Solve the partial differential equation (d u)/(d t) = (d2 u)/(d x2) with boundary conditions u(0, t) = 0 and u(1, t) = 0 using the method of separation of variables.

Step 1: Separate variables.
Assume a solution of the form $u(x, t) = X(x)T(t)$.
Substitute this into the given partial differential equation $\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}$:

$$X(x)T'(t) = X''(x)T(t)$$

Divide by $X(x)T(t)$ to separate the variables:

$$\frac{T'(t)}{T(t)} = \frac{X''(x)}{X(x)}$$

Since the left side depends only on $t$ and the right side depends only on $x$, both sides must be equal to a constant, say $-\lambda$.

$$\frac{T'(t)}{T(t)} = -\lambda \quad \text{and} \quad \frac{X''(x)}{X(x)} = -\lambda$$

This gives two ordinary differential equations:

1. $T'(t) + \lambda T(t) = 0$
2. $X''(x) + \lambda X(x) = 0$

Step 2: Solve the spatial equation $X''(x) + \lambda X(x) = 0$ using the boundary conditions.
The boundary conditions are $u(0, t) = 0$ and $u(1, t) = 0$.
From $u(0, t) = X(0)T(t) = 0$, we get $X(0) = 0$ (assuming $T(t)$ is not identically zero).
From $u(1, t) = X(1)T(t) = 0$, we get $X(1) = 0$.

Consider the possible values for $\lambda$:

• If $\lambda = 0$: $X''(x) = 0 \implies X(x) = Ax + B$. $X(0) = 0 \implies B = 0$. $X(1) = 0 \implies A = 0$. This gives $X(x) = 0$, a trivial solution.
• If $\lambda < 0$: Let $\lambda = -\mu^2$ for $\mu > 0$. $X''(x) - \mu^2 X(x) = 0$. The general solution is $X(x) = A e^{\mu x} + B e^{-\mu x}$. $X(0) = 0 \implies A + B = 0 \implies B = -A$. $X(x) = A(e^{\mu x} - e^{-\mu x}) = 2A \sinh(\mu x)$. $X(1) = 0 \implies 2A \sinh(\mu) = 0$. Since $\mu > 0$, $\sinh(\mu) \neq 0$, so $A = 0$. This again gives a trivial solution.
• If $\lambda > 0$: Let $\lambda = \mu^2$ for $\mu > 0$. $X''(x) + \mu^2 X(x) = 0$. The general solution is $X(x) = A \cos(\mu x) + B \sin(\mu x)$. $X(0) = 0 \implies A \cos(0) + B \sin(0) = 0 \implies A = 0$. So, $X(x) = B \sin(\mu x)$. $X(1) = 0 \implies B \sin(\mu) = 0$. For a non-trivial solution, $B \neq 0$, so $\sin(\mu) = 0$. This implies $\mu = n \pi$ for $n = 1, 2, 3, \dots$. The eigenvalues are $\lambda_n = (n \pi)^2$. The eigenfunctions are $X_n(x) = \sin(n \pi x)$ (we absorb the constant $B$ into the general solution coefficients).

Step 3: Solve the temporal equation $T'(t) + \lambda T(t) = 0$.
For each eigenvalue $\lambda_n = (n \pi)^2$, the equation becomes $T'(t) + (n \pi)^2 T(t) = 0$.
This is a first-order linear ODE.

$$\frac{dT}{dt} = -(n \pi)^2 T$$

Integrating both sides:

$$\int \frac{dT}{T} = \int -(n \pi)^2 dt$$ $$\ln|T| = -(n \pi)^2 t + C'$$ $$T_n(t) = C_n e^{-(n \pi)^2 t}$$

Step 4: Form the general solution.
The general solution is a superposition of all possible solutions:

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