# Partial Differential Equation

I borrowed this book form the library to get a heads up on what i would be doing in PDEs and it had this equation

xUx+yUy=nu

and its solution (according to the back) is: u(x,y)=xnf(y/x)

I did figure out the f(y/x) part:

dx/x=dy/y
ln(x)+c=ln(y)
cx+y
c=y/x

how did the xn come into play

My attempt at a solution yielded a yn. I don't believe that is wrong, but I am wondering why they are multiplied together instead of added.

MathematicalPhysicist
Gold Member
You ofcourse can check your or the book's answer by plugging it back to the equation.
I think that there is no easy way here, the mathematician who find these equations knew already what solution he was seeking and just played with its derivatives to get to this equation.
I don't have time to check your work, but have you plugged your answers back to the eqaution to check that it's valid?

Hootenanny
Staff Emeritus
Gold Member
I borrowed this book form the library to get a heads up on what i would be doing in PDEs and it had this equation

xUx+yUy=nu

and its solution (according to the back) is: u(x,y)=xnf(y/x)

I did figure out the f(y/x) part:

dx/x=dy/y
ln(x)+c=ln(y)
cx+y
c=y/x

how did the xn come into play
The form of the question suggests that you should use the method of characteristics (see http://www.stanford.edu/class/math220a/handouts/firstorder.pdf" [Broken] for more information). Note that the method you have used above to find the characteristics is not generally correct and it is just by 'chance' that you have arrived at the correct characteristic equation (c=y/x).

The link I provided above should give you everything you need, if not, I suggest reading the preceeding chapter of your book.

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