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Find the Green's function for the boundary value problem $$x''-4x=0,\quad x(0)=0,\quad x'(2)=0$$ if the corresponding Cauchy Function is
$$x(t,s)=\frac{1}{4}(e^{2t-2s}-e^{2s-2t})$$

I don't understand why it gave me the Cauchy function and how to evaluate the Cauchy function in order to find Green's function for BVP.

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Apparently, the Cauchy function means a solution of the ODE with initial conditions $x(s)=0$, $x'(s)=1$. In this case it is $x(t,s) = \frac12 \sinh (2t-2s)$, which wouldn't be too hard to find even if not given.

To build Green's function on $(0,2)$, we should start with $0$ at $0$, which any multiple of $x(t,0)$ does. Then, at a point $s$, change the derivative (but not the value) to eventually arrive at $x'(2)=0$. The change of derivative should be by $1$ so that the second derivative is Dirac delta at $s$. Thus, $$g(t,s) = Ax(t,0) + x(t,s)$$ where $A$ is found from the condition $\partial g/\partial t=0$ at $t=2$. Compute: $$A \cosh(4) + \cosh (4-2s) = 0$$ which determines $A$, and with it $g$.

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  • $\begingroup$ why $x'(s)=1$ ? $\endgroup$ – diogenes May 17 '16 at 23:57

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