Question

In these exercises, the boundary value problems involve the same differential equation with different boundary conditions. (a) Obtain the general solution of the differential equation. (b) Apply the boundary conditions, and determine whether the problem has a unique solution, infinitely many solutions, or no solution. If the problem has a solution or solutions, specify them. $$ y^{\prime \prime}+\frac{1}{4} y=1 $$ $$ y(0)=-0 $$ \(y(0)=0, \quad y(\pi)=2\)

Short answer

Answer: The unique solution of the boundary value problem is $y(x) = -4 \cos(\frac{x}{2}) + 2 \sin(\frac{x}{2}) + 4$.

Step-by-step solution

Find the general solution

To find the general solution of the differential equation, we start by finding the complementary function (homogeneous solution) and then the particular integral (non-homogeneous solution). The given differential equation is: $$ y^{\prime \prime}+\frac{1}{4} y=1 $$

Find the complementary function

First, we find the homogeneous solution by solving the homogeneous equation: $$ y^{\prime \prime}+\frac{1}{4} y=0 $$ The characteristic equation of the above differential equation is: $$ m^2 + \frac{1}{4} = 0 $$ This has two roots, which are complex conjugates: $$ m_1 = \frac{i}{2}, \quad m_2 = -\frac{i}{2} $$ So, the complementary function is: $$ y_c(x) = C_1 \cos(\frac{x}{2}) + C_2 \sin(\frac{x}{2}) $$

Find the particular integral

Now, we need to find a particular solution for the non-homogeneous equation. Since the right-hand side of the equation is a constant, let's guess a constant function as a particular solution: $$ y_p(x) = K $$ With this guess, both the first and second derivatives are zero: $$ y_p^{\prime}(x) = y_p^{\prime \prime}(x) = 0 $$ Substituting this into the original equation, we have: $$ 0+\frac{1}{4}K=1 $$ Solving for K, we get: $$ K = 4 $$ So, the particular integral is: $$ y_p(x) = 4 $$

Combine the complementary function and particular integral

Now that we have the complementary function and the particular integral, we can find the general solution by combining them: $$ y(x) = y_c(x) + y_p(x) = C_1 \cos(\frac{x}{2}) + C_2 \sin(\frac{x}{2}) + 4 $$

Apply boundary conditions

Next, we will apply the given boundary conditions to find the solution that satisfies them. $$ y(0) = 0, \quad y(\pi) = 2 $$ Applying the first boundary condition: $$ 0 = C_1 \cos(0) + C_2 \sin(0) + 4 \Rightarrow C_1 = -4 $$ Now, applying the second boundary condition: $$ 2 = (-4) \cos(\frac{\pi}{2}) + C_2 \sin(\frac{\pi}{2}) + 4 \Rightarrow C_2=2 $$

Write down the solution

We have the values of the constants, so we can now write down the solution of the given boundary value problem: $$ y(x) = -4 \cos(\frac{x}{2}) + 2 \sin(\frac{x}{2}) + 4 $$ This problem has a unique solution, which is the function above.

Key concepts

General Solution

To start understanding boundary value problems, we need to grasp the idea of a general solution. In differential equations, the general solution represents the complete set of possible solutions. It includes both the complementary function and the particular integral.
These two parts account for different aspects of the differential equation:

• The complementary function (CF), also known as the homogeneous solution, solves the associated homogeneous equation—this is the equation without any external forces or inputs.
• The particular integral (PI) solves the inhomogeneous equation and represents the steady-state part of the solution that accounts for the specific non-zero right-hand side of the differential equation.

By combining these two components, the general solution can be formed, which satisfies both the internal dynamics of the system and the imposed external conditions.

Complementary Function

The complementary function is a crucial part of solving linear differential equations. For our problem, we found the complementary function by solving the homogeneous equation, which is:
\[ y'' + \frac{1}{4}y = 0 \]
Solving this involves finding the characteristic equation, \( m^2 + \frac{1}{4} = 0 \), and determining its roots. In this case, the roots are complex:

• \( m_1 = \frac{i}{2} \)
• \( m_2 = -\frac{i}{2} \)

Complex roots lead to a solution involving trigonometric functions, leading us to our complementary function:

• \( y_c(x) = C_1 \cos\left(\frac{x}{2}\right) + C_2 \sin\left(\frac{x}{2}\right) \)

These functions reflect oscillatory behavior, which is typical when the roots of the characteristic equation are complex.

Particular Integral

The particular integral is the part of the general solution that addresses the non-homogeneity of the differential equation—which in this case, comes from the constant on the right-hand side of the equation:\[ y'' + \frac{1}{4}y = 1 \]To find the particular integral, we assume a solution that suits the form of the non-homogeneous term. Here, since the non-homogenous term is a constant, we try a constant solution, \( y_p(x) = K \), where

• \( y_p'(x) = 0 \)
• \( y_p''(x) = 0 \)

Substituting these derivatives into the original equation, we solve for \( K \).
This gives:

• \( \frac{1}{4}K = 1 \) implying \( K = 4 \)

Thus, the particular integral is \( y_p(x) = 4 \), a constant solution that balances the equation.

Homogeneous Solution

The process of determining the complementary function involves solving the homogeneous differential equation, which excludes the non-homogeneous part. Therefore, the homogeneous equation for our case is:

• \( y'' + \frac{1}{4}y = 0 \)

For linear differential equations, the homogeneous solution is derived from the characteristic equation, where the roots determine the form of the solution.
In this exercise, the roots are complex, leading to:

• \( m^2 + \frac{1}{4} = 0 \) which results in \( m_1 = \frac{i}{2} \) and \( m_2 = -\frac{i}{2} \)

These complex roots imply the complementary function is expressed in terms of trigonometric functions, namely:

• \( y_c(x) = C_1 \cos\left(\frac{x}{2}\right) + C_2 \sin\left(\frac{x}{2}\right) \)

This solution captures the inherent behavior of the system as oscillatory, without any external forces reflected by the particular integral.