- #1
Larry Cosner
- 5
- 2
In the solution to a differential-equation problem -- proof of the existence of an integrating factor -- the following statements are made regarding a general function "u(xy)" [that is, a function of two variable that depends exclusively on the single factor "x*y"]:
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"Let xy = t
Therefore ux = uy = ut"
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I'm having a hard time seeing this as innately true. The first two functions are the partial derivatives of the function u, with respect to the two variables. If one takes a simple function like u(xy) = 3xy, then the first partial derivatives are:
ux = 3y and uy = 3x; and similarly, if one addresses the substituted variable, ut = 3.
And these are not equal.
Now I certainly see, given t = xy, that dt/dx = y and dt/dy = x. This leads to dx/dt = 1/y and dy/dt = 1/x.
Using this, one can show ut = (du/dx)(dx/dt) = (3y)(1/y) = 3. Similarly for the equivalency of ut = (du/dy)(dy/dt).
But I'm not seeing how this translates into ux = uy = ut?
And it is this exact equivalence that is used further down in the proof (which is why I care).
Thanks in advance for helping me see what I'm missing.
------------------------------------
"Let xy = t
Therefore ux = uy = ut"
-------------------------------------
I'm having a hard time seeing this as innately true. The first two functions are the partial derivatives of the function u, with respect to the two variables. If one takes a simple function like u(xy) = 3xy, then the first partial derivatives are:
ux = 3y and uy = 3x; and similarly, if one addresses the substituted variable, ut = 3.
And these are not equal.
Now I certainly see, given t = xy, that dt/dx = y and dt/dy = x. This leads to dx/dt = 1/y and dy/dt = 1/x.
Using this, one can show ut = (du/dx)(dx/dt) = (3y)(1/y) = 3. Similarly for the equivalency of ut = (du/dy)(dy/dt).
But I'm not seeing how this translates into ux = uy = ut?
And it is this exact equivalence that is used further down in the proof (which is why I care).
Thanks in advance for helping me see what I'm missing.