Here's an idea, perhaps naive? Introduce auxiliary variable $y = x \sqrt{c}$. $$ \left( \begin{array}{cc} E & \sqrt{c} I \\ \sqrt{c} I & -1 \\ \end{array} \right) \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{c} b \\ 0 \end{array} \right). $$$$ \left( \begin{array}{cc} E & \sqrt{c} I \\ \sqrt{c} I & -I \\ \end{array} \right) \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{c} b \\ 0 \end{array} \right). $$ Now an LU decomposition of the larger matrix is $$ \left( \begin{array}{cc} L_E & 0 \\ L_{10} & L_{11} \end{array} \right) \left( \begin{array}{cc} U_E & U_{01} \\ 0 & U_{11} \end{array} \right), $$ where $E = L_E U_E$. When $c$ changes to $d$, the new LU decomposition is given by $$ \left( \begin{array}{cc} L_E & 0 \\ \sqrt{\frac{d}{c}} L_{10} & \tilde L_{11} \end{array} \right) \left( \begin{array}{cc} U_E & \sqrt{\frac{d}{c}} U_{01} \\ 0 & \tilde U_{11} \end{array} \right) $$ where $$ \frac{d}{c} L_{10} U_{01} + \tilde L_{11} \tilde U_{11} = -1. $$$$ \frac{d}{c} L_{10} U_{01} + \tilde L_{11} \tilde U_{11} = -I. $$ But we know $L_{10} U_{01} + L_{11} U_{11} = -1$$L_{10} U_{01} + L_{11} U_{11} = -I$, so substituting $$ \tilde L_{11} \tilde U_{11} = -1 + \frac{d}{c} (1 + L_{11} U_{11}) $$ which is a scalar equation. Pick$$ \tilde L_{11} \tilde U_{11} = -I + \frac{d}{c} (I + L_{11} U_{11}). $$ Find a solution to that and then LU solve (oops ... maybe this is as hard as the original problem?)
