00:01
For this problem on the topic of higher order odees, we are told that for a certain model of one -dimensional fluid flow, the successes value values u -n of the solution that are connected for n greater than equal to 1, with the difference equation c into u -n plus 1 minus u -n -1 is equal to d into u -n plus 1, minus u -u -n -n -1, where c -n -d are positive constants.
00:27
The boundary conditions are given and we want to find a solution to the equation and show that successive values of un have alternating signs if c is greater than d.
00:36
Now we can substitute the trial solution un equal to a lambda to the power n into the recursive relation and we get minus c into lambda n plus 1 minus lambda into n minus 1 plus d into lambda n plus 1 minus 2 lambda to the power n plus lambda plus lambda to the power n minus 1 is equal to 0.
01:14
And so we have d minus c into lambda squared minus 2d lambda plus d plus c equal to 0.
01:28
Now this is a quadratic equation for lambda, which has the solution, lambda is equal to d plus or minus the square root of d squared minus d squared minus c squared, all divided by d minus c, which simply becomes d plus c over d minus c.
01:59
Now the general solution formed by taking a linear superposition of the trial solutions corresponding to the allowed values of lambda is unn equal to a times 1 to the power n plus b into d plus c over d minus c all to the power n which is a plus b times in the bracket we call it mu to the power n.
02:35
Now if we impose the boundary conditions, u0 is equal to zero gives us b equal to minus a.
02:45
Um equal to one gives us a into one minus mu to the power m equal to one...