Solve the following boundary value problem using separation of variables:
∂
2u
∂x2
−
∂u
∂y = u,
satisfying boundary conditions
u(0, y) = u(π, y) = 0 and u(x, 1) = x.

The given boundary value problem is:

\[
\frac{\partial^2 u}{\partial x^2} - \frac{\partial u}{\partial y} = u,
\]
with boundary conditions:
\[
u(0, y) = u(\pi, y) = 0 \quad \text{and} \quad u(x, 1) = x.
\]

### Step 1: Apply separation of variables
We assume that the solution can be written as a product of functions that depend on \(x\) and \(y\) separately:
\[
u(x, y) = X(x) Y(y).
\]
Substitute this into the PDE:
\[
\frac{\partial^2 u}{\partial x^2} - \frac{\partial u}{\partial y} = u \implies X''(x) Y(y) - X(x) Y'(y) = X(x) Y(y).
\]
Dividing both sides by \(X(x)Y(y)\) (assuming \(X(x)Y(y) \neq 0\)) gives:
\[
\frac{X''(x)}{X(x)} - \frac{Y'(y)}{Y(y)} = 1.
\]
Since the left-hand side is a sum of two functions, each depending on a different variable, they must each be equal to a constant. Let’s call this constant \(-\lambda\):
\[
\frac{X''(x)}{X(x)} = \lambda \quad \text{and} \quad -\frac{Y'(y)}{Y(y)} = \lambda + 1.
\]

Thus, we have two ordinary differential equations:
1. For \(X(x)\):
\[
X''(x) = \lambda X(x),
\]
2. For \(Y(y)\):
\[
Y'(y) = -(1 + \lambda) Y(y).
\]

### Step 2: Solve for \(X(x)\)
The equation for \(X(x)\) is:
\[
X''(x) = \lambda X(x).
\]
This is a second-order linear homogeneous differential equation with constant coefficients. The general solution depends on the value of \(\lambda\):
- If \(\lambda = 0\), \(X(x) = A + Bx\).
- If \(\lambda > 0\), \(X(x) = A \cos(\sqrt{\lambda} x) + B \sin(\sqrt{\lambda} x)\).
- If \(\lambda < 0\), \(\lambda = -\mu^2\), then \(X(x) = A \cosh(\mu x) + B \sinh(\mu x)\).

We apply the boundary conditions \(u(0, y) = 0\) and \(u(\pi, y) = 0\). This implies:
\[
X(0) = 0 \quad \text{and} \quad X(\pi) = 0.
\]
From \(X(0) = 0\), we get \(A = 0\) since:
\[
X(0) = A \cos(0) + B \sin(0) = A = 0.
\]
Thus, \(X(x) = B \sin(\sqrt{\lambda} x)\). Applying \(X(\pi) = 0\), we get:
\[
B \sin(\sqrt{\lambda} \pi) = 0.
\]
Since \(B \neq 0\), we must have:
\[
\sqrt{\lambda} \pi = n\pi \quad \text{for} \quad n = 1, 2, 3, \dots,
\]
which gives \(\lambda = n^2\).

Thus, the solution for \(X(x)\) is:
\[
X(x) = B \sin(nx), \quad n = 1, 2, 3, \dots
\]

### Step 3: Solve for \(Y(y)\)
The equation for \(Y(y)\) is:
\[
Y'(y) = -(1 + \lambda) Y(y).
\]
Substituting \(\lambda = n^2\), we get:
\[
Y'(y) = -(1 + n^2) Y(y).
\]
The general solution is:
\[
Y(y) = C e^{-(1 + n^2)y}.
\]

### Step 4: Form the general solution
The general solution for \(u(x, y)\) is:
\[
u(x, y) = \sum_{n=1}^{\infty} B_n \sin(nx) e^{-(1 + n^2)y}.
\]

### Step 5: Apply the boundary condition \(u(x, 1) = x\)
At \(y = 1\), the boundary condition is \(u(x, 1) = x\). Thus:
\[
\sum_{n=1}^{\infty} B_n \sin(nx) e^{-(1 + n^2)} = x.
\]
We express \(x\) as a Fourier sine series:
\[
x = \sum_{n=1}^{\infty} C_n \sin(nx).
\]
The Fourier coefficients \(C_n\) for \(x\) can be computed as:
\[
C_n = \frac{2}{\pi} \int_0^\pi x \sin(nx) \, dx.
\]
Using integration by parts:
\[
C_n = \frac{2}{\pi} \left( \frac{\pi (-1)^{n+1}}{n} \right) = \frac{2 (-1)^{n+1}}{n}.
\]

Thus, we have:
\[
B_n e^{

Solve the following boundary value problem using separation of variables: ∂²u/∂x² - ∂u/∂y = u, satisfying boundary conditions u(0, y) = u(π, y) = 0 and u(x, 1) = x.

Learn how to solve the boundary value problem ∂²u/∂x² - ∂u/∂y = u with boundary conditions using the separation of variables method. Explore Fourier series expansion and eigenvalue problems with detailed step-by-step solutions. Suitable for undergraduate students.

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