Solving a Boundary-Value Problem: y'' + 3y = 9x, y(0) = 0, y(1) + y'(1) = 0

In this article, we will explore the solution to a boundary-value problem involving a second-order linear differential equation. The problem is given as follows:

Solve the boundary-value problem:

y'' + 3y = 9x
y(0) = 0
y(1) + y'(1) = 0

To solve this problem, we will follow these steps:

1. Solve the homogeneous equation: First, we need to find the general solution to the homogeneous equation, which is obtained by setting the right-hand side to zero.
2. Find a particular solution: Next, we need to find a particular solution to the non-homogeneous equation.
3. Apply the boundary conditions: Finally, we will apply the given boundary conditions to determine the constants in the general solution and obtain the final solution to the boundary-value problem.

Let's go through each step in detail:

Step 1: Solve the homogeneous equation

The homogeneous equation is given by:

y'' + 3y = 0

To solve this equation, we can use the characteristic equation method. The characteristic equation is:

r^2 + 3 = 0

The roots of the characteristic equation are:

r1 = √(-3) = ±√3i

Therefore, the general solution to the homogeneous equation is:

y_h(x) = C1 * cos(√3x) + C2 * sin(√3x)

where C1 and C2 are arbitrary constants.

Step 2: Find a particular solution

To find a particular solution to the non-homogeneous equation, we can use the method of undetermined coefficients. Since the right-hand side of the equation is a linear function of x, we can assume a particular solution of the form:

y_p(x) = Ax + B

Substituting this into the original equation, we get:

(Ax + B)'' + 3(Ax + B) = 9x
2A + 3A x + 3B = 9x

Comparing the coefficients, we obtain:

2A + 3A = 9
3B = 0

Solving this system of equations, we get:

A = 3
B = 0

Therefore, the particular solution is:

y_p(x) = 3x

Step 3: Apply the boundary conditions

The boundary conditions are:

y(0) = 0
y(1) + y'(1) = 0

Applying the first boundary condition, we get:

y_h(0) + y_p(0) = 0
C1 = 0

Substituting this into the general solution, we get:

y_h(x) = C2 * sin(√3x)

Applying the second boundary condition, we get:

y_h(1) + y_h'(1) + y_p(1) = 0
C2 * sin(√3) + C2 * √3 * cos(√3) + 3 = 0
C2 * (sin(√3) + √3 * cos(√3)) = -3
C2 = -3 / (sin(√3) + √3 * cos(√3))

Therefore, the final solution to the boundary-value problem is:

y(x) = y_h(x) + y_p(x)
     = C2 * sin(√3x) + 3x
     = -3 / (sin(√3) + √3 * cos(√3)) * sin(√3x) + 3x

This completes the solution to the given boundary-value problem.