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## Homework Statement

Find a solution (Z2) of:

z'' + 2z - 6(tanh(t))

^{2}z = 0

that is linearly independent of Z1 = sech

^{2}

## Homework Equations

## 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 = sech

^{2}(t)v''(t) + 2(-2tanh

^{2}(t)sech

^{2}(t))v'(t)

let y = v'

sech

^{2}(t)y'(t) + 2(-2tanh

^{2}(t)sech

^{2}(t))y(t) = 0

divide both sides by sech

^{2}(t)

y'(t) - 4tanh

^{2}(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..

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