Problem 27 - Advanced Engineering Mathematics

Question

For \(\frac{d i}{d t}+\frac{R}{L} i=\frac{E}{L}\) an integrating factor is \(e^{\int(R / L) d t}=e^{R t / L}\) so that \(\frac{d}{d t}\left[e^{R t / L} i\right]=\frac{E}{L} e^{R t / L}\) and \(i=\frac{E}{R}+c e^{-R t / L}\) for \(-\infty< t<\infty .\) If \(i(0)=i_{0}\) then \(c=i_{0}-E / R\) and \(i=\frac{E}{R}+\left(i_{0}-\frac{E}{R}\right) e^{-R t / L}\).

Step-by-Step Solution

Answer

The solution is \(i = \frac{E}{R} + \left(i_0 - \frac{E}{R}\right) e^{-Rt/L}\).

Step 1: Arrange the differential equation

We start with the given linear first-order differential equation: \(\frac{di}{dt}+\frac{R}{L}i=\frac{E}{L}\).

Step 2: Determine the integrating factor

The integrating factor for a differential equation of the form \(\frac{di}{dt}+P(t)i=Q(t)\) is found by \(\mu(t) = e^{\int P(t) dt}\). From our equation, \(P(t) = \frac{R}{L}\), so the integrating factor is \(e^{\int \frac{R}{L} dt} = e^{Rt/L}\).

Step 3: Multiply through by the integrating factor

Multiply the entire differential equation by the integrating factor \(e^{Rt/L}\): \[ e^{Rt/L} \cdot \left(\frac{di}{dt} + \frac{R}{L}i\right) = e^{Rt/L} \cdot \frac{E}{L} \] This simplifies to:\[ \frac{d}{dt}\left(e^{Rt/L} i\right) = \frac{E}{L} e^{Rt/L} \]

Step 4: Integrate both sides

Integrate both sides with respect to \(t\): \[ \int \frac{d}{dt}(e^{Rt/L} i) dt = \int \frac{E}{L} e^{Rt/L} dt \] This gives:\[ e^{Rt/L} i = \frac{E}{L} \cdot \frac{L}{R} e^{Rt/L} + C \] Simplifying further:\[ e^{Rt/L} i = \frac{E}{R} e^{Rt/L} + C \]

Step 5: Solve for \(i\)

To solve for \(i\), divide by \(e^{Rt/L}\):\[ i = \frac{E}{R} + Ce^{-Rt/L} \]

Step 6: Use the initial condition

We are given the initial condition \(i(0) = i_0\). Substitute \(t=0\) and \(i=i_0\) into the equation:\[ i_0 = \frac{E}{R} + Ce^{0}\] \[ i_0 = \frac{E}{R} + C\] Solve for \(C\):\[ C = i_0 - \frac{E}{R} \]

Step 7: Write the final expression for \(i\)

Substitute \(C = i_0 - \frac{E}{R}\) back into the equation from Step 5:\[ i = \frac{E}{R} + \left(i_0 - \frac{E}{R}\right) e^{-Rt/L}\] This is the general solution using the initial condition.

Key Concepts

Integrating Factor Method

The Integrating Factor Method is an important technique for solving first-order linear differential equations. These equations have the general form \( \frac{di}{dt} + P(t)i = Q(t) \). The idea is to make the left-hand side an exact derivative. To achieve this, we multiply the entire equation by a specially chosen function, known as the integrating factor, denoted by \( \mu(t) \). The integrating factor is calculated as \( \mu(t) = e^{\int P(t) dt} \). This transforms the equation into a form where the left side is the derivative of a product of functions, making it easier to solve. In the provided exercise, the integrating factor \( e^{Rt/L} \) turns the original differential equation into a new one that facilitates straightforward integration. Utilizing the integrating factor ensures that solving the differential equation becomes much more manageable step by step.

Exponential Functions

Exponential functions are a fundamental part of mathematics, crucial for modeling growth and decay processes. These functions, written in the form \( f(t) = a e^{bt} \), exhibit a constant proportional rate of change, characterized by the constant \( b \). The base \( e \), approximately equal to 2.718, is a unique number that maintains particular properties helpful in calculus. Within the exercise, the expression \( e^{Rt/L} \) is an exponential that plays a key role in the Integrating Factor Method. Using exponentials makes it easier to express complex processes, such as charging and discharging in electric circuits, which is the case in this exercise. A critical aspect of exponential functions is that when differentiated, the outcome is a scaled version of the original function, simplifying integration and differentiation tasks.

Initial Conditions

In differential equations, initial conditions are essential to finding a specific solution. They help determine the constants in the general solution of a differential equation, leading to a unique solution that fits particular conditions. The initial condition typically prescribes the value of the unknown function at a specific point, often given as \( i(0) = i_0 \). For this exercise, the initial condition \( i(0) = i_0 \) is used to establish the value of the constant in our solution. By substituting \( t = 0 \), we obtain \( i_0 = \frac{E}{R} + C \). This tells us the value of \( C \) is \( i_0 - \frac{E}{R} \), which ensures the derived function correctly fits the starting situation given by the initial condition.

Linear Differential Equations

Linear differential equations are a category of differential equations characterized by the linear relationship between the function and its derivatives. In a linear first-order differential equation, the unknown function and its derivative appear to the power of one. The general form for such equations is \( \frac{di}{dt} + P(t)i = Q(t) \), where \( P(t) \) and \( Q(t) \) can be any functions of \( t \). The exercise revolves around a linear differential equation, as seen from \( \frac{di}{dt} + \frac{R}{L}i = \frac{E}{L} \). These equations are predominantly used to model linear systems and processes, like electrical circuits, which aligns well with what the equation describes. The solution methods, like the integrating factor method, and interpretations align closely with settings or conditions where such linear relationships hold. Understanding the linear nature allows for streamlined approaches to find solutions and convey important properties about the system being analyzed.