find general solution of:

u_x + x u_y + xy u_z = xyzu

i've tried with this

dx=dy/(x)=dz/(xy)=du/(xyzu) ...but didn't get the correct solution

i appreciate any help...thanks

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- Jan 12th 2010, 02:54 AMtom007partial differential equations
find general solution of:

u_x + x u_y + xy u_z = xyzu

i've tried with this

dx=dy/(x)=dz/(xy)=du/(xyzu) ...but didn't get the correct solution

i appreciate any help...thanks - Jan 12th 2010, 06:35 AMJester
- Jan 12th 2010, 06:55 AMtom007
i don't know how to solve this..i thought F should have 3 components.. i got many different results...

could u pls write the steps, if it's not the problem..

thank you very much Danny! - Jan 12th 2010, 07:09 AMJester
Picking it up from your first post - the characteristic equations are:

$\displaystyle

\frac{dx}{1} = \frac{dy}{x} = \frac{dz}{xy} = \frac{du}{xyz}

$

We'll pick in pairs

1) $\displaystyle

\frac{dx}{1} = \frac{dy}{x} \;\; \text{so}\;\; x^2 - 2y = c_1

$

2) $\displaystyle

\frac{dy}{x} = \frac{dz}{xy} \;\; \text{so}\;\;y^2 - 2z = c_2$

3) $\displaystyle

\frac{dz}{xy} = \frac{du}{xyzu}\;\; \text{so}\;\; ue^{-z^2/2} = c_3

$

The solution $\displaystyle c_3 = F(c_1,c_2)$ gives $\displaystyle ue^{-z^2/2} = F(x^2 - 2y , y^2 - 2z)$ leading to my solution. - Jan 12th 2010, 07:55 AMtom007
wow..that's so simple, i was complicating ..better not to say!!

thanks Danny!! (Sun)

do u maybe know other 2 exercise i've posted? - Jan 12th 2010, 08:07 AMtom007
ey Danny...how could i check this result?

is there any formula? - Jan 12th 2010, 08:39 AMJester
- Jan 13th 2010, 10:31 AMtom007
i know that...but how should i substitute F function?