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I have the PDE listed below and I am not quite sure how to solve it.

$\frac{\partial^2 g}{\partial y^2} - \frac{\partial^2 g}{\partial z^2} - 2 \frac{\partial g}{\partial x}=0$

I have tried separation of variables, however, it does not seem to work. I have been reading about the method of characteristics but it mostly applies to first order PDE. I am unfamiliar with that approach and I am not sure how I may apply it to this problem.

I am more interested in the process of how to solve this problem than the solution but I would gladly accept the solution as well so I can work it out for myself.

I appreciate any assistance.

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  • $\begingroup$ The easiest way is taking a guess as to the structure of $g$, and trying to see if that structure hosts a solution. Other methods are ... not fun. Check out this link. It might give you some ideas as to where to start. $\endgroup$ Mar 24, 2019 at 22:57

1 Answer 1

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Hint

The method "Separation of Variables" works. Define$$g(x,y,z)=f_1(x)f_2(y)f_3(z)$$therefore $$f_1(x)f_2''(y)f_3(z)-f_1(x)f_2(y)f_3''(z)-2f_1'(x)f_2(y)f_3(z)=0$$which yields to$${f_2''(y)\over f_2(y)}-{f_3''(z)\over f_3(z)}-2{f_1'(x)\over f_1(x)}=0$$and hence $${f_2''(y)\over f_2(y)}=k_1\\{f_3''(z)\over f_3(z)}=k_2\\{f_1'(x)\over f_1(x)}={k_1-k_2\over 2}$$

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  • $\begingroup$ Excellent. Thank you. $\endgroup$
    – dsmalenb
    Mar 24, 2019 at 23:05
  • $\begingroup$ Your welcome. Good luck! $\endgroup$ Mar 24, 2019 at 23:08

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