1. ## 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!

2. (d^2u/dx^2)-u+3u^2
That's not an equation. Where does the equals sign go?

3. 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)

4. If $\displaystyle u = AB\,x\,\textrm{sech}^2\,{x}$ evaluate $\displaystyle \frac{d^2u}{dx^2}$.

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

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

5. 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.

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

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

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

7. 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...

8. 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!

9. 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.