Problem 47
Question
Writing the differential equation as \(\frac{d E}{d t}+\frac{1}{R C} E=0\) we see that an integrating factor is \(e^{t / R C}\). Then $$\begin{aligned} \frac{d}{d t}\left[e^{t / R C} E\right] &=0 \\ e^{t / R C} E &=c \\ E &=c e^{-t / R C} \end{aligned}$$ From \(E(4)=c e^{-4 / R C}=E_{0}\) we find \(c=E_{0} e^{4 / R C} .\) Thus, the solution of the initial-value problem is $$E=E_{0} e^{4 / R C} e^{-t / R C}=E_{0} e^{-(t-4) / R C}$$
Step-by-Step Solution
Verified Answer
The solution is: \( E = E_0 e^{-(t-4)/R C} \).
1Step 1: Write the Differential Equation
The differential equation is given as \( \frac{d E}{d t} + \frac{1}{R C} E = 0 \). This is a first-order linear differential equation that we need to solve.
2Step 2: Identify the Integrating Factor
An integrating factor \( \mu(t) \) for the differential equation can be written as \( e^{t/R C} \). Multiplying both sides of the equation by this integrating factor helps in simplifying it.
3Step 3: Applying the Integrating Factor
Multiplying the entire differential equation by \( e^{t/R C} \), we transform it into \( \frac{d}{dt}(e^{t/R C} E) = 0 \). This step converts our differential equation into an integrable form.
4Step 4: Integrate Both Sides
Integrate \( \frac{d}{d t}(e^{t / R C} E) = 0 \) with respect to \( t \), resulting in \( e^{t/R C} E = c \), where \( c \) is a constant of integration.
5Step 5: Solve for \( E \)
Rewrite the equation \( e^{t/R C} E = c \) to solve for \( E \). Therefore, \( E = c e^{-t/R C} \).
6Step 6: Apply Initial Condition
Given the condition \( E(4) = E_0 \), substitute \( t = 4 \) into the equation: \( E_0 = c e^{-4/R C} \). Solve for \( c \), resulting in \( c = E_0 e^{4/R C} \).
7Step 7: Substitute Constant Back into Equation
Substitute \( c = E_0 e^{4/R C} \) back into \( E = c e^{-t/R C} \) to find the particular solution. Thus, \( E = E_0 e^{4/R C} e^{-t/R C} = E_0 e^{-(t-4)/R C} \).
Key Concepts
Integrating FactorFirst-order Linear Differential EquationInitial-value Problem
Integrating Factor
One of the powerful techniques to solve a first-order linear differential equation is the use of an integrating factor. An integrating factor is a function, often denoted by \( \mu(t) \), that we multiply every term of the original differential equation by. This manipulation transforms the equation into a form that can be easily integrated.
Consider the differential equation \( \frac{dE}{dt} + \frac{1}{RC}E = 0 \). By identifying the integrating factor as \( e^{t/RC} \), we achieve an important simplification. Seeing this factor, we multiply through the equation, so that the left side becomes a clear derivative: \( \frac{d}{dt}(e^{t/RC}E) \).
This outcome is crucial because it tells us that we can integrate immediately, reducing the original differential problem to a more straightforward algebraic one. The step of multiplying by an integrating factor not only eases integration but also ensures that the resulting expression is straightforward and solvable.
Consider the differential equation \( \frac{dE}{dt} + \frac{1}{RC}E = 0 \). By identifying the integrating factor as \( e^{t/RC} \), we achieve an important simplification. Seeing this factor, we multiply through the equation, so that the left side becomes a clear derivative: \( \frac{d}{dt}(e^{t/RC}E) \).
This outcome is crucial because it tells us that we can integrate immediately, reducing the original differential problem to a more straightforward algebraic one. The step of multiplying by an integrating factor not only eases integration but also ensures that the resulting expression is straightforward and solvable.
First-order Linear Differential Equation
Differential equations come in various forms, and understanding each type is key to solving them. A first-order linear differential equation is one of the simplest yet very important types.
The general form of a first-order linear differential equation is \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( y \) is the dependent variable, \( x \) is the independent variable, and \( P(x) \) and \( Q(x) \) are given functions.
In the problem \( \frac{dE}{dt} + \frac{1}{RC}E = 0 \), we see that it fits this pattern, with \( E \) acting as \( y \), \( t \) as \( x \), \( P(t) = 1/RC \), and a zero constant function \( Q(t) = 0 \).
Solving such equations involves recognizing the structure and applying methods like integrating factors to simplify and solve for the unknown function \( y(t) \). The manageable form of these equations gives them wide applications in physics, engineering, and various scientific fields, where they model processes like exponential growth and decay.
The general form of a first-order linear differential equation is \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( y \) is the dependent variable, \( x \) is the independent variable, and \( P(x) \) and \( Q(x) \) are given functions.
In the problem \( \frac{dE}{dt} + \frac{1}{RC}E = 0 \), we see that it fits this pattern, with \( E \) acting as \( y \), \( t \) as \( x \), \( P(t) = 1/RC \), and a zero constant function \( Q(t) = 0 \).
Solving such equations involves recognizing the structure and applying methods like integrating factors to simplify and solve for the unknown function \( y(t) \). The manageable form of these equations gives them wide applications in physics, engineering, and various scientific fields, where they model processes like exponential growth and decay.
Initial-value Problem
In mathematics, solving differential equations often comes with constraints known as initial conditions, leading to what's called an initial-value problem. This problem not only requires finding a suitable function that satisfies the differential equation but also adheres to specific conditions at a given point.
For instance, in this exercise, the initial condition is \( E(4) = E_0 \). What this implies is that at \( t = 4 \), the energy \( E \) has a specific value, \( E_0 \). Such an initial condition is crucial because it determines the particular solution to the differential equation from the infinite number of solutions possible.
By substituting the initial conditions into the general solution, \( E = c e^{-t/RC} \), we solve for the constant \( c \), making the solution particular to our problem. Therefore, the initial-value problem leads us to a final form of \( E = E_0 e^{-(t-4)/RC} \), fully determined by the initial condition placed on the system.
For instance, in this exercise, the initial condition is \( E(4) = E_0 \). What this implies is that at \( t = 4 \), the energy \( E \) has a specific value, \( E_0 \). Such an initial condition is crucial because it determines the particular solution to the differential equation from the infinite number of solutions possible.
By substituting the initial conditions into the general solution, \( E = c e^{-t/RC} \), we solve for the constant \( c \), making the solution particular to our problem. Therefore, the initial-value problem leads us to a final form of \( E = E_0 e^{-(t-4)/RC} \), fully determined by the initial condition placed on the system.
Other exercises in this chapter
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