That's not an equation. Where does the equals sign go?(d^2u/dx^2)-u+3u^2
I would also point out that you might not be able to uniquely solve for A and B. But you just need to find nonzero values that work. Hence, if you find an equation for A in terms of B, just pick a value for B and determine the corresponding A. You're not asked to find unique nontrivial values, just values.
So, f(x)=Asech^2(Bx), then f'(x)=-2ABsech^2(Bx)tanh(Bx), so f''(x)=4AB^2sech^2(Bx)tanh^2(Bx)-2AB^2sech^4(Bx)
So the whole equation is 4AB^2sech^2(Bx)tanh^2(Bx)-2AB^2sech^4(Bx)-Asech^2(Bx)+3A^2sech^4(Bx)
But I'm stuck on where to go, I've tried various relationships but can't get anything to work. Any more help would be extremely appreciated!
Nevermind, I looked at it and realised I could take out Asech^2(Bx) as a common factor, then use 4B^2tanh^2(Bx)=1-4B^2sech^2(Bx) which allowed me to take out a further sech^2(Bx) common factor to get: Asech^4(Bx)(-6B^2+3A) and so -6b^2+3A=0 so A=2, B=1 is a solution. Thanks for your help.