# Thread: Method of characteristics and general solution for partial differential equations

1. ## Method of characteristics and general solution for partial differential equations

Find the general solution of the linear equation , x2ux + y2uy = (x+y)u .

I found out the first constant C1 = x-1 - y-1 ,

using dx/x2 = dy/y2 = du/(x+y)u. (1)

According to my book, the second constant C2 = (x -y)/u , using (dx -dy)/ (x2-y2) = du/(x+y)u.

I don't understand how did they derive (dx -dy)/(x2 - y2) from equation (1)?

2. ## Re: Method of characteristics and general solution for partial differential equations

(x2 - y2) = (x-y)* (x+y) and then (dx -dy)/(x-y)(x+y)=du/(x+y)u.

simplify (x+y) in the left side with (x+y) in the right side

Then (dx -dy)/(x-y) = du/u.

=> d(x-y)/(x-y) = du/u.

Integrating => ln(x-y) = ln U + ln C
=> ln[(x-y)/u] = ln C

=>(x-y)/u = c