non-linear differential equations!

byrnesj1

1. The problem statement, all variables and given/known data

Show that w(t) = tanh(t) solves the nonlinear problem:

w''(t)+2w(t)-2w³(t) = 0
t ε ℝ

2. Relevant equations
[itex]\frac{d^2tanh(t)}{dt^2}[/itex] = -2tanh(t)sech²(t) = [itex]\frac{-8sinh(2t)cosh^2(t)}{(cosh(2t)+1)^3}[/itex]

tanh(t) = [itex]\frac{sinh(2t)}{cosh(2t)+1}[/itex]


tanh(t)³ = [itex]\frac{sinh^3(2t)}{(cosh(2t)+1)^3}[/itex]




3. The attempt at a solution
plug and chug? I'm not good at hyperbolic functions.

Any ideas?

byrnesj1

yeah.. I keep getting stuck.

currently at:

[itex]\frac{[-6sinh(2t)cosh^2(t)+4cosh(2t)sinh(2t)+2sinh(2t)-2sinh^3(2t)]}{(cosh(2t)+1)^3}[/itex]

HallsofIvy

It's very hard to decipher what you have written. Yes, if w= tanh(t) then [itex]w''= -2sech^2(t)tanh(t)[/itex] but it is not clear where you get the rest of that from. Just replacing w'' with [itex]-2sech^2(t)tanh(t)[/itex] and w with tanh(t), the left side becomes
[itex]-2sech^2(t)tanh(t)+ 2tanh(t)- 2tanh^3(t)[/itex]
Factor 2 tanh(t) out of that to get
[itex]2tanh(t)(1- sech^2(t)- tanh^2(t))[/itex]

Now, what is [itex]tanh^2(t)+ sech^2(t)[/itex]?

byrnesj1

Thank you! Also I was attempting to simplify in terms of sinh(t) and cosh(t).. I don't know why.