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Math Help - general soltn by using mthd of characteristic

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    Exclamation 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.
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  2. #2
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    Quote Originally Posted by bogazichili View Post
    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

     <br />
r_x^2 + 10 r_x r_y + 9 r_y^2 = 0<br />

    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

     <br />
r = R(x - y),\;\; s = S(9x-y)<br />

    from which we choose simple

     <br />
r = x - y,\;\;\; y = 9x - y<br />

    Under this change of variables we obtain

     <br />
64 u_{rs} = \frac{9r}{8} - \frac{s}{8}.<br />

    This we can integrate twice.
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