Question

Determine the solution of each of the following boundary-value problems: (a) \(y^{\prime \prime}+y=1\), (c) \(y^{\prime \prime}=\sin x\), \(y(0)=0, \quad y(1)=0 .\) \(y(0)=0, \quad y(1)+2 y^{\prime}(1)=0 .\) (b) \(y^{\prime \prime}+4 y=e^{x}\), \(y(0)=0, \quad y^{\prime}(1)=0\)

Short answer

Question: For each of the boundary-value problems (a), (b), and (c), find the unique solution that satisfies both the differential equation and boundary conditions. Answer: (a) \(y(x) = -\cos{x} + \left(\frac{\cos{1} - 1}{\sin{1}}\right) \sin{x} + 1\) (b) \(y(x) = -\frac{1}{5}\cos{2x} + \frac{1}{10\cos{2}}\left(\frac{2}{5}\sin{2} - \frac{1}{5}e\right) \sin{2x} + \frac{1}{5}e^x\) (c) \(y(x) = \frac{1}{2}\left(\sin{1} + \cos{1} - 1\right) x - \sin{x}\)

Step-by-step solution

Problem (a)

First, the problem is given by \(y^{\prime \prime} + y = 1\). We begin by finding the homogeneous solution, \(y_h\), by solving the following equation: \(y_h^{\prime \prime} + y_h = 0\). With characteristic equation \(r^2 + 1 = 0\), the roots are \(r_1 = i\) and \(r_2 = -i\). Therefore, the homogeneous solution is given by: \(y_h(x) = C_1 \cos{x} + C_2 \sin{x}\) Next, we find the particular solution, \(y_p\), by guessing the form \(y_p(x) = A\). Substituting this guess into the original equation yields \(A = 1\). Thus, the particular solution is: \(y_p(x) = 1\). The general solution is given by \(y(x) = y_h(x) + y_p(x)\), so we have: \(y(x) = C_1 \cos{x} + C_2 \sin{x} + 1\) Now, we apply the boundary conditions:\(y(0) = 0\) and \(y(1) = 0\). This gives us: \(y(0) = C_1 + 1 = 0 \implies C_1 = -1\) \(y(1) = -\cos{1} + C_2 \sin{1} + 1 = 0\) Solve for \(C_2\), we get: \(C_2 = \frac{\cos{1} - 1}{\sin{1}}\) The solution for problem (a) is: \(y(x) = -\cos{x} + \left(\frac{\cos{1} - 1}{\sin{1}}\right) \sin{x} + 1\).

Problem (b)

The problem is given by \(y^{\prime \prime} + 4y = e^{x}\). We begin by finding the homogeneous solution, \(y_h\), by solving the following equation: \(y_h^{\prime \prime} + 4y_h = 0\). With characteristic equation \(r^2 + 4 = 0\), the roots are \(r_1 = 2i\) and \(r_2 = -2i\). Therefore, the homogeneous solution is given by: \(y_h(x) = C_1 \cos{2x} + C_2 \sin{2x}\) Next, we find the particular solution, \(y_p\), by guessing the form \(y_p(x) = Ae^x\). Substituting this guess into the original equation and solving for A, we get: \(y_p(x) = \frac{1}{5}e^x\). The general solution is given by \(y(x) = y_h(x) + y_p(x)\), so we have: \(y(x) = C_1 \cos{2x} + C_2 \sin{2x} + \frac{1}{5}e^x\) Now, we apply the boundary conditions:\(y(0) = 0\) and \(y^{\prime}(1) = 0\). This gives us: \(y(0) = C_1 + \frac{1}{5} = 0 \implies C_1 = -\frac{1}{5}\) The derivative of \(y(x)\) is: \(y^{\prime}(x) = -2C_1 \sin{2x} + 2C_2 \cos{2x} + \frac{1}{5}e^x\) Using the second boundary condition, \(y^{\prime}(1) = 0\), we get: \(0 = -\frac{2}{5}\sin{2} + 2C_2 \cos{2} + \frac{1}{5}e\) Solving for \(C_2\), we obtain: \(C_2 = \frac{1}{10\cos{2}}\left(\frac{2}{5}\sin{2} - \frac{1}{5}e\right)\) The solution for problem (b) is: \(y(x) = -\frac{1}{5}\cos{2x} + \frac{1}{10\cos{2}}\left(\frac{2}{5}\sin{2} - \frac{1}{5}e\right) \sin{2x} + \frac{1}{5}e^x\).

Problem (c)

The problem is given by \(y^{\prime \prime} = \sin x\). We begin by finding the homogeneous solution, \(y_h\), by solving the following equation: \(y_h^{\prime \prime} = 0\). With the characteristic equation \(r^2 = 0\), the roots are \(r_1 = r_2 = 0\). Therefore, the homogeneous solution is given by: \(y_h(x) = C_1 + C_2 x\) Next, we find the particular solution, \(y_p\), by guessing the form \(y_p(x) = A\cos x + B\sin x\). Taking the second derivative and substituting into the original equation, we get: \(y_p(x) = -A\sin x + B\cos x\) Comparing with the right-hand side of the original equation, we deduce \(A = 0\) and \(B = -1\). Thus, the particular solution is: \(y_p(x) = -\sin x\) The general solution is given by \(y(x) = y_h(x) + y_p(x)\), so we have: \(y(x) = C_1 + C_2 x - \sin{x}\) Now, we apply the boundary conditions:\(y(0) = 0\) and \(y(1) + 2y^{\prime}(1) = 0\). This gives us: \(y(0) = C_1 = 0\) The derivative of \(y(x)\) is: \(y^{\prime}(x) = C_2 - \cos{x}\) Using the second boundary condition, we get: \(0 = y(1) + 2y^{\prime}(1) = 1 - \sin{1} + 2(C_2 - \cos{1})\) Solving for \(C_2\), we obtain: \(C_2 = \frac{1}{2}\left(\sin{1} + \cos{1} - 1\right)\) The solution for problem (c) is: \(y(x) = \frac{1}{2}\left(\sin{1} + \cos{1} - 1\right) x - \sin{x}\).

Key concepts

Homogeneous Solutions

When dealing with boundary value problems, finding the homogeneous solution is a critical first step. The homogeneous solution, denoted by \( y_h \), is derived from the associated homogeneous equation. This equation arises by setting the non-homogeneous part (or the forcing function) to zero. Let's look at some key points:

• Consider a differential equation in the form \( y'' + ay = f(x) \). The homogeneous equation is \( y'' + ay = 0 \).
• To solve the homogeneous equation, assume a solution of the form \( y = e^{rx} \), which leads us to the characteristic equation.
• The roots of the characteristic equation determine the form of the homogeneous solution.
• For real and distinct roots, the solution will be a linear combination of exponentials.
• If roots are complex, like in problem (a), the solution involves trigonometric functions such as sine and cosine.

Understanding the homogeneous solution helps set the foundation for constructing the complete solution to a differential equation.
This complete solution includes contributions from both the homogeneous and particular solutions.

Characteristic Equation

The characteristic equation plays a pivotal role in solving homogeneous differential equations. It's an algebraic equation derived by plugging in a trial solution \( y = e^{rx} \) into the homogeneous differential equation.

• For a second-order differential equation like \( y'' + ay = 0 \), assume \( y = e^{rx} \). This substitution yields the characteristic equation \( r^2 + a = 0 \).
• The roots of the characteristic equation can be real, complex, or repeated.
• Real roots result in exponential solutions, whereas complex roots, such as those encountered in the original problems, lead to solutions involving trigonometric functions.
• The characteristic roots determine the basis of solutions for the homogeneous equation.
• In problem (a), with complex roots \( r = i \) and \( r = -i \), the solution formed by \( y_h(x) = C_1 \cos x + C_2 \sin x \) reflects the oscillatory nature typical of complex roots.

This characteristic equation approach is fundamental for finding whether the system exhibits exponential growth, decay, or oscillatory behaviors.

Particular Solutions

Finding a particular solution to a boundary value problem involves determining a function that satisfies the entire differential equation, not just the homogeneous part. Here’s a simple guide to understanding how particular solutions work:

• The particular solution \( y_p \) complements the homogeneous solution to form the general solution.
• There are several methods to find \( y_p \): the method of undetermined coefficients, variation of parameters, and sometimes simple guessing.
• In problems (a) and (b), undetermined coefficients are used by proposing functions whose forms reflect the non-homogeneous part (like choosing \( y_p(x) = A \) for a constant forcing function, or \( y_p(x) = Ae^x \) to match \( e^x \)).
• By substituting \( y_p \) back into the original equation, you can solve for any unknown constants to ensure that the equation holds true.
• The versatility in choosing or guessing an appropriate form for \( y_p \) lies in understanding the nature of the non-homogeneous function \( f(x) \).

The particular solution is often the key component in adjusting the complete solution to meet specific boundary conditions, which can otherwise be unattainable with only the homogeneous part.