# general soltn by using mthd of characteristic

• May 7th 2010, 12:55 AM
bogazichili
general soltn by using mthd of characteristic
u_xx + 10u_xy + 9u_yy = y

how can we find this equation's general solution by using the method of characteristics? ı m new in this field.
• May 7th 2010, 07:00 AM
Jester
Quote:

Originally Posted by bogazichili
u_xx + 10u_xy + 9u_yy = y

how can we find this equation's general solution by using the method of characteristics? ı m new in this field.

First we will determine the type. Then general form of linear PDEs is

$A u_{xx} + B u_{xy} + C u_{yy} + \text{l.o.t.s} = 0$

$B^2 - 4 A C$ gives the type. Here $B^2 - 4 A C = 64 > 0$ so it's hyperbolic which means we can transform it to standard form, i.e. $u_{rs} + \text{l.o.t.s} = 0$. After a change of variables and choosing our target form, we find that $r$ and $s$ must satisfy

$
r_x^2 + 10 r_x r_y + 9 r_y^2 = 0
$

or $\left(r_x + r_y \right) \left(r_x + 9 r_y\right) = 0$ (choose one for $r$ and the other for $s$).

so $r$ satisfies $r_x + r_y = 0$ and $s$ satisfies $s_x + 9 s_y = 0$. Each are easily solved giving

$
r = R(x - y),\;\; s = S(9x-y)
$

from which we choose simple

$
r = x - y,\;\;\; y = 9x - y
$

Under this change of variables we obtain

$
64 u_{rs} = \frac{9r}{8} - \frac{s}{8}.
$

This we can integrate twice.