# 2nd order non-linear homogeneous differential equation

• byrnesj1
In summary, the homework statement is to find a solution of z'' + 2z - 6(tanh(t))2z = 0 that is linearly dependent of Z1 = sech2. Integration can be done using the integrating factor.f

## Homework Statement

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..

Last edited:

## Homework Statement

Find a solution (Z2) of:
z'' + 2z - 6(tanh(t))2z = 0

that is linearly dependent 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..

Check the derivative of sech2(t).

ehild

ahh that works beautifully. thanks ehild.