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^{