Thread: PDE

**harveyo** · January 26th 2010, 10:32 PM

$ u(x,y) = B (3x - 2y) $

B is an arbitary function

find partial differentiation of u. Not including B

**mr fantastic** · January 27th 2010, 03:51 AM

Originally Posted by harveyo

$ u(x,y) = B (3x - 2y) $

B is an arbitary function

find partial differentiation of u. Not including B

$\frac{\partial u}{\partial x} = 3B$ .... (1)

$\frac{\partial u}{\partial y} = -2B$ .... (2)

At this level you should be able to combine these two equations in such a way as to eliminate B.

**harveyo** · January 27th 2010, 04:14 AM

i had

$ dy/dx = \partial x / \partial y = 3B (3x-2y)/ 2B (3x-2y) = -3/2 $

was this right?

**harveyo** · January 27th 2010, 04:18 AM

Originally Posted by harveyo

i had

$ dy/dx = \partial x / \partial y = 3B (3x-2y)/ 2B (3x-2y) = -3/2 $

was this right?

i have used the wrong formula here?

i should solve the above simultaneously?

**dedust** · January 27th 2010, 04:45 AM

Originally Posted by mr fantastic

$\frac{\partial u}{\partial x} = 3B$ .... (1)

$\frac{\partial u}{\partial y} = -2B$ .... (2)

At this level you should be able to combine these two equations in such a way as to eliminate B.

$2\frac{\partial u}{\partial x} = 6B$ .... (1)

$3\frac{\partial u}{\partial y} = -6B$ .... (2)

hence

$2\frac{\partial u}{\partial x} + 3\frac{\partial u}{\partial y} = 0$

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