Problem 42 - Advanced Engineering Mathematics

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).] \(\begin{aligned} y^{\prime \prime}-2 y^{\prime}+10 y &=g(x), y(0)=0, y^{\prime}(0)=0, \text { where } \\ g(x) &=\left\\{\begin{array}{ll}20, & 0 \leq x \leq \pi \\ 0, & x>\pi\end{array}\right.\end{aligned}\)

Step-by-Step Solution

Answer

Solve for separate intervals, ensuring solutions have continuous derivatives at \(x = \pi\). Splitting leads to unique solutions on each interval.

Step 1: Identify the Type of Problem

The problem is an initial-value problem with a second-order linear differential equation, where the input function \(g(x)\) is piecewise, having a discontinuity at \(x = \pi\). This requires solving the equation on separate intervals.

Step 2: Define Initial Intervals and Conditions

We can split the solution into two intervals: 1. Interval \( [0, \pi] \) where \( g(x) = 20 \).2. Interval \( (\pi, \infty) \) where \( g(x) = 0 \).We need to ensure smooth joining of solutions at \( x = \pi\).

Step 3: Solve on Interval [0, \pi]

For \(0 \leq x \leq \pi\), the differential equation becomes: \( y'' - 2y' + 10y = 20 \).First, solve the homogeneous equation:\( y'' - 2y' + 10y = 0 \).The characteristic equation is \( r^2 - 2r + 10 = 0 \) with roots \( r = 1 \pm 3i \). Thus, the general solution to the homogeneous equation is:\( y_h = e^x (C_1 \cos(3x) + C_2 \sin(3x)) \).Find particular solution \( y_p \) by setting it as a constant since \( g(x) \) is constant:\( y_p = 2 \).Thus, the general solution is:\( y = y_h + y_p = e^x (C_1 \cos(3x) + C_2 \sin(3x)) + 2 \).Applying initial conditions \( y(0) = 0 \) and \( y'(0) = 0 \) helps find \( C_1 \) and \( C_2 \).\( y(0) = C_1 + 2 = 0 \implies C_1 = -2 \).Differentiate:\( y' = e^x (C_1 \cos(3x) + C_2 \sin(3x)) + e^x (-3C_1 \sin(3x) + 3C_2 \cos(3x)) \).\( y'(0) = -2 + 3C_2 = 0 \implies C_2 = \frac{2}{3} \). Thus, the solution on \([0, \pi]\) is:\( y(x) = 2e^x ( -\cos(3x) + \frac{1}{3} \sin(3x)) + 2 \).

Step 4: Solve on Interval (\pi, ∞)

For \(x > \pi\), the equation is: \( y'' - 2y' + 10y = 0 \).The general solution is given by:\( y(x) = e^x (C_3 \cos(3x) + C_4 \sin(3x)) \).We need to match this smoothly with the solution from \([0, \pi]\) at \(x = \pi\). This gives conditions:\( y(\pi) \) and \( y'(\pi) \) should be continuous.

Step 5: Ensure Continuity at x = π

We find \( y(\pi) \) and \( y'(\pi) \) from solutions on both intervals. Calculate these from the solution on \([0, \pi]\) to find values:\( y(\pi) = 2e^\pi ( -\cos(3\pi) + \frac{1}{3} \sin(3\pi)) + 2 \).This value should match \( y(\pi) = e^\pi (C_3 \cos(3\pi) + C_4 \sin(3\pi)) \) from \((\pi, \infty)\).Perform similar matching for \( y'(\pi) \). Solve for \( C_3 \) and \( C_4 \) using these equations.

Key Concepts

Second-Order Linear Differential Equation

A second-order linear differential equation is a type of equation that involves the second derivative of a function. Such equations are fundamental in describing systems with acceleration, such as mechanical and electrical systems. In the context of the problem, the equation given is:

• \(y'' - 2y' + 10y = g(x)\)

This equation is referred to as 'linear' because it can be expressed as a linear combination of the unknown function \(y\), its first derivative \(y'\), and its second derivative \(y''\). Here, the highest derivative is the second, making it a second-order equation. Dealing with such equations often involves finding a solution that satisfies both the equation and any initial conditions provided, like \(y(0)=0\) and \(y'(0)=0\). The combination of known functions and their derivatives forms the complete solution to the given problem. Intervals need to be considered separately due to the discontinuous input function \(g(x)\).

Piecewise Function

In mathematics, a piecewise function is a function composed of multiple sub-functions, each of which applies to a certain interval of the main function’s domain. These sub-functions can have different expressions.

• In this problem, the input function \(g(x)\) is defined as a piecewise function:

\[g(x) =\begin{cases} 20, & 0 \leq x \leq \pi \ 0, & x > \pi \end{cases}\]The piecewise nature here introduces a discontinuity at \(x = \pi\). This discontinuity means that the differential equation needs to be solved separately for each of the two intervals: \([0, \pi]\) and \((\pi, \infty)\). For \(0 \leq x \leq \pi\), \(g(x) = 20\), simplifying the equation to a non-homogeneous form with a constant. For \(x > \pi\), \(g(x) = 0\), simplifying to a homogeneous equation.

Continuity Conditions

Continuity conditions are crucial for linking solutions across intervals to ensure there is no sudden jumps or gaps at the point where the intervals meet. In this exercise, the problem requires solving two sets of differential equations on two overlapping intervals, and they must be joined smoothly at the point \(x = \pi\).

• First, solve the equation for each interval, say Interval 1 (\([0, \pi]\)) and Interval 2 (\((\pi, \infty)\)).
• Then, ensure that the solution from Interval 1 (ending at \(x = \pi\)) aligns perfectly with the beginning of Interval 2 (starting from \(x = \pi\)).

This requires that both the value of \(y\) and its first derivative \(y'\) are continuous at \(x = \pi\). Calculations include:

• The continuous value of \(y(\pi)\) which is obtained from both intervals.
• The continuous value of \(y'(\pi)\) from both overlapping intervals.

By solving these conditions simultaneously, we ensure there are no abrupt changes at the join.

Characteristic Equation

The characteristic equation comes into play when solving the homogeneous part of a second-order differential equation. It is essentially an algebraic relation derived from the differential equation itself, used to find the general solution of the homogeneous equation.

• For the differential equation \(y'' - 2y' + 10y = 0 \), the characteristic equation is obtained by assuming a solution of the form \(y = e^{rx}\):

\[r^2 - 2r + 10 = 0\]To solve this characteristic equation, find its roots using the quadratic formula. The roots are:

• \(r = 1 \pm 3i\)

These complex roots lead to a general solution for the homogeneous equation of the form:

• \(y_h = e^x (C_1 \cos(3x) + C_2 \sin(3x))\)

The characteristic equation, therefore, helps identify the type of functions (exponential or trigonometric) that form the basis of the solution. Understanding the characteristic roots is pivotal to constructing the complete solution and ensuring it satisfies any initial or boundary conditions.