partial differential equations

**tom007** · January 12th 2010, 02:54 AM

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

**Danny** · January 12th 2010, 06:35 AM

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?

**tom007** · January 12th 2010, 06:55 AM

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!

**Danny** · January 12th 2010, 07:09 AM

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.

**tom007** · January 12th 2010, 07:55 AM

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

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

**tom007** · January 12th 2010, 08:07 AM

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

is there any formula?

**Danny** · January 12th 2010, 08:39 AM

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.

**tom007** · January 13th 2010, 10:31 AM

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

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