How can I solve the ODE with boundary conditions? I have solved it without boundary conditions. I have no idea with such boundary conditions.


$${\partial ^2 Y(X) \over \partial ^2 X} = \beta Y(X)$$

Boundary Conditions:

$${\partial \ln Y(X) \over \partial X} = \begin{cases} a < 0,X = k \\[5pt] b , X = 0 \\[5pt] c > 0,X = - k \end{cases} $$

where $a,b,c \in \mathbb{R}$ are given and $X$, $Y$ and $\beta$ are unkown.

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    $\begingroup$ Is $\beta$ an unknown? Is it, perhaps, something that has to be determined upon imposing the condition at $x = 0$? And more importantly, what have you tried? $\endgroup$ – Dmoreno Jun 28 '17 at 2:23
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    $\begingroup$ Note that if $a,b,c$ and $\beta$ are given constants, your problem is then overdetermined. How could you impose a third boundary condition when the problem is second-order? $\endgroup$ – Dmoreno Jun 28 '17 at 2:26
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    $\begingroup$ Why has this been given two votes up? The OP showed no effort what so ever. $\endgroup$ – Mattos Jun 28 '17 at 2:29
  • $\begingroup$ $\beta$ is an unknown. Because I don't how to solve ODE with such boundary condition, I treat it as temporary "know" to simplify the problem. There is no restriction on b where $X = 0$. @Dmoreno $\endgroup$ – XJ.C Jun 28 '17 at 2:43
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    $\begingroup$ I don't know if I'm getting this right. The constant $\beta$ is an unknown, which makes sure that all 3 boundary conditions are satisfied, that is, it's an 'eigenvalue' of the problem. But you say that $a,b$ and $c$ are real and fixed (given), right? Where are you having troubles specifically? Solve the ODE as usual, impose the boundary conditions at $X = \pm k$ and write down what you would get for the third one. Maybe that gives you a condition for $\beta$... $\endgroup$ – Dmoreno Jun 28 '17 at 2:50

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