a partial differentiation problem

**lollol** · January 23rd 2010, 01:39 AM

Hi guys, i ve got a problem and i've never seen this type of question before..could u help me out?

thank you

**shawsend** · January 23rd 2010, 04:41 AM

That is in the form of a solution to a first order PDE with constant coefficients:

$a u_x+bu_y+cu=f(x,y)$

in which you let $w=bx-ay$ and $z=y$ and $v(w,z)=u(x,y)$ then express the PDE in u in terms of v. If I then solve the equation:

$2u_x+3u_y=0$

using this method, I get $3v_z=0$ or $v(w,z)=C(w)$ where $C(w)$ is an arbitrary function of $w=3x-2y$ that is, the function $u(x,y)=C(3x-2y)$ solves my PDE like $u(x,y)=3x-2y$ , $u(x,y)=\sin(3x-2y)$ , etc.

My favorite Basic PDE book is "Basic Partial Differential Equations" by Bleecker and Csordas which goes over this subject nicely I think. It's an easy read and I recommend it if you're studying the subject.

**lollol** · January 23rd 2010, 05:24 PM

thanks for the reply, but i dont quite understand how to express pde in terms of v?

**Danny** · January 24th 2010, 06:21 AM

Originally Posted by lollol

thanks for the reply, but i dont quite understand how to express pde in terms of v?

Let's try another way. If $u(x,y) = \phi(3x-2y)$ then taking an $x$ and $y$ derivative gives

$u_x = 3\phi'(3x-2y)$ and $u_y = -2\phi'(3x-2y)$ .

Now, what linear combination

$a u_x + b u_y = 0$ identically (i.e. no $\phi'$ )?

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