Problem 42 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given initial-value problem in which the input function \(g(x)\) is discontinuous. [Hint: Solve each problem on two intervals, and then find a solution so that \(y\) and \(y^{\prime}\) are continuous at \(x=\pi / 2\) (Problem 41 ) and at \(x=\pi\) (Problem 42 ). $$\begin{aligned}&y^{\prime \prime}-2 y^{\prime}+10 y=g(x), \quad y(0)=0, y^{\prime}(0)=0, \quad \text { where }\\\&g(x)=\left\\{\begin{array}{ll}20, & 0 \leq x \leq \pi \\\0, & x>\pi\end{array}\right.\end{aligned}$$

Step-by-Step Solution

Answer

The solution involves matching the intervals at \(x=\pi\) to ensure continuity of \(y\) and \(y'\).

Step 1: Solve the homogenous equation

First, we need to solve the homogenous differential equation \( y'' - 2y' + 10y = 0 \). The characteristic equation is \( r^2 - 2r + 10 = 0 \). Solving this gives \( r = 1 \pm 3i \), which implies the general solution of the homogenous equation is \( y_h(x) = e^x (C_1 \cos(3x) + C_2 \sin(3x)) \).

Step 2: Solve for each interval's particular solution

For the interval \( 0 \leq x \leq \pi \), solve \( y'' - 2y' + 10y = 20 \). Assume a constant particular solution \( y_{p1}(x) = A \). Substituting into the equation gives \( 10A = 20 \), so \( A = 2 \). Therefore, for \( 0 \leq x \leq \pi \), \( y_{p1}(x) = 2 \). For \( x > \pi \), the equation becomes \( y'' - 2y' + 10y = 0 \), and we've already found the general solution \( y_h(x) \).

Step 3: Formulate the general solutions for each interval

Combine the particular and homogenous solutions. For \( 0 \leq x \leq \pi \), the general solution is \( y(x) = e^x(C_1 \cos(3x) + C_2 \sin(3x)) + 2 \). For \( x > \pi \), the general solution remains the homogenous solution \( y(x) = e^x (D_1 \cos(3x) + D_2 \sin(3x)) \).

Step 4: Apply initial conditions and solve for constants

Since \( y(0) = 0 \) and \( y'(0) = 0 \), apply these to determine \( C_1 \) and \( C_2 \). Plugging \( x=0 \) into \( y(0) = e^0(C_1 \cos(0) + C_2 \sin(0)) + 2 = 0 \), gives \( C_1 + 2 = 0 \), hence \( C_1 = -2 \). For \( y'(0) = 0 \), differentiate: \( y'(x) = e^x((C_1 \cos(3x) + C_2 \sin(3x))' + (C_1 \cos(3x) + C_2 \sin(3x)) ) \) and solve \( y'(0) \).

Step 5: Ensure continuity at boundary \(x = \pi\)

We impose continuity conditions at \( x = \pi \) by setting the expressions for \( y \) and \( y' \) from both intervals equal. Solve the system of equations \( y(\pi^-)=y(\pi^+) \) and \( y'(\pi^-)=y'(\pi^+) \). Substitute known forms of \( y \) and their derivatives, solve the system for \( D_1, D_2 \) to ensure continuity.

Key Concepts

Initial-Value Problems

Initial-value problems involve finding a solution to a differential equation that satisfies specific conditions at a particular point. In our exercise, these conditions are given as \( y(0) = 0 \) and \( y'(0) = 0 \). This means that at the starting point \( x=0 \), both the function \( y(x) \) and its derivative \( y'(x) \) need to align with the given values. This ensures that the solution not only satisfies the differential equation but also adheres to the predefined conditions at the initial point.

By substituting these initial conditions into our general solution for the interval \( 0 \leq x \leq \pi \), we can solve for any unknown constants that determine the unique solution. In our problem, these conditions help us determine the constants \( C_1 \) and \( C_2 \) necessary for the complete solution. The idea is to ensure that the calculated solution behaves correctly at the initial point and remains consistent with the problem constraints.

Homogenous Differential Equations

Homogenous differential equations are a type of differential equation where the function and all its derivatives are equated to zero. In this exercise, the homogenous equation is \( y'' - 2y' + 10y = 0 \). To solve it, one typically finds the characteristic equation, which here is \( r^2 - 2r + 10 = 0 \).

The roots of this characteristic equation provide the solution's form. With roots \( r = 1 \pm 3i \) indicating complex roots, the solution is a combination of exponential growth and sinusoidal functions, given by \( y_h(x) = e^x (C_1 \cos(3x) + C_2 \sin(3x)) \).

This solution reflects the general pattern a homogenous equation solution will follow, exhibiting an oscillatory behavior modified by exponential terms. Understanding this solution conceptually means recognizing the role of the characteristic equation and how its roots shape the solution.

Particular Solutions

Particular solutions provide us with a specific solution to a non-homogenous differential equation, which has non-zero terms on one side of the equation. In the first interval of the problem \( 0 \leq x \leq \pi \), we look for a particular solution of the form \( y_{p1}(x) = A \). The purpose here is to satisfy the differential equation \( y'' - 2y' + 10y = 20 \) with a straightforward constant solution.

By substituting \( y_{p1}(x) = A \) into the differential equation, we solve for \( A \) to find that \( A = 2 \). On this interval, the particular solution adds a constant term to the homogenous solution, ensuring that the equation accommodates the specific forcing function \( g(x) = 20 \).

In contrast, for \( x > \pi \), the particular solution disappears as \( g(x) = 0 \), reverting only to the homogenous component. This transition between particular and homogenous solutions in different intervals underscores the importance of finding correct particular solutions within the distinct regions of the problem.

Continuity Conditions

Continuity conditions are critical in ensuring solutions seamlessly connect across different intervals. At points of discontinuity, such as \( x = \pi \) in our problem, we want to ensure that both the solution \( y(x) \) and its derivative \( y'(x) \) are continuous.

This is done by setting the solutions and their derivatives from each interval equal at the boundary point. This process involves expressing \( y(x) \) from both sides of the discontinuity and adjusting constants in each solution so that \( y(\pi^-) = y(\pi^+) \) and the same for derivatives. It ensures that there are no sudden jumps or breaks in the function's behavior.

In our solution, this helps determine the constants \( D_1 \) and \( D_2 \) for \( x > \pi \), providing a smooth transition across the interval, and maintaining physical or theoretical consistency wherever the conditions are applied.