Question

Either solve the given boundary value problem or else show that it has no solution. \(y^{\prime \prime}+4 y=\cos x, \quad y(0)=0, \quad y(\pi)=0\)

Short answer

Answer: The solution to the given boundary value problem is \(y(x) = -\frac{1}{3}\cos(2x) + C_2\sin(2x) + \frac{1}{3}\cos x\).

Step-by-step solution

Solving the non-homogeneous ODE

First, we need to solve the given non-homogeneous ODE, which is: \(y^{\prime\prime} + 4y = \cos x\) We can rewrite the above equation as: \(y^{\prime\prime} + 4y = 0 + \cos x\) Here, the homogeneous part of the given equation is: \(y^{\prime\prime} + 4y = 0\). The solution to this equation, denoted by \(y_h\), has the general form: \(y_h = C_1\cos(2x) + C_2\sin(2x)\), where \(C_1\) and \(C_2\) are arbitrary constants. Next, we need to find a particular solution \(y_p\) for the non-homogeneous equation. We guess the form of \(y_p\) as: \(y_p = A\cos x + B\sin x\), where \(A\) and \(B\) are constants to be determined. We find the first and second derivatives of \(y_p\): \(y_p^{\prime} = -A\sin x + B\cos x\), \(y_p^{\prime\prime} = -A\cos x - B\sin x\) Now, we can substitute these expressions into the original non-homogeneous ODE: \((-A\cos x - B\sin x) + 4(A\cos x + B\sin x) = \cos x\) Comparing coefficients of \(\cos x\) and \(\sin x\) on both sides, we get the following system of equations: \((-A + 4A) = 1\) \((- B + 4B) = 0\) Solving this system, we find that \(A = \frac{1}{3}\) and \(B = 0\). So, the particular solution is \(y_p = \frac{1}{3}\cos x\). Now, we can combine the homogeneous and particular solutions to get the general solution of the given ODE: \(y(x) = y_h + y_p = C_1\cos(2x) + C_2\sin(2x) + \frac{1}{3}\cos x\)

Applying the boundary conditions

We have the boundary conditions: \(y(0) = 0\) and \(y(\pi) = 0\). Now we'll apply these conditions to the general solution we found, \(y(x)\), to see if it satisfies the given boundary conditions. For the condition \(y(0) = 0\), we substitute \(x = 0\) in the equation for \(y(x)\) and get: \(y(0) = C_1\cos(0) + C_2\sin(0) + \frac{1}{3}\cos(0) = C_1 + \frac{1}{3} = 0\) Thus, \(C_1 = -\frac{1}{3}\). For the condition \(y(\pi) = 0\), we substitute \(x = \pi\) in the equation for \(y(x)\) and get: \(y(\pi) = (-\frac{1}{3})\cos(2\pi) + C_2\sin(2\pi) + \frac{1}{3}\cos(\pi) = -\frac{1}{3} + \frac{1}{3} = 0\) Thus, the solution satisfies both of the given boundary conditions. Therefore, the solution of the given boundary value problem is: \(y(x) = -\frac{1}{3}\cos(2x) + C_2\sin(2x) + \frac{1}{3}\cos x\)

Key concepts

Understanding Non-Homogeneous ODEs

Ordinary Differential Equations (ODEs) are fundamental to the study of mathematical modeling, where we describe how a quantity changes with respect to another. Specifically, a non-homogeneous ODE includes a term that doesn't depend on the function we're trying to find. In our exercise, the ODE given is
\(y'' + 4y = \text{cos}(x) \).
The right side of this equation, \( \text{cos}(x) \), makes it non-homogeneous because it isn't zero and doesn't involve the function \(y\) or its derivatives. To tackle such an equation, we divide our approach into two parts: finding the homogeneous solution and finding the particular solution.
These two solutions will be combined to form the general solution of the non-homogeneous ODE. The homogeneous solution addresses the part of the ODE without the non-homogeneous term, and the particular solution specifically accounts for the non-zero term on the right side. Understanding this division helps in simplifying complex problems and finding the particular constants that satisfy the given boundary conditions.

The Homogeneous Equation

In the context of our exercise, we focus on the homogeneous equation when we remove the non-homogeneous term, resulting in \( y'' + 4y = 0 \).
This equation represents a scenario with no external forces acting on the system. The solutions form a vector space, and their combination, through superposition, gives us the complete set of solutions for the homogeneous part. In this case, the solutions are of the form \( C_1\text{cos}(2x) + C_2\text{sin}(2x) \), where \( C_1 \) and \( C_2 \) are constants that will be determined by boundary conditions.

• \( C_1 \) and \( C_2 \) allow an infinite number of solutions for the homogeneous equation.
• The sine and cosine functions reflect the periodic nature of the solutions.

The idea behind solving a homogeneous ODE is to find a basis set of solutions upon which any other solution can be built by linear combination.

Finding the Particular Solution

The particular solution is necessary to address the non-homogeneous part of the ODE, and constructing it can be more art than science. We predict a form that the solution could take, which often reflects the non-homogeneous term. In our exercise,
\(y_p = A\text{cos}(x) + B\text{sin}(x)\)
, was chosen as an educated guess, considering that the non-homogeneous term was a cosine function.

• The goal is to find values for \(A\) and \(B\) that will satisfy the original non-homogeneous ODE.
• Once \(A\) and \(B\) are determined, the particular solution will represent how the system behaves due to the non-zero term on the right side of the ODE.

It was found that \(A = \frac{1}{3}\) and \(B = 0\), which meant that the particular influence due to the cosine term in the non-homogeneous ODE only altered the amplitude of the function, not its frequency or phase. Upon combining the particular solution with the homogeneous solution under the boundary conditions, the constants turn into precise values, painting a complete picture of the system's behavior across the domain.