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.

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- May 7th 2010, 12:55 AMbogazichiligeneral 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 AMJester
First we will determine the type. Then general form of linear PDEs is

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

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

$\displaystyle

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

$

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

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

$\displaystyle

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

$

from which we choose simple

$\displaystyle

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

$

Under this change of variables we obtain

$\displaystyle

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

$

This we can integrate twice.