How do you solve 40000y''-729y=0?
Assume a solution to the homogeneous equation of the form, y = e^(mx).
Then y = m.e^(mx), y'' = m^2.e^(mx)
The DE now is,
40000.m^2.e^(mx) - 729.e^(mx) = 0
200^2.m^2.e^(mx) - 27^2.e^(mx) = 0
200^2.m^2 = 27^2
m = +/- 27/200
The solution now is: y = A.e^(27x/200) + B.e^(-27x/200)