Problem 39 - Calculus

Question

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

Step-by-Step Solution

Answer

The solution is \( R = -\frac{b}{a} + Ce^{at} \).

Step 1: Identify the type of differential equation

The given differential equation \( \frac{dR}{dt} = aR + b \) is a first-order linear differential equation. We need to solve it by separating variables or using an integrating factor.

Step 2: Rearrange the equation

Write the equation in the standard linear form \( \frac{dR}{dt} - aR = b \) to recognize the components for applying an integrating factor.

Step 3: Calculate the integrating factor

The integrating factor \( \mu(t) \) for this equation is given by \( e^{\int -a \ dt} = e^{-at} \). This factor will be used to multiply through the equation.

Step 4: Multiply the equation by the integrating factor

Multiply each term in \( \frac{dR}{dt} - aR = b \) by \( e^{-at} \) to obtain \( e^{-at} \frac{dR}{dt} - ae^{-at} R = b e^{-at} \).

Step 5: Recognize the derivative of a product

The left-hand side of the equation \( e^{-at} \frac{dR}{dt} - ae^{-at} R \) can be rewritten as the derivative of a product: \( \frac{d}{dt} \left( e^{-at} R \right) \).

Step 6: Integrate both sides

Integrate both sides with respect to \( t \): \( \int \frac{d}{dt} \left( e^{-at} R \right) \ dt = \int b e^{-at} \ dt \). The left side integrates to \( e^{-at} R \), and the right side integrates to \( -\frac{b}{a} e^{-at} + C \), where \( C \) is the constant of integration.

Step 7: Solve for \( R \)

Isolate \( R \) by multiplying both sides by \( e^{at} \): \( R = -\frac{b}{a} + Ce^{at} \). This is the general solution of the differential equation.

Key Concepts

Integrating Factor

In order to solve a first-order linear differential equation like \( \frac{dR}{dt} = aR + b \), we often use an integrating factor. This technique transforms the equation into a form that is easier to solve by integration. An integrating factor is a function, usually denoted as \( \mu(t) \), that when multiplied throughout a differential equation helps in simplifying the process of arriving at a solution.

To find the integrating factor, you first need the equation in the form \( \frac{dR}{dt} + P(t)R = Q(t) \). In our example, rearrange the equation to \( \frac{dR}{dt} - aR = b \) so \( P(t) = -a \). The integrating factor \( \mu(t) \) is determined by \( e^{\int P(t) \, dt} \). Hence, for our equation, it becomes \( e^{-at} \). This integrating factor converts the original differential equation into a product of a derivative, simplifying the solving process.

General Solution

The general solution is what we aim to find when solving a differential equation. It represents a family of solutions containing a constant of integration, reflecting various possible specific solutions of the equation.

In the step-by-step solution, after applying the integrating factor \( e^{-at} \), the equation is transformed into \( \frac{d}{dt}(e^{-at} R) = b e^{-at} \). Integrating both sides gives us \( e^{-at} R = -\frac{b}{a} e^{-at} + C \). The term \( C \) is crucial as it represents the constant of integration, allowing us to express in terms of arbitrary initial conditions or boundary values.

To find \( R \), multiply through by \( e^{at} \), yielding the general solution:

• \( R = -\frac{b}{a} + Ce^{at} \)

This expression illustrates all potential scenarios for \( R \) given any initial conditions.

Separation of Variables

Separation of variables is another approach to solving differential equations, but it's distinct from the process we used here involving an integrating factor. In cases where it applies, it involves rewriting the equation such that each variable and its differential are on opposite sides of the equation. While the given differential equation \( \frac{dR}{dt} = aR + b \) doesn't lend itself directly to separation of variables without adjustments, the technique can still sometimes be helpful, especially if the structure of the equation is simpler, such as a pure proportionate relationship between \( R \) and \( t \).Separation of variables is notably useful when you have:

• Equations in the form \( \frac{dy}{dx} = g(y)h(x) \)
• When each side can be integrated independently

Here, each step in obtaining the general solution was more straightforward using the integrating factor method given the specific equation structure.