Originally Posted by

**TheProphet** I am attempting to show that

$\displaystyle y'' -10y' -11y = 0 $, with $\displaystyle y(0) = 1 $ and $\displaystyle y'(0) = -1 $ is ill-conditioned.

I assumed a solution of the form $\displaystyle y(x) = e^{rx} $ which (of course) lead me to $\displaystyle r^{2} - 10r - 11 = 0 $. Solutions to this equation are $\displaystyle r_{1} = 11 $ and $\displaystyle r_{2} = -1 $.

Thus, $\displaystyle y(x) = c_{1} e^{11x} + c_{2} e^{-x} $. Now we use the boundary conditions given, leading to

$\displaystyle \begin{pmatrix} 1 & 1 \\ 11 & -1 \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.

How to go from here? The determinant of the coefficient matrix isn't particularly small, either.