)/h(y)
=> h(y) dy/dx = g(x)
Now, we integrate both sides,
int h(y) dy/dx dx = int g(x) dx
But the left hand side is the same as
int h(y) dy by substitution rule of integration.
Therefore,
int h(y) dy = int g(x) dx
Proceed with solving now, no abuse since the substitution rule is provable. QED