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Math Help - PDE - Difficult

  1. #1
    Member Altair's Avatar
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    PDE - Difficult

    y^2u_{yy} + 2yu_y - 2u = 0

    The bad thing here is that the variable with respect to which partial differentiation is being done also appears in the PDE. How to get this one done?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Since the independent variable x is not at all present, the PDE equation is equivalent to the linear homogeneus DE...

    y^{2}\cdot u^{''} + 2\cdot y\cdot u^{'} - 2\cdot u =0 (1)

    ... the solution of which is...

    u(x,y) = c_{1} (x) \cdot u_{1} (y) + c_{2} (x) \cdot u_{2} (y) (2)

    ... where u_{1}(y), u_{2} (y) are two linearly independent solutions of (1) and c_{1} (x), c_{2} (x) two arbitrary functions of the x only...

    Kind regards

    \chi \sigma
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