Find a solution (Z2) of:
z'' + 2z - 6(tanh(t))2z = 0
that is linearly independent of Z1 = sech2
The Attempt at a Solution
reduction of order gives you
v''(t)(Z1(t))+v'(t)(2 * Z1'(t)) + v(t)(Z1''(t)+p(t)Z1'(t)) = 0
however the third term on the LHS can be dropped since we know that Z1 is a solution to the original problem.
v''(t)(Z1(t))+v'(t)(2 * Z1'(t)) = 0 = sech2(t)v''(t) + 2(-2tanh2(t)sech2(t))v'(t)
let y = v'
sech2(t)y'(t) + 2(-2tanh2(t)sech2(t))y(t) = 0
divide both sides by sech2(t)
y'(t) - 4tanh2(t)y(t) = 0
from here would I use integrating factor, or should I have done exact equations for the step before this?
using integrating factor
μ(t) = e(4tanh(t)-4t)
y = e-(4tanh(t)-4t)
v = ∫e-(4tanh(t)-4t)dt
can any1 point me in the correct direction? I also don't know how to integrate the last part..