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

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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- May 14th 2009, 09:42 PMAltairPDE - Difficult
$\displaystyle y^2u_{yy} + 2yu_y - 2u = 0$

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? - May 14th 2009, 10:31 PMchisigma
Since the independent variable $\displaystyle x$ is not at all present, the PDE equation is equivalent to the linear homogeneus DE...

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

... the solution of which is...

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$