Results 1 to 5 of 5

Math Help - PDE

  1. #1
    Junior Member
    Joined
    Dec 2009
    Posts
    31

    PDE

     <br />
u(x,y) = B (3x - 2y)<br />

    B is an arbitary function

    find partial differentiation of u. Not including B
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by harveyo View Post
     <br />
u(x,y) = B (3x - 2y)<br />

    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2009
    Posts
    31
    i had

     <br />
dy/dx = \partial x / \partial y = 3B (3x-2y)/ 2B (3x-2y) = -3/2<br />

    was this right?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Dec 2009
    Posts
    31
    Quote Originally Posted by harveyo View Post
    i had

     <br />
dy/dx = \partial x / \partial y = 3B (3x-2y)/ 2B (3x-2y) = -3/2<br />

    was this right?
    i have used the wrong formula here?

    i should solve the above simultaneously?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Nov 2009
    Posts
    263
    Quote Originally Posted by mr fantastic View Post
    \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
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum