To solve the differential equation $x y'' + y = 0$, with boundary conditions $y(1) = 1$, $y(2) = 2$, and step size $h = 0.25$, using the finite difference method, we'll proceed step-by-step.

Finite Difference Approximation:

For the equation $x y'' + y = 0$, we need to approximate the second derivative $y''$ using finite differences. At each point $x_i$, the second derivative can be approximated as: $y''(x_i) \approx \frac{y_{i+1} - 2y_i + y_{i-1}}{h^2}$ where $y_i$ is the value of $y$ at $x_i$, and $h = 0.25$ is the step size.

Applying to the Equation:

Substituting the finite difference approximation of $y''$ into the original equation gives us a system of equations at each point: $x_i \left( \frac{y_{i+1} - 2y_i + y_{i-1}}{h^2} \right) + y_i = 0$ or $\frac{x_i}{h^2} (y_{i+1} - 2y_i + y_{i-1}) + y_i = 0$

Mesh Points:

The boundary values are given at $x_1 = 1$ and $x_n = 2$. With $h = 0.25$, the points in between will be:

• $x_1 = 1$
• $x_2 = 1.25$
• $x_3 = 1.5$
• $x_4 = 1.75$
• $x_5 = 2$

At $x_1 = 1$, we have the boundary condition $y_1 = 1$. At $x_5 = 2$, we have the boundary condition $y_5 = 2$.

System of Equations:

We now set up the system of linear equations for the unknowns $y_2, y_3, y_4$. Using the finite difference method, we will form equations at each interior point $x_2, x_3, x_4$.

For $x_2 = 1.25$: $\frac{1.25}{0.25^2} (y_3 - 2y_2 + y_1) + y_2 = 0$ Simplifying: $80 (y_3 - 2y_2 + 1) + y_2 = 0 \quad \text{or} \quad 80y_3 - 159y_2 + 80 = 0$

For $x_3 = 1.5$: $\frac{1.5}{0.25^2} (y_4 - 2y_3 + y_2) + y_3 = 0$ Simplifying: $96 (y_4 - 2y_3 + y_2) + y_3 = 0 \quad \text{or} \quad 96y_4 - 191y_3 + 96y_2 = 0$

For $x_4 = 1.75$: $\frac{1.75}{0.25^2} (y_5 - 2y_4 + y_3) + y_4 = 0$ Simplifying: $112 (2 - 2y_4 + y_3) + y_4 = 0 \quad \text{or} \quad -224y_4 + 112y_3 + 112 + y_4 = 0 \quad \text{or} \quad 113y_4 - 112y_3 = 224$

Matrix Form:

We can write the system of equations for $y_2, y_3, y_4$ as: