## non-linear differential equations!

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

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

Any ideas?
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 yeah.. I keep getting stuck. currently at: $\frac{[-6sinh(2t)cosh^2(t)+4cosh(2t)sinh(2t)+2sinh(2t)-2sinh^3(2t)]}{(cosh(2t)+1)^3}$
 Recognitions: Gold Member Science Advisor Staff Emeritus 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 $tanh^2(t)+ sech^2(t)$?

## non-linear differential equations!

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