Problem 34 - Calculus

Question

Solve the differential equations in Problems \(34-43 .\) Assume \(a, b,\) and \(k\) are nonzero constants. $$\frac{d R}{d t}=k R$$

Step-by-Step Solution

Answer

The solution is \( R(t) = Ce^{kt} \), where \( C \) is a constant.

Step 1: Identify the Type of Differential Equation

The given differential equation \( \frac{dR}{dt} = kR \) is a first-order linear differential equation, specifically a separable differential equation.

Step 2: Separate the Variables

Rewrite the differential equation to separate the variables \(R\) and \(t\):\[ \frac{1}{R} dR = k dt \]

Step 3: Integrate Both Sides

Integrate both sides of the equation:\[ \int \frac{1}{R} \, dR = \int k \, dt \] This gives:\[ \ln |R| = kt + C \]where \(C\) is the constant of integration.

Step 4: Solve for R

To solve for \(R\), exponentiate both sides to remove the natural log:\[ |R| = e^{kt + C} \]This implies:\[ R = e^{C} e^{kt} \]Set \( e^{C} \) as a new constant \( C' \), then the solution is:\[ R(t) = C'e^{kt} \]

Step 5: Simplify the Solution

Since \( C' \) is simply a constant, rename it to \( C \) (for simplicity), where \( C = e^{C} \). The general solution is:\[ R(t) = Ce^{kt} \]

Key Concepts

First-Order Linear Differential Equations

First-order linear differential equations are a fundamental part of calculus and differential equations. These equations are characterized by derivatives, which represent the rate of change of a function concerning another variable. In our exercise, the equation \( \frac{dR}{dt} = kR \) is identified as a first-order linear differential equation because it involves the first derivative \( \frac{dR}{dt} \). Such equations often take the form:\[ \frac{dy}{dx} + Py = Q \]where \( P \) and \( Q \) may be constants or functions of \( x \). In our case, it is a special (and simpler) form because the right-hand side is proportional to the function \( R \), and \( P \) and \( Q \) are constants. Understanding the basic structure of these equations is crucial as they frequently appear in modeling real-life problems in science and engineering.

Separable Differential Equations

Separable differential equations are a specific instance of first-order differential equations where the equation can be manipulated algebraically to isolate all instances of one variable on one side of the equation and all instances of the other variable on the opposite side. For the given problem, \( \frac{dR}{dt} = kR \), we can "separate" variables by arranging terms like this:\[ \frac{1}{R} \, dR = k \, dt \]This separation allows for each side of the equation to be integrated with respect to its own variable. Separable equations are relatively easier to solve due to this straightforward variable separation process. By performing these algebraic manipulations, we transform the differential equation into integrable parts, allowing us to solve for the function.

Integration

Integration is the mathematical process used to find functions from their derivatives. When solving separable differential equations, integration plays a pivotal role in finding the solution. In our step-by-step solution, once the equation \( \frac{1}{R} \, dR = k \, dt \) was achieved, we integrated both sides:\[ \int \frac{1}{R} \, dR = \int k \, dt \]This gives:\[ \ln |R| = kt + C \]where \( C \) is the constant of integration. Integration enables us to recover a function from its derivative, thus providing the general solution to the differential equation. We then use algebraic techniques to isolate the function \( R \) and express it explicitly.

Constant of Integration

The constant of integration, denoted as \( C \), arises whenever we perform an indefinite integral. In solving differential equations, it represents the family of solutions that can result from different initial conditions. After integrating \( \ln |R| = kt + C \), we solve for \( R \) using the method of exponentiation:\[ |R| = e^{kt+C} \]which simplifies to:\[ R = e^{C} e^{kt} \]By setting \( e^{C} \) as a new constant, \( C' \), the solution becomes \( R(t) = C'e^{kt} \). Simplifying the notation, we can let \( C = C' \), and the general solution is \( R(t) = Ce^{kt} \). This constant reflects the inherent uncertainty of the original conditions, enabling the delivery of a comprehensive solution that incorporates all potential particular solutions.