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$$ \frac{\partial^2{u(x,y)}}{\partial y^2} -x^2 {u(x,y)} = 0$$

If i have something like this^. is it okay to use the standard technique for solving second order homogeneous differential equations?

i.e. $ r^2 = x^2 $

Which leads to $${u}(x,y) = A e^{x y} + B e^{-x y}$$

Just wondering because u is in terms of x but the differentials are with respect to y.

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    This is not a PDE, since only derivatives in $y$ appear. Here, $x$ can be treated as a parameter, and yes you can treat it as an ODE and use the general solution you wrote. – 2017-02-25
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    Cool, cheers man! – 2017-02-25

1 Answers 1

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What if

$$u(x,y)=e^{xy}?$$

The first partial wrt $y$ is

$$xe^{xy}$$

and the second one is then

$$x^2e^{xy}=x^2u(x,y).$$