Using second order finite difference method, solve the boundary value problem  ,  ,  ,   

To solve the boundary value problem \( y'' + 5y' + 4y = 1 \) with the boundary conditions \( y(0) = 0 \) and \( y(1) = 0 \) using the second-order finite difference method with \( h = \frac{1}{4} \), we'll discretize the domain and approximate the derivatives.

Let's denote \( y_i \) as the approximation of \( y(x_i) \) where \( x_i = ih \), \( i = 0, 1, 2, ..., N \), and \( h \) is the step size.

The second-order finite difference approximation for the second derivative is:

\[ y''(x_i) \approx \frac{y_{i+1} - 2y_i + y_{i-1}}{h^2} \]

The first derivative approximation can be obtained similarly.

Given the boundary conditions, we can write the system of equations for each interior point \( i = 1, 2, ..., N-1 \) as:

\[ \frac{y_{i+1} - 2y_i + y_{i-1}}{h^2} + 5\frac{y_{i+1} - y_{i-1}}{2h} + 4y_i = 1 \]

Now, we'll solve this system of equations to find \( y_i \) for \( i = 1, 2, ..., N-1 \), using the boundary conditions \( y_0 = 0 \) and \( y_N = 0 \).

Let's denote \( N = \frac{1}{h} \) to represent the number of intervals.

Given \( h = \frac{1}{4} \), we have \( N = 4 \). Now, let's solve the system of equations for \( y_1, y_2, y_3 \).

To solve the boundary value problem using the second-order finite difference method with \( h = \frac{1}{4} \), let's first set up the system of equations for the interior points \( i = 1, 2, 3 \):

1. For \( i = 1 \):
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