Question

Show that the problem $$ y^{\prime \prime}+\pi^{2} y=\pi^{2} x, \quad y(0)=1, \quad y(1)=0 $$ has the solution $$ y=c_{1} \sin \pi x+\cos \pi x+x $$ Also show that this solution cannot be obtained by splitting the problem as suggested in Problem \(15,\) since neither of the two subsidiary problems can be solved in this case.

Short answer

#Answer# (c) The given function \(y(x) = c_1 \sin(\pi x) + \cos(\pi x) + x\) is indeed a solution to the given ODE and satisfies the boundary conditions when \(c_1=0\), resulting in \(y(x) = \cos(\pi x) + x\). However, when trying to split the problem into two subsidiary problems, as suggested in Problem 15, no solution for the subsidiary problems can be found. Thus, the given solution can't be obtained by splitting the problem in this way.

Step-by-step solution

Check if the given function is the solution of the ODE and satisfies the boundary conditions

To check if the given function y(x) is a solution to the ODE and satisfies the boundary conditions, we need to find its first and second derivatives, then substitute them into the ODE and the boundary conditions. Given function: \(y(x) = c_1 \sin(\pi x) + \cos(\pi x) + x\) First derivative: \(y'(x) = \pi c_1 \cos(\pi x) - \pi \sin(\pi x) + 1\) Second derivative: \(y''(x) = -\pi^2 c_1 \sin(\pi x) - \pi^2 \cos(\pi x)\) Now, substitute these derivatives into the ODE: \(y''(x) + \pi^2 y(x) = -\pi^2 c_1 \sin(\pi x) - \pi^2 \cos(\pi x) + \pi^2 (c_1 \sin(\pi x) + \cos(\pi x) + x) = \pi^2 x\) This simplifies to: \(\pi^2 x = \pi^2 x\) Thus, the given function y(x) solves the ODE for any value of \(c_1\). Now let's check the boundary conditions: \(y(0) = c_1 \sin(\pi \cdot 0) + \cos(\pi \cdot 0) = 1 \Rightarrow c_1 = 0\) This means \(y(x) = \cos(\pi x) + x\) Now check the second boundary condition: \(y(1) = \cos(\pi \cdot 1) + 1 = 0\) which holds true. Therefore, y(x) does satisfy the ODE and the given boundary conditions.

Analyze splitting the problem and show the solution cannot be obtained in this way

Problem 15 suggests splitting the given problem into two subsidiary problems. Let's split the ODE and boundary conditions: First subsidiary problem: \(y_1''(x) + \pi^2 y_1(x) = 0\) \(y_1(0) = 1\), \(y_1(1) = 0\) Second subsidiary problem: \(y_2''(x) + \pi^2 y_2(x) = \pi^2 x\) \(y_2(0) = 0\), \(y_2(1) = 0\) Now let's try to solve the first subsidiary problem: For a homogeneous linear ODE, the solution has the form: \(y_1(x) = A\cos(\pi x) + B\sin(\pi x)\) Insert the boundary conditions to determine the values of A and B: \(1 = y_1(0) = A\cos(\pi \cdot 0) + B\sin(\pi \cdot 0) \Rightarrow A = 1\) \(0 = y_1(1) = \cos(\pi \cdot 1) + B\sin(\pi \cdot 1)\) For B=0, we get an incorrect value for \(y_1(1)\). Therefore, the first subsidiary problem cannot be solved. Similarly, the second subsidiary problem cannot be solved. This is because it is an inhomogeneous linear ODE with boundary conditions that are not consistent with the given source term (\(\pi^2 x\)). Therefore, it is impossible to satisfy both boundary conditions given the source term. Hence, we've shown that the given solution for the original problem cannot be obtained by splitting the problem as suggested in Problem 15.

Key concepts

Second Order Linear Differential Equations

Second order linear differential equations are equations that involve the second derivative of a function. These equations have the general form \(a(x)y'' + b(x)y' + c(x)y = f(x)\), where \(a(x)\), \(b(x)\), and \(c(x)\) are functions of \(x\), and \(f(x)\) is a given function called the forcing function or source term.
- **Order**: Refers to the highest derivative present in the equation. Here, the second derivative \(y''\) gives it the name "second order."
- **Linearity**: The equation is linear if \(y\), \(y'\), and \(y''\) are not multiplied together or appear as any power other than one. This means the unknown function and its derivatives appear to the power of one.
In the given exercise, the equation \(y'' + \pi^2 y = \pi^2 x\) represents a second order linear differential equation. Here, \(f(x) = \pi^2 x\), and the terms involving \(y\) and its derivatives are linear, making the equation linear. Such equations are prevalent in physics and engineering, where they are used to model phenomena such as mechanical vibrations and electrical circuits. Understanding these equations involves finding solutions that satisfy both the differential equation and any specified boundary or initial conditions.

Inhomogeneous Differential Equations

Inhomogeneous differential equations, also known as non-homogeneous equations, are those where the right-hand side (RHS) is not equal to zero. This RHS function is referred to as the "forcing function" or "non-homogeneous term."
In the context of the exercise, the equation \(y'' + \pi^2 y = \pi^2 x\) is inhomogeneous due to the term \(\pi^2 x\) on the RHS. Solving such equations typically involves two steps:

• First, find the general solution to the associated homogeneous equation \(y'' + \pi^2 y = 0\). This involves solving for solutions where the RHS is zero.
• Next, find a particular solution to the inhomogeneous equation by considering the presence of non-zero terms on the RHS.

These solutions are then combined to form the general solution of the inhomogeneous equation. A particular solution is tailored to satisfy the non-zero RHS, while the general solution of the homogeneous part can be adjusted with constants to match boundary conditions.

Boundary Conditions

Boundary conditions are additional constraints imposed on differential equations to make them useful in real-world applications. They specify the values of the solution at specific points and are necessary for determining the particular solution of a differential equation.
In the provided problem, we have two boundary conditions: \( y(0) = 1 \) and \( y(1) = 0 \). This means:

• At \( x = 0 \), the value of \( y \) is 1.
• At \( x = 1 \), the value of \( y \) is 0.

These conditions are crucial because they directly influence the values of the arbitrary constants in the solutions of differential equations. Solving the given problem involved ensuring that any solution meets these conditions, which is often essential for applications such as determining physical displacements, velocities in dynamics, or potentials in electronic circuits. Boundary conditions help narrow down possible solutions to one that realistically represents the situation being modeled.