Originally Posted by
Ray
instead of finding a solution of d^2y/dx^2 + b dy/dx + cy = 0 which is the generic 2nd order ode problem, you're given the general solution instead.
substitute y=e^(kx), finding values of k so that y is a solution of a d^2y/dx^2 + b dy/dx + cy = 0, as the auxillary equation of this is ak^(2) + bk + c = 0 where y=e^(kx). it's literally like solving a quadratic
so for 1)
the general solution is: y_(cf)(x) = Ae^(4x) + Be^(5x), the two solutions are y=e^(4x) and y=^(5x) and the auxillary equation can be factorised as (k-4)(k-5) = 0 and so the required values of k are 4 and 5. now finding the auxilary equation from these two 'roots': k^2 - 9k + 20 = 0 and hence d^2y/dx^2 - 9* dy/dx + 20y = 0
another method is by differentiating the general solution but i prefer this method but it's all essentially the same shit
try do the same for 2), it's a little easier than the first