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Chapter 9: Problem 32

a. Show that the solution of the equation $$\frac{d i}{d t}+\frac{R}{L} i=\frac{V}{L}$$ is $$i=\frac{V}{R}+C e^{-(R / L) x}$$ b. Then use the initial condition \(i(0)=0\) to determine the value of \(C .\) This will complete the derivation of Equation \((7)\) . c. Show that \(i=V / R\) is a solution of Equation \((6)\) and that \(i=C e^{-(R / L) x}\) satisfies the equation $$\frac{d i}{d t}+\frac{R}{L} i=0$$

Short Answer

Expert verified
The value of \( C \) is \(-\frac{V}{R}\).

Step by step solution

01

Solve the Homogeneous Equation

Start by considering the homogeneous part of the equation: \( \frac{d i}{d t} + \frac{R}{L} i = 0 \). This is a linear equation with constant coefficients. The solution is of the form \( i_h = C e^{ax} \), where \( a = -\frac{R}{L} \). Thus, the homogeneous solution is \( i_h = C e^{-\frac{R}{L} x} \).
02

Find Particular Solution

To find a particular solution \( i_p \), assume \( i_p = \frac{V}{R} \). Differentiating gives \( \frac{d i_p}{d t} = 0 \). Substitute \( i_p \) into the original equation: \( 0 + \frac{R}{L} \frac{V}{R} = \frac{V}{L} \). The equation holds true, confirming \( i_p = \frac{V}{R} \) is a particular solution.
03

General Solution

The general solution of the differential equation is the sum of the homogeneous and particular solutions: \[ i = i_h + i_p = C e^{-\frac{R}{L} x} + \frac{V}{R} \].
04

Apply Initial Condition

Use the initial condition \( i(0) = 0 \) to find \( C \). Substitute \( x = 0 \) into the general solution: \( 0 = Ce^{0} + \frac{V}{R} \), which simplifies to \( 0 = C + \frac{V}{R} \). Thus, \( C = -\frac{V}{R} \).
05

Substitute Value of C

Substitute \( C = -\frac{V}{R} \) back into the general solution: \[ i = \frac{V}{R} - \frac{V}{R} e^{-\frac{R}{L} x} \].
06

Verify Particular Solution 1

To verify \( i = \frac{V}{R} \), differentiate: \( \frac{d i}{d t} = 0 \). Check: \( 0 + \frac{R}{L} \frac{V}{R} = \frac{V}{L} \), which holds, confirming it satisfies the original equation.
07

Verify Homogeneous Solution

For \( i = C e^{-\frac{R}{L} x} \), differentiate: \( \frac{d i}{d t} = -\frac{R}{L} C e^{-\frac{R}{L} x} \). Substitute into the equation: \( -\frac{R}{L} C e^{-\frac{R}{L} x} + \frac{R}{L} C e^{-\frac{R}{L} x} = 0 \). Both terms cancel, verifying that it satisfies the homogeneous equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Differential Equations
In mathematics, linear differential equations play a pivotal role in describing systems with known rates of change. A differential equation is called "linear" when the dependent variable and all its derivatives appear to the power of one, without any products of the dependent variable with itself or with its derivatives. For example, the linear differential equation \( \frac{d i}{d t} + \frac{R}{L} i = \frac{V}{L} \) describes the relationship between the electric current \( i \), resistance \( R \), inductance \( L \), and voltage \( V \).

  • The general form of a first-order linear differential equation is \( \frac{dy}{dt} + P(t)y = Q(t) \).
  • In our equation, \( P(t) = \frac{R}{L} \) and \( Q(t) = \frac{V}{L} \).
  • Methods for finding solutions often involve finding the homogeneous solution and particular solution.
Particular Solution
Identifying a particular solution is a crucial step in solving linear differential equations. This solution accounts for the non-homogeneous part of the equation. For our equation \( \frac{d i}{d t} + \frac{R}{L} i = \frac{V}{L} \), the particular solution is found by assuming a constant solution where the derivative \( \frac{d i_p}{d t} = 0 \).

  • Assume \( i_p = \frac{V}{R} \), leading to \( \frac{d i_p}{d t} = 0 \).
  • Substitute \( i_p \) back into the equation to verify: \( 0 + \frac{R}{L} \cdot \frac{V}{R} = \frac{V}{L} \).
  • This value of \( i_p \) satisfies the original differential equation, confirming that \( i_p = \frac{V}{R} \) is indeed a particular solution.
The particular solution is key because it directly represents a specific solution without involving arbitrary constants, reflecting the system's steady state.
Homogeneous Solution
The homogeneous solution deals with the differential equation when no external forces or inputs are considered, meaning the right-hand side is zero. For the equation \( \frac{d i}{d t} + \frac{R}{L} i = 0 \), the solution reflects the natural behavior of the system.

  • To find the homogeneous solution, consider the equation without the external input: \( \frac{d i}{d t} + \frac{R}{L} i = 0 \).
  • Assume the solution is of the form \( i_h = Ce^{ax} \), where \( a = -\frac{R}{L} \).
  • Thus, \( i_h = C e^{-(R/L)x} \), which describes exponentially decaying behavior.
The homogeneous solution captures the transient behavior of the system and depends on the initial conditions, eventually giving way to the steady-state behavior represented by the particular solution.
Initial Conditions
Initial conditions are critical in determining the specific values of constants in solutions to differential equations. They ensure that the solution fits the real-world scenario from a specific starting point. Given the general solution \( i = \frac{V}{R} + C e^{-(R/L) x} \), if \( i(0) = 0 \), then substituting these conditions allows solving for \( C \).

  • Substitute the initial condition into the general solution: \( 0 = \frac{V}{R} + C e^{0} \).
  • This simplifies to \( 0 = \frac{V}{R} + C \), thus \( C = -\frac{V}{R} \).
  • The initial conditions effectively "pin down" the value of \( C \), giving the complete solution adapted to the particular situation.
By applying the initial condition, we transition from a generic form to one that meets the specified requirements, ensuring accuracy and relevance in real-life applications.

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Most popular questions from this chapter

Use a CAS to explore graphically each of the differential equations in Exercises \(21-24 .\) Perform the following steps to help with your explorations. a. Plot a slope field for the differential equation in the given \(x y-\) window. b. Find the general solution of the differential equation using your CAS DE solver. c. Graph the solutions for the values of the arbitrary constant \(C=-2,-1,0,1,2\) superimposed on your slope field plot. d. Find and graph the solution that satisfies the specified initial condition over the interval \([0, b] .\) e. Find the Euler numerical approximation to the solution of the initial value problem with 4 subintervals of the \(x\) -interval and plot the Euler approximation superimposed on the graph produced in part (d). f. Repeat part (e) for \(8,16,\) and 32 subintervals. Plot these three Euler approximations superimposed on the graph from part (e). g. Find the error \((y \text { exact })-y(\text { Euler })\) at the specified point \(x=b\) for each of your four Euler approximations. Discuss the improvement in the percentage error. $$ \begin{array}{l}{\text { A logistic equation } y^{\prime}=y(2-y), \quad y(0)=1 / 2} \\ {0 \leq x \leq 4, \quad 0 \leq y \leq 3 ; \quad b=3}\end{array} $$

Solve the differential equations. \(x y^{\prime}+3 y=\frac{\sin x}{x^{2}}, \quad x>0\)

The spread of information Sociologists recognize a phenome- non called social diffusion, which is the spreading of a piece of information, technological innovation, or cultural fad among a population. The members of the population can be divided into two classes: those who have the information and those who do not. In a fixed population whose size is known, it is reasonable to assume that the rate of diffusion is proportional to the number who have the information times the number yet to receive it. If \(X\) denotes the number of individuals who have the information in a population of \(N\) people, then a mathematical model for social diffusion is given by $$ \frac{d X}{d t}=k X(N-X) $$ where \(t\) represents time in days and \(k\) is a positive constant. a. Discuss the reasonableness of the model. b. Construct a phase line identifying the signs of \(X^{\prime}\) and \(X^{\prime \prime}\) . c. Sketch representative solution curves. d. Predict the value of \(X\) for which the information is spreading most rapidly. How many people eventually receive the information?

In Exercises \(1-6,\) use Euler's method to calculate the first three approximations to the given initial value problem for the specified increment size. Calculate the exact solution and investigate the accuracy of your approximations. Round your results to four decimal places. $$ y^{\prime}=2 x e^{x^{2}}, \quad y(0)=2, \quad d x=0.1 $$

In Exercises \(1-8\) a. Identify the equilibrium values. Which are stable and which are unstable? b. Construct a phase line. Identify the signs of \(y^{\prime}\) and \(y^{\prime \prime}\) c. Sketch several solution curves. $$ \frac{d y}{d x}=y^{2}-2 y $$
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