Hi, I'm trying to work out the following:

Find non-trivial values of A and B for which u=Asech^2xBx is a solution of (d^2u/dx^2)-u+3u^2.

Any help would be really appreciated!

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- January 22nd 2011, 02:44 AMAlanC877Hyperbolic Inverse Equation
Hi, I'm trying to work out the following:

Find non-trivial values of A and B for which u=Asech^2xBx is a solution of (d^2u/dx^2)-u+3u^2.

Any help would be really appreciated! - January 22nd 2011, 03:02 AMAckbeetQuote:

(d^2u/dx^2)-u+3u^2

- January 22nd 2011, 03:20 AMAlanC877
- January 22nd 2011, 03:36 AMProve It
If evaluate .

Then substitute this all into .

You should be able to solve for . - January 22nd 2011, 03:38 AMAckbeet
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.

- January 22nd 2011, 03:48 AMAlanC877
- January 22nd 2011, 04:04 AMProve It
- January 23rd 2011, 03:02 PMAlanC877
Thank you so much for all your help so far but I'm still a little stuck.

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)

And 3u^2=3A^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! - January 23rd 2011, 03:23 PMAlanC877
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.