# PDE

• Jan 26th 2010, 10:32 PM
harveyo
PDE
$\displaystyle u(x,y) = B (3x - 2y)$

B is an arbitary function

find partial differentiation of u. Not including B
• Jan 27th 2010, 03:51 AM
mr fantastic
Quote:

Originally Posted by harveyo
$\displaystyle u(x,y) = B (3x - 2y)$

B is an arbitary function

find partial differentiation of u. Not including B

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

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

At this level you should be able to combine these two equations in such a way as to eliminate B.
• Jan 27th 2010, 04:14 AM
harveyo

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

• Jan 27th 2010, 04:18 AM
harveyo
Quote:

Originally Posted by harveyo

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

i have used the wrong formula here?

i should solve the above simultaneously?
• Jan 27th 2010, 04:45 AM
dedust
Quote:

Originally Posted by mr fantastic
$\displaystyle \frac{\partial u}{\partial x} = 3B$ .... (1)

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

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

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

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

hence

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