1. ## partial 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

2. Originally Posted by tom007
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
Your method is correct! I got

$u = e^{z^2/2}F(x^2-2y,y^2-2z)$.

What did you get?

3. 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!

4. Originally Posted by tom007
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!
Picking it up from your first post - the characteristic equations are:

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

We'll pick in pairs

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

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

3) $
\frac{dz}{xy} = \frac{du}{xyzu}\;\; \text{so}\;\; ue^{-z^2/2} = c_3
$

The solution $c_3 = F(c_1,c_2)$ gives $ue^{-z^2/2} = F(x^2 - 2y , y^2 - 2z)$ leading to my solution.

5. wow..that's so simple, i was complicating ..better not to say!!
thanks Danny!!

do u maybe know other 2 exercise i've posted?

6. ey Danny...how could i check this result?

is there any formula?

7. Originally Posted by tom007
ey Danny...how could i check this result?

is there any formula?
The easiest way to check is to substitute the solution into the PDE itself and check that it's satisfied.

8. i know that...but how should i substitute F function?