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. $$ \begin{aligned} &y^{\prime \prime}+\frac{1}{4} y=1 \\ &y^{\prime}(0)=-2, \quad y(\pi)=0 \end{aligned} $$

Short answer

Question: Find the unique solution to the boundary value problem given by the following differential equation and boundary conditions: $$ y^{\prime \prime}+\frac{1}{4}y=1, \quad y^{\prime}(0)=-2, \quad y(\pi)=0 $$ Answer: The unique solution to the boundary value problem is: $$ y(x) = -8\cos{\frac{x}{2}} - 4\sin{\frac{x}{2}} + 4 $$

Step-by-step solution

Solve the homogeneous differential equation

To solve the homogeneous part of the ODE, we consider the following equation without the constant term on the right-hand side: $$ y^{\prime \prime}+\frac{1}{4}y=0 $$ We will look for the solutions of the form \(y = e^{rt}\), where \(r\) is a constant. Substituting this into the above equation, we get the auxiliary equation: $$ r^2 + \frac{1}{4}=0 $$ Solving this quadratic equation, we find the roots: $$ r = \pm \frac{\sqrt{-1}}{2}i = \pm \frac{i}{2} $$ Hence, the homogeneous general solution \(y_h\) is given by: $$ y_h(x) = C_1 e^{\frac{i}{2}x} + C_2 e^{-\frac{i}{2}x} $$ We can turn this into a real function using Euler's identity: \(e^{ix} = \cos{x} + i\sin{x}\). Therefore, the real form of the homogeneous general solution is: $$ y_h(x) = A\cos{\frac{x}{2}} + B\sin{\frac{x}{2}} $$ where \(A = C_1 + C_2\) and \(B = i(C_1 - C_2)\) are real constants.

Solve the non-homogeneous differential equation

Now, we need to find a particular solution \(y_p\) for the non-homogeneous part of the ODE. Since the right-hand side is a constant, we assume a particular solution of the form $$ y_p = K $$ where \(K\) is a constant. Substituting this into the non-homogeneous differential equation, we get: $$ 0+\frac{1}{4}K=1 $$ Solving for \(K\), we find: $$ K = 4 $$ Thus, the particular solution \(y_p(x) = 4\).

Combine homogeneous and particular solutions

The general solution \(y(x)\) can be expressed as the sum of the homogeneous and particular solutions: $$ y(x) = y_h(x) + y_p(x) = A\cos{\frac{x}{2}} + B\sin{\frac{x}{2}} + 4 $$

Apply the boundary conditions

We will now use the given boundary conditions to find the constants \(A\) and \(B\): 1. \(y^{\prime}(0)=-2\): Take the derivative of \(y(x)\) with respect to \(x\): $$ y^{\prime}(x) = -\frac{A}{2}\sin{\frac{x}{2}} + \frac{B}{2}\cos{\frac{x}{2}} $$ Substituting \(x=0\), we get: $$ -\frac{A}{2}\cdot\sin{0} + \frac{B}{2}\cdot\cos{0} = -2 \\ B = -4 $$ 2. \(y(\pi) = 0\): Substituting \(x = \pi\) and \(B=-4\) into the general solution \(y(x)\): $$ A\cdot\cos{\frac{\pi}{2}} - 4\cdot\sin{\frac{\pi}{2}} + 4 = 0 $$ Solving for \(A\), we find: $$ A = - 8 $$

Write the final solution

We found \(A = -8\) and \(B = -4\). Substituting these values into the general solution, we obtain the unique solution to the boundary value problem: $$ y(x) = -8\cos{\frac{x}{2}} - 4\sin{\frac{x}{2}} + 4 $$

Key concepts

Homogeneous Differential Equation

A homogeneous differential equation is a type of differential equation that represents a system or process in which the external force or input is zero. In mathematical terms, it appears without a non-zero constant or function on the right-hand side. The homogeneous differential equation from the exercise given is:\[y^{\prime \prime} + \frac{1}{4}y = 0\]To solve this kind of equation, you search for solutions that inherently involve exponential functions of the form \( y = e^{rt} \), where \( r \) are the roots of the characteristic or auxiliary equation derived from the original differential expression. For this particular equation, solving the auxiliary equation:\[r^2 + \frac{1}{4} = 0\] yields complex roots \( r = \pm \frac{i}{2} \). This indicates the solution involves imaginary numbers, leading to a general solution:\[ y_h(x) = C_1 e^{\frac{i}{2}x} + C_2 e^{-\frac{i}{2}x} \]By applying Euler's formula, we transform it into a more practical real-number function involving trigonometric terms:\[y_h(x) = A\cos{\frac{x}{2}} + B\sin{\frac{x}{2}}\]In which \( A \) and \( B \) are real constants determined by the boundary conditions.

Non-Homogeneous Differential Equation

A non-homogeneous differential equation includes a term that is non-zero, reflecting the presence of an external influence or force. In our exercise, the non-homogeneous equation is given as:\[y^{\prime \prime} + \frac{1}{4} y = 1\]Here, the constant 1 on the right-hand side represents the non-homogeneous part. Such equations require a different approach than homogeneous ones, as the general solution comprises two parts:

• Homogeneous Solution: The solution to the associated homogeneous equation, \(y_h(x)\).
• Particular Solution: A specific solution, \(y_p(x)\), that satisfies the non-homogeneous equation.

For simple non-homogeneous terms like constants, one often assumes a particular solution of the same type. In this case, we assumed a constant particular solution, \(y_p = K\), solving for \(K\) as follows:\[0 + \frac{1}{4}K = 1\]Solving gives \(K = 4\), so the particular solution is \(y_p(x) = 4\). This constant is then added to the homogeneous solution to form the general solution to the differential equation.

Euler's Formula

Euler's formula is a fundamental equation in complex analysis. It links complex exponential functions to trigonometric functions and is particularly useful in real-world applications. Euler's formula is expressed as:\[e^{ix} = \cos{x} + i\sin{x}\]In differential equations involving complex roots, such as our case with roots \(r = \pm \frac{i}{2} \), Euler's formula helps convert solutions from complex exponentials into practical real-valued trigonometric functions. Using this formula, the homogeneous solution:\[y_h(x) = C_1 e^{\frac{i}{2}x} + C_2 e^{-\frac{i}{2}x}\]is rewritten in terms of sine and cosine:\[y_h(x) = A\cos{\frac{x}{2}} + B\sin{\frac{x}{2}}\]This transformation is crucial for working with real-valued variables, especially when applying boundary conditions that stipulate specific real-world constraints or outputs.

Particular Solution

In solving non-homogeneous differential equations, finding a particular solution involves determining a specific function that, when substituted into the original equation, satisfies the non-homogeneous term. This is different from the homogeneous solution, which represents the natural response of the system.For our exercise, given the equation:\[y^{\prime \prime} + \frac{1}{4} y = 1\]we assumed the simplest form of particular solution when the non-homogeneous term is a constant, by setting \(y_p = K\). Substituting back into the equation allows us to solve for \(K\). Here we found:\[\frac{1}{4}K = 1\]Solving yields \(K = 4\), thus the particular solution is \(y_p(x) = 4\). The combined general solution for the equation includes both this particular solution and the homogeneous solution \(y_h(x)\):\[y(x) = y_h(x) + y_p(x)\]This approach often involves guessing the form of \(y_p(x)\) based on the function type on the right-hand side of the equation, iteratively refining until the non-homogeneous term is exactly matched.