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Given $x''-x=0$, where $x(t_1)=a$, $x(t_2)=b$ for $t_1\neq t_2$ and $a,b$ real, we must show that the previous boundary value problem has a unique solution.

Calculating the ODE we get $x(t)=c_1e^t+c_2e^{-t}$ for $c_1,c_2$ real. Am I to assume it is the nature of the functions $c_1e^t ,\ c_2e^{-t}$ both being continuous and monotonic individually that imply uniqueness of the solution?

Thanks :)

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You can write $x(t_1)=a$ and $x(t_2)=b$ as a matrix equation

$$ \left(\begin{array}{cc} e^{t_1} & e^{-t_1} \\ e^{t_2} & e^{-t_2} \end{array}\right) \cdot \left(\begin{array}{c} c_1 \\ c_2 \end{array}\right) = \left(\begin{array}{c} a \\ b \end{array}\right) $$

This has a unique solution if and only if the determinant of the matrix is zero, i.e.

$$e^{t_1}e^{-t_2}-e^{t_2}e^{-t_1} = e^{t_1-t_2}-e^{-(t_1-t_2)}=2\sinh(t_1-t_2)=0.$$

But the hyperbolic sin of $t$ is known to be zero only when $t=0$, forcing $t_1=t_2$ in your equation.

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