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Thread: 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

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

    $\displaystyle B^2 - 4 A C$ 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.
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