Solving a 3D PDE with given initial conditions and characteristics"

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The discussion focuses on solving the partial differential equation (PDE) defined by the equation x∂u/∂x + xy∂u/∂y + z∂u/∂z = 0 with the initial condition u(1, y, z) = yz. The characteristics of this PDE are derived as dx/dt = x, dy/dt = xy, and dz/dt = z. The solution involves expressing the variables in terms of a parameter t, leading to the relationships x = t, y = BeAet, and z = Cet/A. The final solution is an arbitrary function of the constants derived from the characteristics.

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b]1. Homework Statement [/b]
Find the characteristics, and then the solution, of the partial differential equation

x\frac{\partial u}{\partial x}+xy\frac{\partial u}{\partial y}+z\frac{\partial u}{\partial z}=0


given that u(1, y, z)=yz


Homework Equations





The Attempt at a Solution



Found this question on an old exam and am not quite sure what to do.
Initialy i tried to take the z derivative to the other side and still take dy/dy=y to be the characteristics, leaving me to solve

x\frac{\partial u}{\partial x}+z\frac{\partial u}{\partial z}=0.

but i think that's wrong.
Could someone please point me in the right direction.
thanks
[
 
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Are you clear on what "characteristics" are for a partial differential equation?

The characteristics for this equation are given by
dx/dt= x, dy/dt= xy, and dz/dt= z. From the first and third, x= Aet and z= Cet. The second equation is dy/dt= Aety which separates as dy/y= Aetdt and integrates as ln(y)= Aet+ B' or y= BeAet= Bex. Also, z= Cet and, from x= Aet, et= x/A so z= Cex/A. That is, taking x= t as parameter, the charateristics are given by x= t, y= Bet, and z= Cet/A.
 
Ok thanks. So would the solution of the PDE still be just an arbitary function of the constant combination of variables? how do i combine all the characteristics to form a single solution?
 

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