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eaglefield | 3 months 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.
d4rkn0d3z|3 months 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
tptacek|3 months ago
I find this whole topic very gratifying because Leibniz notation seems very arbitrary and I'm glad it's not just me. :)
leephillips|3 months ago