91Ó°ÊÓ

A first order linear differential equation is an equation that can be expressed in the form: \(y' + p(x)y = g(x)\). Here, \( y' \) is the derivative of \( y \) with respect to \( x \), \( p(x) \) is a function of \( x \) that acts as a coefficient for \( y \), and \( g(x) \) is a function of \( x \) that represents the non-homogeneous part of the equation. First order linear differential equations are fundamental in mathematics, as they appear in many scientific disciplines to model dynamic systems. In the given problem, the equation provided is \( y' + y = \frac{1}{1+2e^{x}} \). This fits the standard form where \( p(x) = 1 \) and \( g(x) = \frac{1}{1+2e^{x}} \). By understanding these components, solving the equation involves using methods like finding the homogeneous and particular solutions.

The homogeneous solution of a first order linear differential equation is found when the equation is set to zero, i.e., \( y' + p(x)y = 0 \). This approach highlights the behavior of the system without any external influences, represented by \( g(x) \). For \( y' + y = 0 \), the solution involves finding an integrating factor, typically expressed as \( e^{∫p(x)dx} \). In our example, this factor is \( e^{x} \). By multiplying the entire equation by this factor, we obtain \( e^{x}y' + e^{x}y = 0 \), which simplifies to \((e^{x}y)' = 0\). Integrating both sides, we find that \( e^{x}y = C \), where \( C \) is the constant of integration. Thus, \( y = Ce^{-x} \) is the homogeneous solution \( y_c \).

The particular solution of a differential equation involves finding a solution specific to the non-homogeneous part \( g(x) \) of the equation \( y' + p(x)y = g(x) \). This solution is not general and caters specifically to the form of \( g(x) \). To find the particular solution in our exercise, we use the method of integrating factors again. The equation becomes \((e^{x} y)' = e^{x}\frac{1}{1+2e^{x}}\). By integrating both sides, we derive: \(e^{x} y = -\frac{1}{2} \ln{(|1+2 e^{x}|)} + K\). Solving for \( y \), we find the particular solution \( y_p = -\frac{1}{2e^{x}} \ln{(1+2e^{x})} \). This solves for the portion of \( y \) that responds directly to the \( g(x) \) term, giving us a full picture of how both the natural exponential decay and enhancement represented by \( g(x) \) are factored into the system.

The integrating factor is a crucial concept in solving first order linear differential equations. It is particularly useful when dealing with equations that cannot be solved directly by basic integration. An integrating factor for an equation in the form \( y' + p(x)y = g(x) \) is usually given by \( e^{∫p(x)dx} \). This factor, when multiplied through the equation, transforms it into a perfect derivative, which can then be easily integrated. In our specific problem, the integrating factor is \( e^{x} \) since \( p(x) = 1 \). By multiplying both sides of \( y' + y = \frac{1}{1+2 e^{x}} \) by \( e^{x} \), the equation becomes \((e^{x}y)'\), a perfect derivative of the product \( e^{x}y \), which simplifies the process of finding solutions. Thus, the integrating factor is key to converting challenging equations into manageable forms by leveraging properties of exponential functions.