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

thank you

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- Jan 23rd 2010, 01:39 AMlollola partial differentiation problem
Hi guys, i ve got a problem and i've never seen this type of question before..could u help me out? (Crying)

thank you - Jan 23rd 2010, 04:41 AMshawsend
That is in the form of a solution to a first order PDE with constant coefficients:

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

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

$\displaystyle 2u_x+3u_y=0$

using this method, I get $\displaystyle 3v_z=0$ or $\displaystyle v(w,z)=C(w)$ where $\displaystyle C(w)$ is an arbitrary function of $\displaystyle w=3x-2y$ that is, the function $\displaystyle u(x,y)=C(3x-2y)$ solves my PDE like $\displaystyle u(x,y)=3x-2y$, $\displaystyle u(x,y)=\sin(3x-2y)$, 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. - Jan 23rd 2010, 05:24 PMlollol
thanks for the reply, but i dont quite understand how to express pde in terms of v?(Hi)

- Jan 24th 2010, 06:21 AMJester
Let's try another way. If $\displaystyle u(x,y) = \phi(3x-2y)$ then taking an $\displaystyle x$ and $\displaystyle y$ derivative gives

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

Now, what linear combination

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