Problem 41 - Advanced Engineering Mathematics

Question

In Problems 41 and 42, 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).] \(y^{\prime \prime}+4 y=g(x), y(0)=1, y^{\prime}(0)=2\), where $$ g(x)=\left\\{\begin{array}{ll} \sin x, & 0 \leq x \leq \pi / 2 \\ 0, & x>\pi / 2 \end{array}\right. $$

Step-by-Step Solution

Answer

Solve on intervals [0, π/2] and (π/2, ∞), ensuring continuity at π/2.

Step 1: Solve the Differential Equation on Interval [0, π/2]

On the interval \(0 \leq x \leq \pi/2\), the input function is \(g(x) = \sin x\). The differential equation becomes:\[ y'' + 4y = \sin x. \]The complementary solution to \(y'' + 4y = 0\) is \(y_c = C_1 \cos(2x) + C_2 \sin(2x)\).To find the particular solution, use undetermined coefficients and try \(y_p = A \sin x + B \cos x\). Substitute into the differential equation to obtain:\[ -A \sin x - B \cos x + 4A \sin x + 4B \cos x = \sin x. \]By equating coefficients, determine \(A = 1/3\), \(B = 0\). Thus, the particular solution is \(y_p = \frac{1}{3} \sin x\).The general solution on this interval is:\[ y_1(x) = C_1 \cos(2x) + C_2 \sin(2x) + \frac{1}{3} \sin x. \]Apply initial conditions \(y(0) = 1\) and \(y'(0) = 2\) to find \(C_1\) and \(C_2\).

Step 2: Solve the Differential Equation on Interval (π/2, ∞)

On the interval \(x > \pi/2\), the input function is \(g(x) = 0\). The differential equation becomes:\[ y'' + 4y = 0. \]The general solution is:\[ y_2(x) = D_1 \cos(2x) + D_2 \sin(2x). \]There are no additional input functions to consider here, which simplifies the equation to its complementary form.

Step 3: Ensure Continuity of y and y' at x = π/2

To ensure the solution is valid for all \(x\), enforce the conditions that \(y_1(\pi/2) = y_2(\pi/2)\) and \(y_1'(\pi/2) = y_2'(\pi/2)\).Calculate \(y_1(\pi/2)\) and \(y_1'(\pi/2)\) from \(y_1(x)\) and enforce these values to be equal to \(y_2(\pi/2)\) and \(y_2'(\pi/2)\) from \(y_2(x)\).By solving these two equations, determine the constants \(D_1\) and \(D_2\) such that the continuity conditions are satisfied.

Key Concepts

Initial-Value Problem

An initial-value problem involves finding a function that satisfies a given differential equation, subject to specified initial conditions. These initial conditions typically provide specific values for the function and its derivatives at a certain point. In mathematical terms, an initial-value problem can be expressed as a differential equation, usually second-order in our context, with conditions like:

• Function value: \( y(x_0) = y_0 \)
• Derivative value: \( y'(x_0) = y_0' \)

Given this setup, you are tasked with determining a solution that not only satisfies these conditions but also adheres to the overall equation within its domain. In the original problem, we see this clearly with the conditions \(y(0)=1\) and \(y'(0)=2\). These conditions guide us in finding the constants in the solution, ensuring it's specific to the scenario described. Thus, an initial-value problem challenges us to not only solve a differential equation but also to make our solution personally meaningful at initial points.

Discontinuous Input Function

In differential equations, an input function can often be discontinuous. This means that the function abruptly changes values or behavior at certain points. Dealing with discontinuous input functions can be tricky, but it's an important skill.
For example, consider the function \(g(x)\) given in the exercise:

• \( g(x) = \sin x \) for \(0 \leq x \leq \pi/2\)
• \( g(x) = 0 \) for \(x > \pi/2\)

Here, the function changes from \(\sin x\) to \(0\) exactly at \(x = \pi/2\). Discontinuous functions are often handled by solving the differential equation separately in each interval where the function is continuous, as exemplified in the original problem by solving over \([0, \pi/2]\) and \((\pi/2, \infty)\).
The main goal when working with these functions is to ensure continuity and coherence in your solution across these change-points, which often requires matching conditions at these points, such as continuity in the function and its first derivative.

Undetermined Coefficients

The method of undetermined coefficients is used to find a particular solution to a non-homogeneous differential equation. This technique leverages the form of the non-homogeneous term, here termed \(g(x)\), and assumes a form for the particular solution.
The power of this method lies in its ability to turn an initially complex equation into a simpler algebraic problem by assuming a particular form for the solution that "looks like" the non-homogeneous term. For instance, in the original exercise, our differential equation had \(g(x)\) as \(\sin x\). So, we guessed:

• \( y_p = A \sin x + B \cos x \)

We then substituted this assumed form into the equation, allowing us to solve for the constants \(A\) and \(B\). Here, the values turned out to be \(A = 1/3\) and \(B = 0\), giving a particular solution of \(y_p = \frac{1}{3} \sin x\).
The method of undetermined coefficients works well with periodic functions like sines and cosines, exponentials, and polynomials, making it a go-to approach when the non-homogeneous term fits these profiles.