# When do I use y=Acoshx+Bsinhx?

1. Dec 5, 2014

### yuze

When I have:
y''-yu^2=0

Why can't I just solve it using y=Ae^ux+Be^-ux ?
Is there any specific reason for me to use y=Acoshx + Bsinhx?

Also, for questions such as ut=uxx-u with BC u(0,t)=0 and u(L,t)=u1
When I am setting up u=v+w do I have to set V as Acoshx+Bsinhx?
I saw the solutions do that, but why can't I set V as u1x/L in this case?

2. Dec 5, 2014

### yuze

When I am separating variables, my understanding is that I can use either +lambda^2 or -lambda^2 depending on how I want my Xn or Tn to look like based on what boundary conditions I have. Is this line of thinking correct?

For example:
u=XT
X''/X=T''/T=+/-lambda^2

Sorry for so many questions at once.

Last edited: Dec 5, 2014
3. Dec 5, 2014

### LCKurtz

If you have $X'' - \mu^2 X=0$ you may use either $X = Ae^{\mu x} + Be^{-\mu x}$ or $X= A\cosh (\mu x) + B\sinh(\mu x)$. The advantage of one over the other is in solving for the constants with the boundary conditions. For example if you have $X(0) = c$ and $X'(0)= d$, in the first case you would have$$X(0) = Ae^0 + Be^0 = A +B = c$$and in the second case you would have$$X(0)= A\cosh(0) + B\sinh(0) = A\cdot 1 + B\cdot 0 = A = c$$So you already know $A$ and the second boundary condition will be easy to solve for $B$. In the exponential case you have two linked equations in two unknowns and it will be slightly more work to solve for the constants.

4. Dec 5, 2014

### yuze

I see, thanks!!