2nd order non-linear homogeneous differential equation

[byrnesj1]

Homework Statement

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

that is linearly independent of Z1 = sech²

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 = 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..

 

[ehild]

Homework Statement

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

that is linearly dependent of Z1 = sech²

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 = 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

 

[byrnesj1]

ahh that works beautifully. thanks ehild.