Problem 32 - Advanced Engineering Mathematics

Question

For \(y^{\prime}+y=f(x)\) an integrating factor is \(e^{x}\) so that $$y e^{x}=\left\\{\begin{array}{ll} e^{x}+c_{1}, & 0 \leq x \leq 1 \\ -e^{x}+c_{2}, & x>1 \end{array}\right.$$ If \(y(0)=1\) then \(c_{1}=0\) and for continuity we must have \(c_{2}=2 e\) so that $$y=\left\\{\begin{array}{ll} 1, & 0 \leq x \leq 1 \\ 2 e^{1-x}-1, & x>1 \end{array}\right.$$

Step-by-Step Solution

Answer

The piecewise solution for the differential equation is \( y=1 \) for \( 0\leq x \leq 1 \) and \( y=2e^{1-x}-1 \) for \( x>1 \).

Step 1: Identify the Differential Equation

The given differential equation is \( y^{\prime} + y = f(x) \). We are tasked to solve for \( y \) using the integrating factor method.

Step 2: Determine the Integrating Factor

We are given that the integrating factor is \( e^{x} \). The purpose of using the integrating factor is to make the left-hand side of the differential equation into the derivative of a product. This helps us integrate both sides easily.

Step 3: Multiply Equation by the Integrating Factor

Multiply each term of the differential equation \( y^{\prime} + y = f(x) \) by the integrating factor \( e^x \). This transforms the equation into \( e^x y^{\prime} + e^x y = e^x f(x) \).

Step 4: Express Left-Hand Side as a Single Derivative

The left-hand side is now \( \frac{d}{dx}(e^x y) \), since \( \frac{d}{dx}(e^x y) = e^x y^{\prime} + e^x y \). Hence, the equation becomes \( \frac{d}{dx}(e^x y) = e^x f(x) \).

Step 5: Integrate Both Sides

Integrate both sides with respect to \( x \):\[ \int \frac{d}{dx}(e^x y) \, dx = \int e^x f(x) \, dx \]The left-hand side integration results in \( e^x y \), and the solution depends on the form \( \int e^x f(x) \, dx \), provided in the piecewise defined equation.

Step 6: Use Initial Condition \( y(0)=1 \)

For \( 0 \leq x \leq 1 \), the equation is \( e^x y = e^x + c_1 \). Given \( y(0) = 1 \), substitute into the equation: \( e^0 \cdot 1 = e^0 + c_1 \), which simplifies to \( 1 = 1 + c_1 \). Thus, \( c_1 = 0 \).

Step 7: Check Continuity at \( x=1 \) and Find \( c_2 \)

Continuity at \( x=1 \) requires that \( y \) is continuous across the boundary conditions. - For \( 0 \leq x \leq 1 \), \( y = 1 \) at \( x = 1 \).- For \( x > 1 \), \( e^x y = -e^x + c_2 \), leading to \( y = 2e^{1-x} - 1 \) for \( x = 1 \).Equating the functions at \( x=1 \) gives \( 2e^{1-1} - 1 = 1 \), means \( 2e - 1 = 1 \), implying \( c_2 = 2e \).

Step 8: Write the Piecewise Solution

The solution of the differential equation is:\[ y = \begin{cases} 1, & 0 \leq x \leq 1 \ 2e^{1-x} - 1, & x > 1 \end{cases} \]

Key Concepts

Integrating Factor

The integrating factor is a crucial tool in solving first-order linear differential equations. It's used to simplify these equations and make them easier to solve. The integrating factor is typically denoted as \( \mu(x) \), and it transforms a linear differential equation into a form that can be easily integrated. For an equation of the form \( y' + P(x)y = Q(x) \), the integrating factor is obtained as \( \mu(x) = e^{\int P(x) \, dx} \).

In this exercise, we have that \( P(x) = 1 \), therefore, the integrating factor becomes \( e^x \). By multiplying the entire differential equation \( y' + y = f(x) \) by \( e^x \), the left-hand side becomes a single derivative, \( \frac{d}{dx}(e^x y) \). This conversion enables us to integrate both sides of the equation seamlessly to find \( y \).

• Step: Multiply the equation by the integrating factor \( e^x \).
• Result: The equation simplifies to a total derivative, enabling easy integration.

Piecewise Functions

Piecewise functions are used to define a function by different expressions for different intervals of the domain. They are especially useful in mathematics to model situations where a function's behavior changes over certain ranges.

In the given problem, the solution to the differential equation is expressed as a piecewise function:\[y = \begin{cases} 1, & 0 \leq x \leq 1 \2e^{1-x} - 1, & x > 1 \\end{cases}\]

This piecewise function reflects how the behavior and continuity of \( y \) change at \( x=1 \). The first part, \( y = 1 \), holds for \( 0 \leq x \leq 1 \), while the second part governs \( y \) for \( x > 1 \), ensuring the continuation and proper solution of the differential equation across these segments.

• Part 1: \( y = 1 \) for interval \( 0 \leq x \leq 1 \).
• Part 2: \( y = 2e^{1-x} - 1 \) for \( x > 1 \).

Initial Condition

An initial condition is a value that defines the specific solution of a differential equation from a family of solutions. It provides a reference point—usually given as \( y(x_0) = y_0 \)—to determine a unique solution by eliminating arbitrary constants.

In this exercise, the given initial condition is \( y(0) = 1 \). This is used to find the constant \( c_1 \) in the piecewise function. By substituting \( x = 0 \) into the equation for \( y \), we get \( e^0 \, y = e^0 + c_1 \), leading to \( 1 = 1 + c_1 \) and thus \( c_1 = 0 \).

Having a correct initial condition ensures the solution obtained reflects the actual requirement given in the problem statement and also maintains the continuity of the piecewise function at \( x = 1 \).

• Initial condition: \( y(0) = 1 \).
• Result: Solving for \( c_1 \) gives \( c_1 = 0 \).