Methods of obtaining numerical solutions to diff. equation of second order is the same as for the first. The difference will be one additional member of iteration \(Y(x_{-1})\), which we will describe below, but for now solve the following equation: \[\ddot Y = f(x)\,Y + a \qquad (1)\] let us Introduce for better perception the following simplification: \(Y_i = Y(x_i), Y_{i-1} = Y(x_{i-1})\), and present the equation in the form of approximate values: \[{{Y_i - Y_{i-1} \over \Delta x} - {Y_{i-1} - Y_{i-2} \over \Delta x} \over \Delta x} = {Y_i - 2Y_{i-1} + Y_{i-2} \over \Delta x^2} = f(x_i)\,Y_i + a \qquad (2)\] From it immediately gives: \[Y_i = {2Y_{i-1} - Y_{i-2} + a\,\Delta x^2 \over 1 - f(x_i)\,\Delta x^2} \qquad (3)\] Since iteration starts with one, it remains to determine the initial members of \(Y_0\) and \(Y_{-1}\). Of methods for diff. equations of the first order we already know that \(Y_0 = Y(0)\), but the second member will deal in more detail. In Mat. the analysis, it appears as \(\dot Y(0)\), and for our method this means that the approximate values looks like this: \[\dot Y(0) = {Y_{0} - Y_{-1} \over \Delta x} \qquad (4)\] where did we get this member: \[Y_{-1} = Y_{0} - \dot Y(0)\,\Delta x = Y(0) - \dot Y(0)\,\Delta x \qquad (5)\] Therefore, the final equation can be written as: \[Y_i = {2Y_{i-1} - Y_{i-2} + a\,\Delta x^2 \over 1 - f(x_i)\,\Delta x^2}, \quad Y_0 = Y(0), \quad Y_{-1} = Y(0) - \dot Y(0)\,\Delta x \qquad (6)\]

On the next page is a table of optimized algorithms for the numerical solution of some differential. equations.