# Hyperbolic Inverse Equation

• Jan 22nd 2011, 01:44 AM
AlanC877
Hyperbolic 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!
• Jan 22nd 2011, 02:02 AM
Ackbeet
Quote:

(d^2u/dx^2)-u+3u^2
That's not an equation. Where does the equals sign go?
• Jan 22nd 2011, 02:20 AM
AlanC877
Quote:

Originally Posted by Ackbeet
That's not an equation. Where does the equals sign go?

Apologies, it's =0

((d^2u/dx^2)-u+3u^2=0)
• Jan 22nd 2011, 02:36 AM
Prove It
If $\displaystyle \displaystyle u = AB\,x\,\textrm{sech}^2\,{x}$ evaluate $\displaystyle \displaystyle \frac{d^2u}{dx^2}$.

Then substitute this all into $\displaystyle \displaystyle \frac{d^2u}{dx^2} - u + 3u^2 = 0$.

You should be able to solve for $\displaystyle \displaystyle A,B$.
• Jan 22nd 2011, 02:38 AM
Ackbeet
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.
• Jan 22nd 2011, 02:48 AM
AlanC877
Quote:

Originally Posted by Prove It
If $\displaystyle \displaystyle u = AB\,x\,\textrm{sech}^2\,{x}$ evaluate $\displaystyle \displaystyle \frac{d^2u}{dx^2}$.

Then substitute this all into $\displaystyle \displaystyle \frac{d^2u}{dx^2} - u + 3u^2 = 0$.

You should be able to solve for $\displaystyle \displaystyle A,B$.

Sorry again, I made a stupid mistake when entering the question. u=Asech^2(Bx). Would I just follow the same method as you outlined?
• Jan 22nd 2011, 03:04 AM
Prove It
Quote:

Originally Posted by AlanC877
Sorry again, I made a stupid mistake when entering the question. u=Asech^2(Bx). Would I just follow the same method as you outlined?

Yes, and it should simplify the problem greatly...
• Jan 23rd 2011, 02:02 PM
AlanC877
Quote:

Originally Posted by Prove It
Yes, and it should simplify the problem greatly...

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!
• Jan 23rd 2011, 02:23 PM
AlanC877
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.