- #1

Beer-monster

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[tex]\frac{\partial{\psi}}{\partial{x}} = k(x+y)[/tex]

Anyone know any good tutorials or webpages for these sorts of equation? I'm a bit rusty with them

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- Thread starter Beer-monster
- Start date

- #1

Beer-monster

- 296

- 0

[tex]\frac{\partial{\psi}}{\partial{x}} = k(x+y)[/tex]

Anyone know any good tutorials or webpages for these sorts of equation? I'm a bit rusty with them

- #2

mathmike

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- #3

HallsofIvy

Science Advisor

Homework Helper

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Since x and y are independent, all you can do is integrate, with respect to x, treating y as a constant:

[tex]\psi(x,y)= \frac{k}{2}x^2+ yx+ f(y)[/tex]

Since the partial derivative wrt x is taken treating y as a constant, f(y) could be any function of y alone- its derivative will be 0.

- #4

Beer-monster

- 296

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Although isn't the second term kyx as it too is multiplied by k? Or am I being incredibly dense, it does happen a lot

- #5

Beer-monster

- 296

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Thanks for the help anyway

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