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I am trying to understand the method of finding the general solution to the 2nd order hyperbolic partial differential equation

$$v_{\xi\eta} - v = 0$$

described in this answer.

I am able to derive the equation

$$p\hat{v}_{\xi}-\hat{v}=v_{\xi}(\xi,0)$$

where $\hat{v}(\xi,p)$ is the Laplace transform of $v(\xi,\eta)$ taken on the 2nd variable.

However, I get stuck at the following step:

integrate the equation $p\hat{v}_{\xi}-\hat{v}=v_{\xi}(\xi,0) \overset{\rm not}{=}f(\xi)$   to find that $$ \hat{v}(\xi,p)=\int\limits_0^{\xi}f(s) \frac{e^{\frac{\xi-s}{p}}}{p}\,ds+ c(p)e^{\frac{\xi}{p}}. $$

I have puzzled over it quite a bit, and searched for explanations on google, but so far without making progress. Integrate the equation how? What are one or more of the intermediate steps?

Math Keeps Me Busy
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  • Making the ansatz $v = \cosh(\xi + \eta)$ or $v = \sinh(\xi + \eta)$ works. You can see this because the equation is linear, so we can guess $v = e^{\lambda_{1} x + \lambda_{2} y}$ – Matthew Cassell Jul 04 '21 at 01:10
  • @mattos Thank you, but I'm not looking for solutions to the pde, but seeking to understand how the formula for the general solution was derived. – Math Keeps Me Busy Jul 04 '21 at 01:14
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    Oh sorry, I didn't read the question properly. I'll look at the other question when I get a chance. Though to solve $$p \hat{v}{\xi} - \hat{v} = v{\xi}(\xi, 0)$$ use the integrating factor $e^{- \xi/p}$ to get $$(e^{- \xi/p} \hat{v}){\xi} = e^{- \xi/p} v{\xi}(\xi, 0) \implies e^{- \xi/p} \hat{v} = \int e^{- \xi/p} v_{\xi}(\xi, 0) + C \implies \dots$$ – Matthew Cassell Jul 04 '21 at 01:20
  • It looks like I made a typo and missed a factor of $p^{-1}$ attached to the $v_{\xi}(\xi,0)$ term after applying the integrating factor. So it should actually be $$(e^{- \xi/p} \hat{v}){\xi} = p^{-1} e^{- \xi/p} v{\xi}(\xi, 0) \implies \dots$$ – Matthew Cassell Jul 04 '21 at 07:50
  • Thank you, that clarifies a great deal. Is the $\overset{\rm not}{=}f(\xi)$ in the original answer a typo? Should it have been $\widetilde{=}f(\xi)$ (or $\equiv f(\xi)$) ? – Math Keeps Me Busy Jul 04 '21 at 11:15

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