HARD - derivation of general solution to K Backward Equation

hi all

so this question is quite tough to answer.

the fundamental solution of the kolmogorov backward equation is:

dp/dt = Lp = 1/2 D^2(x) d2p/dx2 + M(x)dp/dx

where D is the variance and M is the mean.

given a stationary distribution Lp = 0 so if we have a function T(x)

1/2 D^2(x) T''(x) + M(x)T'(x) = 0

a solution to this is the harmonic function

now i want to know how is

T'(x) = C exp(-int_xo^x 2M(s)/D^2(s) ds )

and

T(X) = int^x exp(-int_xo^x 2M(y)/D^2(y) dy ) du


My kingdom for latex..... i hope someone can help...

perhaps i have to do this

1/2 D^2(x) T''(x) + M(x)T'(x) = 0

h = T'(x)

h'/h = 2M(x) / D^2(x)


maybe? whats the solution of this?

cheers all