Deriving soln. of $U_{xx}-sU=1$, where $U=L(u(x,t))=U(x,s)$ 
$U_{xx}-sU=-1$, $U(0,s)=U(1,s)=0$

Where $U(x,s)$ is the Laplace transform of $u(x,t)$
I 'm wondering how to show that:

The solution is, therefore
  $U(x,s)=\frac{1}{s(e^{\sqrt{s}}-e^{-\sqrt{s}})}[e^{\sqrt{s}}-e^{-\sqrt{s}}+(e^{\sqrt{s}}-1)e^{\sqrt{s}x}-(e^{\sqrt{s}}-1)e^{-\sqrt{s}x}]$ 
  (97)

Page 27, PDE equations, Analytical solution techniques, J. Kevorkian.
I tried following the methods in the preceding pages, but don't see how they got it.
 A: We'll treat the $s$ as a constant, so this is a constant coefficient linear ODE with independent variable $x$ and unknown function $U(x,s)$. This type of equation can be solved by complimentary and particular solutions. The complimentary solution should solve the associated homogeneous equation
$$
V_{xx}-sV=0,
$$
which is now a constant coefficient linear homogeneous ODE, solvable by trying a guessed form of a solution $V=e^{\lambda x}$, often called an ansatz:
$$
V_{xx}-sV= (\lambda^2 - s)e^{\lambda x} = 0.
$$
This equation is satisfied when $\lambda=\pm\sqrt{s}$, leading to two solutions, any linear combination of which will satisfy the homogeneous equation. So, we have
$$
V(x,s) = c_1e^{-x\sqrt{s}} + c_2e^{x\sqrt{s}} = a(s)e^{-x\sqrt{s}} + b(s)e^{x\sqrt{s}}
$$
Note that the two "constants" need only be constants in the variable $x$, so a general $s$ dependency is allowed, leading to the most general form of the complimentary solution.
Now, we need to find the particular solution, which solves
$$
W_{xx}-sW=-1
$$
The standard rule of finding particular solutions says you should try a general form of the right-hand side (called the method of undetermined coefficients); in this case, the right-hand side is a constant in $x$, so an unknown constant is what should be tried as a solution, $W(x,s) = c(s)$, which is an unknown constant in $x$ but is allowed to vary in $s$. We plug that into the equation above
$$
\frac{\partial^2 c(s)}{\partial x^2} - s c(s) = -1
$$
$$
-sc(s) = -1 \;\Rightarrow\; c(s) = \frac{1}{s}
$$
So now we have the particular solution, $W(x,s) = \frac{1}{s}$. The full solution is the complimentary plus the particular solution, $U = V + W$,
$$
U(x,s) = a(s)e^{-x\sqrt{s}} + b(s)e^{x\sqrt{s}} + \frac{1}{s}
$$
Now, we have to fix those two free parameters $a(s)$ and $b(s)$ by using the given boundary values,  $U(0,s)=U(1,s)=0$. So we plug the form above into these two conditions:
$$
U(0,s) = a(s) + b(s) + \frac{1}{s} = 0
$$
$$
U(1,s) = a(s)e^{-\sqrt{s}} + b(s)e^{\sqrt{s}} + \frac{1}{s} = 0
$$
These are two equations in two unknowns, so solve the first one for $b(s)$
$$
b(s) = -a(s) - \frac{1}{s}
$$
and plug into the second one
$$
a(s)e^{-\sqrt{s}} + \left(-a(s) - \frac{1}{s}\right)e^{\sqrt{s}} + \frac{1}{s} = 0
$$
$$
a(s)e^{-\sqrt{s}} - a(s)e^{\sqrt{s}} - \frac{1}{s}e^{\sqrt{s}} + \frac{1}{s} = 0
$$
$$
a(s)\left(e^{-\sqrt{s}} - e^{\sqrt{s}}\right) - \frac{1}{s}\left(e^{\sqrt{s}} - 1\right) = 0,
$$
which can be solved for $a(s)$ as
$$
a(s) = \frac{e^{\sqrt{s}} - 1}{s \left(e^{-\sqrt{s}} - e^{\sqrt{s}}\right)}
$$
Now we plug that back into the equation for $b(s)$ to get
$$
b(s) = -a(s) - \frac{1}{s} = -\frac{e^{\sqrt{s}} - 1}{s \left(e^{-\sqrt{s}} - e^{\sqrt{s}}\right)} - \frac{1}{s}
$$
and simplify this
$$
b(s) = -\frac{e^{\sqrt{s}} - 1}{s \left(e^{-\sqrt{s}} - e^{\sqrt{s}}\right)} - \frac{e^{-\sqrt{s}} - e^{\sqrt{s}}}{s\left(e^{-\sqrt{s}} - e^{\sqrt{s}}\right)}
$$
$$
b(s) = \frac{-e^{\sqrt{s}} + 1 - e^{-\sqrt{s}} + e^{\sqrt{s}}}{s \left(e^{-\sqrt{s}} - e^{\sqrt{s}}\right)}
$$
$$
b(s) = \frac{1 - e^{-\sqrt{s}}}{s \left(e^{-\sqrt{s}} - e^{\sqrt{s}}\right)}
$$
Now that $a$ and $b$ have been fixed by the boundary condition, we have the solution
$$
U(x,s) = \frac{e^{\sqrt{s}} - 1}{s \left(e^{-\sqrt{s}} - e^{\sqrt{s}}\right)} e^{-x\sqrt{s}} + \frac{1 - e^{-\sqrt{s}}}{s \left(e^{-\sqrt{s}} - e^{\sqrt{s}}\right)} e^{x\sqrt{s}} + \frac{1}{s}
$$
Further simplification is required to get that in the same form as your reference
$$
U(x,s) = \frac{e^{\sqrt{s}} - 1}{s \left(e^{-\sqrt{s}} - e^{\sqrt{s}}\right)} e^{-x\sqrt{s}} + \frac{1 - e^{-\sqrt{s}}}{s \left(e^{-\sqrt{s}} - e^{\sqrt{s}}\right)} e^{x\sqrt{s}} + \frac{e^{-\sqrt{s}} - e^{\sqrt{s}}}{s\left(e^{-\sqrt{s}} - e^{\sqrt{s}}\right)}
$$
$$
U(x,s) = \frac{1}{s \left(e^{-\sqrt{s}} - e^{\sqrt{s}}\right)}
\left[ 
\left(e^{\sqrt{s}} - 1\right) e^{-x\sqrt{s}} + \left(1 - e^{-\sqrt{s}}\right) e^{x\sqrt{s}} + e^{-\sqrt{s}} - e^{\sqrt{s}}
\right]
$$
This is getting awfully close to what you wrote, but I suspect you have at least one sign error when you copied the solution from the book, and it's possible that I dropped a minus sign somewhere. Nevertheless, the method I provided is how to solve the problem and I encourage you to carefully follow the method and check all the signs to arrive at the final answer.
A: Typo now corrected in my previous answer, which was the solving of $U_{xx}-sU(x,s)=1$ instead of :
$$U_{xx}-sU(x,s)=-1$$
Since the derivative is with respect to only one variable $x$, this isn't a PDE, but an ODE of the kind :
$$y''-sy(x)=-1\qquad\text{where } s \text{ is a parameter.}$$
The general solution is commonly obtained :
$\quad y(x)=c_1 e^{\sqrt{s}\,x}+c_2 e^{-\sqrt{s}\,x}+\frac{1}{s}.\quad$
$$U(x,s)=c_1 e^{\sqrt{s}\,x}+c_2 e^{-\sqrt{s}\,x}+\frac{1}{s}.$$
With the boundary conditions $U(0,s)=U(1,s)=0$ one have to solve the system :
$$\begin{cases}
c_1 +c_2 +\frac{1}{s}=0\\
c_1 e^{\sqrt{s}}+c_2 e^{-\sqrt{s}}+\frac{1}{s}=0
\end{cases}$$
$c_2=-c_1-\frac{1}{s}$ put into the second equation leads to :
$\quad c_1=-\frac{1}{s}\:\frac{e^{-\sqrt{s}}-1}{e^{-\sqrt{s}}-e^{\sqrt{s}}}$
and then : $\qquad c_2= \frac{1}{s}\:\frac{e^{-\sqrt{s}}-1}{e^{-\sqrt{s}}-e^{\sqrt{s}}} -\frac{1}{s} =-\frac{1}{s}\:\frac{1-e^{\sqrt{s}}}{e^{-\sqrt{s}}-e^{\sqrt{s}}}$
The result is :
$$U(x,s)=-\frac{1}{s}\:\frac{e^{-\sqrt{s}}-1}{e^{-\sqrt{s}}-e^{\sqrt{s}}} e^{\sqrt{s}\,x}-\frac{1}{s}\:\frac{1-e^{\sqrt{s}}}{e^{-\sqrt{s}}-e^{\sqrt{s}}} e^{-\sqrt{s}\,x}+\frac{1}{s}$$
$$U(x,s)=-\frac{1}{s\,(e^{-\sqrt{s}}-e^{\sqrt{s}}) } \left( (e^{-\sqrt{s}}-1)e^{\sqrt{s}\,x} + (1-e^{\sqrt{s}})e^{-\sqrt{s}\,x} -e^{-\sqrt{s}}+e^{\sqrt{s}}\right)$$
This result is exactly the same as the result of rajb245.
But, this is not consistent with the result given in the book. Moreover, the solution of the book, which satisfies the PDE, doesn't satisfies the conditions $U(0,s)=U(1,s)=0$. I suspect a mess from the conditions specified before the Laplace transform (conditions for $u(x,t)$ ) to the conditions rewritten after the Laplace transform (conditions for $U(x,s)$ ). Since "unseen_rider" didn't show this part of the calculus, one cannot say more about the origin of the discrepancy.
