PDE - Difficult

**Altair** · May 14th 2009, 09:42 PM

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

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?

**chisigma** · May 14th 2009, 10:31 PM

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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