For the general solution of the differential equation in X use A and B for your constants and list the functions in alphabetical order, for example $y = A\cos (x) + B \sin(x)$. For the differential equation in T use the C and D.For the variable $\lambda$ type the word lambda and type alpha for $\alpha$,otherwise treat them as you would any other variable.

Use the prime notation for derivatives, so the derivative of $X$ is written as $X^\prime$. Do NOT use $X^\prime(x)$

The longitudinal displacement u(x,t) of a vibrating elastic bar can be modeled by a wave equation with free-end conditions

$$\displaystyle a^2\frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 u}{\partial t^2}, 0< x < L, t > 0$$ $$\displaystyle \frac{\partial u}{\partial x}\Big\vert_{x=0} = 0,\hskip 10pt \frac{\partial u}{\partial x}\Big\vert_{x=L} = 0, t > 0$$ $$\displaystyle u(x,0) = x \hskip 10pt \frac{\partial u}{\partial t}\Big\vert_{t=0}=-2,0<x<L$$

PART 1: Separate

We look for solutions of the form $u(x,t) = X(x)T(t)$.
Assuming solutions of this form the PDE can be written as

Then we can separate the PDE,placing constants in the DE into T:
= = $-\lambda$ ($\lambda$ is typed as lambda)

The boundary conditions in the PDE translate into initial conditions for the differential equation in X:
Since these differential equations are independent of each other, they can be separated
DE in X: $= 0$
IC's in X:

DE in T: $= 0$

This leads us to a Sturm-Louiville problem in X. In each case the general solution in X is written with constants A and B.

PART 2: Solve for X

Case 1: $\lambda = 0$
$X(x) =$
Using this formula

Making X a constant function. We get the constant solution another way, so we will discard this case.

Case 2: $\lambda = -\alpha^2$
$X(x) =$
$X^\prime(0) =$ $= 0$
That simplifies X down to . Using the simplified X:
$X^\prime(L) =$ $= 0$

Therefore $A =$ and $X =$

Case 3: $\lambda = \alpha^2$
$X(x) =$
$X^\prime(0) =$ $= 0$
So X simplifies to .
Using the simplified X:
$X^\prime(L) =$ $= 0$
Which leads us to the eigenvalues $\lambda =$
Note If n = 0 we get the constant case from $\lambda = 0$, including it would have been redundant.


PART 3: Solve for T


So the only case we need consider is the $\lambda = \alpha^2$ case. Therefore the differential equation in T is now
$= 0$
which has the general solution (using C and D as the unknown constants)
$T =$

PART 4: Find the series solution

Since the constant from the $\cos\!\left(\frac{n\pi x}{L}\right)$ term would be redundant, functions of the form $u(x,t) = X(x)T(t) =$, for $n\geq 0$, are solutions of the PDE $$\displaystyle a^2\frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 u}{\partial t^2}, 0< x < L, t > 0$$ $$\displaystyle \frac{\partial u}{\partial x}\Big\vert_{x=0} = 0,\hskip 10pt \frac{\partial u}{\partial x}\Big\vert_{x=L} = 0, t > 0$$ To add in the initial conditions we will have to use a series solution. For the series we relabel the coefficients so that $C = A_n$ and $D=B_n$
$u(x,t) = \displaystyle\sum\limits_{n=0}^\infty \Bigg( A_n$ $+ B_n$ $\Bigg)$

Using the initial conditions we can compute

$A_0$

$=$

-------------

$\displaystyle\int$

$L$ $0$

$dx$

$\hskip 12pt$

$=$

$A_n$

$=$

-------------

$\displaystyle\int$

$L$ $0$

$dx$

$\hskip 12pt$

$=$

$B_n$

$=$

-------------

$\displaystyle\int$

$L$ $0$

$dx$

$\hskip 12pt$

$=$

PART 5: Finally!

$u(x,t) =$+$\displaystyle\sum\limits_{n=1}^\infty$