91Ó°ÊÓ

A first-order linear differential equation involves a derivative of a dependent variable with respect to an independent variable. In this case, the equation \( \frac{dR}{dt} = kR \) describes how the rate of change of \( R \) with respect to time \( t \) is proportional to \( R \) itself. Here, \( R \) is the function we want to find, and \( k \) is a constant that affects the rate of change. Understanding the relationship is key because the solution will demonstrate how \( R \) behaves over time. Such equations often model systems which experience either exponential growth or decay, depending on the sign of \( k \). For example, if \( k \) is positive, this could describe population growth, while if \( k \) is negative, it might model radioactive decay.

Separation of variables is a technique used to solve differential equations where variables are rearranged so that all terms involving the dependent variable are on one side of the equation, and all terms involving the independent variable are on the other. In our differential equation \( \frac{dR}{dt} = kR \), this process involves rewriting the equation as \( \frac{1}{R}dR = k \, dt \). This effectively separates the variables \( R \) and \( t \), allowing each side to be integrated independently. Separation of variables is a powerful method particularly suitable for solving first-order equations like this one, providing a straightforward path to finding a general solution.

Integration is the mathematical process used to find the function \( R(t) \) from its derivative \( \frac{dR}{dt} \). After separating variables, we integrate each side of the equation. When we integrate \( \int \frac{1}{R} \, dR \), we get \( \ln |R| \), which describes how \( R \) accumulates over time relative to its initial value. For the right side, integrating \( \int k \, dt \) gives \( kt + C \), where \( C \) is an integration constant. This constant is crucial as it accounts for all initial conditions of the problem. Together, these integrations provide the implicit solution \( \ln |R| = kt + C \), foundational for reaching the explicit form of \( R(t) \).

The final step in solving \( \ln |R| = kt + C \) is to isolate \( R \), resulting in an exponential growth solution. By exponentiating both sides, we transform the equation into \( R = e^{kt + C} \). This can be rewritten as \( R = e^C \cdot e^{kt} \). By letting \( A = e^C \), with \( A \) being a positive constant, the solution simplifies to \( R(t) = A e^{kt} \). This is the general solution that describes exponential growth or decay, depending on \( k \). If \( k > 0 \), \( R(t) \) represents growth; if \( k < 0 \), it demonstrates decay. Such exponential functions are crucial in modeling real-world processes including population dynamics and financial growth.