Question

Solve the initial-value problems. $$ (x+1) \frac{d y}{d x}+y=f(x), \text { where } f(x)=\left\\{\begin{array}{ll} x, & 0 \leq x<3, \\ 3, & x \geq 3, \end{array} \quad y(0)=1 / 2\right. $$

Short answer

The solution to the given initial-value problem is \[ y(x) = \begin{cases} (x-1) + \frac{3}{2}e^{-x}, & 0 \leq x < 3, \\ 3 + \left(1 - \frac{3}{2}e^{-3}\right)e^{-x}, & x \geq 3. \end{cases} \]

Step-by-step solution

Identify the general form of the first-order linear differential equation

The given equation is \[ (x+1) \frac{d y}{d x} + y = f(x), \] which is a first-order linear differential equation of the form \[ a(x)\frac{dy}{dx}+b(x)y = f(x). \]

Calculate the integrating factor

In this case, the coefficient of \(y\) is \(b(x) = 1\). The integrating factor is given by the exponential function of the integral of the coefficient \(b(x)\), which is \[ \mu(x) = e^{\int b(x) dx} = e^{\int 1 dx} = e^{x}. \]

Multiply the given equation by the integrating factor

Multiplying the given equation by the integrating factor, we obtain \[ e^x(x+1) \frac{dy}{dx} + e^x y = e^x f(x), \] which simplifies to \[ \frac{d}{dx}(ye^x) = e^x f(x). \]

Integrate both sides of the above equation with respect to \(x\)

Integrating both sides, we get \[ \int \frac{d}{dx}(ye^x) dx = \int e^x f(x) dx. \] Now, the left side integrates to \(ye^x\), whereas the right side depends on the piecewise-defined function \(f(x)\). Therefore, we must consider both the cases of \(f(x)\) for the right side.

Consider cases of the piecewise-defined function

For the case when \(0 \leq x < 3\), \(f(x) = x\), so our equation becomes \[ ye^x = \int xe^x dx + C_1, \] where \(C_1\) is the constant of integration. For the case when \(x \geq 3\), \(f(x) = 3\), so our equation becomes \[ ye^x = \int 3e^x dx + C_2, \] where \(C_2\) is the constant of integration.

Solve for y(x) in both cases

For the case when \(0 \leq x < 3\), we can use integration by parts to find the integral, obtaining \[ ye^x = (x-1)e^x + C_1, \] which implies \[ y(x) = (x-1) + C_1e^{-x}. \] For the case when \(x \geq 3\), we have \[ ye^x = 3e^x + C_2, \] which implies \[ y(x) = 3 + C_2e^{-x}. \]

Use the initial condition to find C_1

We are given that \(y(0) = \frac{1}{2}\). Using this in the solution for the \(0\le x<3\) case, we get \[ \frac{1}{2} = (0-1) + C_1e^{0}. \] This implies that \(C_1 = \frac{3}{2}\).

Find C_2 by matching the functions at x = 3

To ensure the continuity of the solution, we set the functions equal at \(x=3\) to find \(C_2\): \[ (3-1) + \frac{3}{2}e^{-3} = 3 + C_2e^{-3}. \] Solving for \(C_2\), we get \[ C_2 = 1 - \frac{3}{2}e^{-3}. \]

Write the final piecewise solution

Combining both solutions for the different cases of \(f(x)\), we get the final solution for the initial-value problem: \[ y(x) = \begin{cases} (x-1) + \frac{3}{2}e^{-x}, & 0 \leq x < 3, \\ 3 + \left(1 - \frac{3}{2}e^{-3}\right)e^{-x}, & x \geq 3. \end{cases} \]

Key concepts

First-Order Linear Differential Equation

A first-order linear differential equation is a foundational concept in calculus that pertains to equations of the form \[a(x)\frac{dy}{dx}+b(x)y = g(x)\]. In this general notation, \(a(x)\) and \(b(x)\) represent functions of the independent variable, usually \(x\), and \(g(x)\) is the known function or the 'forcing' function. The essence of solving such an equation lies in finding the function \(y\) that satisfies this relation for a given \(g(x)\).

Understanding the process of solving such equations is critical, as they often describe a variety of physical phenomena, such as growth and decay processes, electrical circuits, and motion under resistance. In our textbook exercise, identifying the provided equation as a first-order linear differential equation is the first step. Students should aim to familiarize themselves with identifying the structure of these equations as this forms the basis for applying subsequent solution methods.

Integrating Factor

An integrating factor is a mathematical tool that simplifies the process of solving first-order linear differential equations. It is essentially a function, commonly denoted as \(\mu(x)\), which when multiplied by the original differential equation, transforms it into an exact differential that is easier to integrate.

The integrating factor is often found by taking the exponential of the integral of \(b(x)\), where \(b(x)\) is the coefficient of \(y\) in the differential equation. This methodology is extremely powerful as it helps reduce complex differential equations into integrable forms. For instance, in the textbook example, multiplying the given equation by \(e^x\) simplifies it notably, allowing the function \(y(x)\) to be found after integration. The integrating factor not only paves the way for solving the equation but also highlights the significance of the exponential function in the realm of differential equations.

Piecewise-Defined Functions

A piecewise-defined function is a type of function that has different expressions for different intervals of the independent variable. These are particularly useful when modeling situations where a system behaves differently under various conditions. For example, the cost of shipping might vary depending on the weight of the package, or a physical object might have different modes of motion under different forces.

In the context of differential equations, the solution may also change form based on the intervals defined by the piecewise function. As seen in the provided exercise, the \(f(x)\) function alters its expression depending on the range of \(x\). Solving the differential equation involves integrating separately for each part of the piecewise function and applying the initial condition to determine any constants of integration. Furthermore, the solutions for each interval must be matched up to ensure the overall solution remains continuous, adding another layer of complexity to the problem-solving process. This is why piecewise-defined functions can present unique challenges and why understanding them is indispensable for students tackling real-world problems.