Non-linear differential equations

[byrnesj1]

Homework Statement

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

w''(t)+2w(t)-2w³(t) = 0
t ε ℝ

Homework Equations

$\frac{d^2tanh(t)}{dt^2}$ = -2tanh(t)sech²(t) = $\frac{-8sinh(2t)cosh^2(t)}{(cosh(2t)+1)^3}$

tanh(t) = $\frac{sinh(2t)}{cosh(2t)+1}$


tanh(t)³ = $\frac{sinh^3(2t)}{(cosh(2t)+1)^3}$

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:

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

 

[HallsofIvy]

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

Now, what is $tanh^2(t)+ sech^2(t)$?

 

[byrnesj1]

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