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Calculus Volume 2

4.5 First-order Linear Equations

Calculus Volume 24.5 First-order Linear Equations
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  1. Preface
  2. 1 Integration
    1. Introduction
    2. 1.1 Approximating Areas
    3. 1.2 The Definite Integral
    4. 1.3 The Fundamental Theorem of Calculus
    5. 1.4 Integration Formulas and the Net Change Theorem
    6. 1.5 Substitution
    7. 1.6 Integrals Involving Exponential and Logarithmic Functions
    8. 1.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  3. 2 Applications of Integration
    1. Introduction
    2. 2.1 Areas between Curves
    3. 2.2 Determining Volumes by Slicing
    4. 2.3 Volumes of Revolution: Cylindrical Shells
    5. 2.4 Arc Length of a Curve and Surface Area
    6. 2.5 Physical Applications
    7. 2.6 Moments and Centers of Mass
    8. 2.7 Integrals, Exponential Functions, and Logarithms
    9. 2.8 Exponential Growth and Decay
    10. 2.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  4. 3 Techniques of Integration
    1. Introduction
    2. 3.1 Integration by Parts
    3. 3.2 Trigonometric Integrals
    4. 3.3 Trigonometric Substitution
    5. 3.4 Partial Fractions
    6. 3.5 Other Strategies for Integration
    7. 3.6 Numerical Integration
    8. 3.7 Improper Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  5. 4 Introduction to Differential Equations
    1. Introduction
    2. 4.1 Basics of Differential Equations
    3. 4.2 Direction Fields and Numerical Methods
    4. 4.3 Separable Equations
    5. 4.4 The Logistic Equation
    6. 4.5 First-order Linear Equations
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  6. 5 Sequences and Series
    1. Introduction
    2. 5.1 Sequences
    3. 5.2 Infinite Series
    4. 5.3 The Divergence and Integral Tests
    5. 5.4 Comparison Tests
    6. 5.5 Alternating Series
    7. 5.6 Ratio and Root Tests
    8. Key Terms
    9. Key Equations
    10. Key Concepts
    11. Chapter Review Exercises
  7. 6 Power Series
    1. Introduction
    2. 6.1 Power Series and Functions
    3. 6.2 Properties of Power Series
    4. 6.3 Taylor and Maclaurin Series
    5. 6.4 Working with Taylor Series
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  8. 7 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 7.1 Parametric Equations
    3. 7.2 Calculus of Parametric Curves
    4. 7.3 Polar Coordinates
    5. 7.4 Area and Arc Length in Polar Coordinates
    6. 7.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 4.5.1. Write a first-order linear differential equation in standard form.
  • 4.5.2. Find an integrating factor and use it to solve a first-order linear differential equation.
  • 4.5.3. Solve applied problems involving first-order linear differential equations.

Earlier, we studied an application of a first-order differential equation that involved solving for the velocity of an object. In particular, if a ball is thrown upward with an initial velocity of v0v0 ft/s, then an initial-value problem that describes the velocity of the ball after tt seconds is given by

dvdt=−32,v(0)=v0.dvdt=−32,v(0)=v0.

This model assumes that the only force acting on the ball is gravity. Now we add to the problem by allowing for the possibility of air resistance acting on the ball.

Air resistance always acts in the direction opposite to motion. Therefore if an object is rising, air resistance acts in a downward direction. If the object is falling, air resistance acts in an upward direction (Figure 4.24). There is no exact relationship between the velocity of an object and the air resistance acting on it. For very small objects, air resistance is proportional to velocity; that is, the force due to air resistance is numerically equal to some constant kk times v.v. For larger (e.g., baseball-sized) objects, depending on the shape, air resistance can be approximately proportional to the square of the velocity. In fact, air resistance may be proportional to v1.5,v1.5, or v0.9,v0.9, or some other power of v.v.

A diagram of a baseball with an arrow above it pointing up and an arrow below it pointing down. The upper arrow is labeled “air resistance –kv” and the lower arrow is labeled “g = -9.8 m/sec ^ 2.”
Figure 4.24 Forces acting on a moving baseball: gravity acts in a downward direction and air resistance acts in a direction opposite to the direction of motion.

We will work with the linear approximation for air resistance. If we assume k>0,k>0, then the expression for the force FAFA due to air resistance is given by FA=kv.FA=kv. Therefore the sum of the forces acting on the object is equal to the sum of the gravitational force and the force due to air resistance. This, in turn, is equal to the mass of the object multiplied by its acceleration at time tt (Newton’s second law). This gives us the differential equation

mdvdt=kvmg.mdvdt=kvmg.

Finally, we impose an initial condition v(0)=v0,v(0)=v0, where v0v0 is the initial velocity measured in meters per second. This makes g=9.8m/s2.g=9.8m/s2. The initial-value problem becomes

mdvdt=kvmg,v(0)=v0.mdvdt=kvmg,v(0)=v0.
4.13

The differential equation in this initial-value problem is an example of a first-order linear differential equation. (Recall that a differential equation is first-order if the highest-order derivative that appears in the equation is 1.)1.) In this section, we study first-order linear equations and examine a method for finding a general solution to these types of equations, as well as solving initial-value problems involving them.

Definition

A first-order differential equation is linear if it can be written in the form

a(x)y+b(x)y=c(x),a(x)y+b(x)y=c(x),
4.14

where a(x),b(x),a(x),b(x), and c(x)c(x) are arbitrary functions of x.x.

Remember that the unknown function yy depends on the variable x;x; that is, xx is the independent variable and yy is the dependent variable. Some examples of first-order linear differential equations are

(3x24)y+(x3)y=sinx(sinx)y(cosx)y=cotx4xy+(3lnx)y=x34x.(3x24)y+(x3)y=sinx(sinx)y(cosx)y=cotx4xy+(3lnx)y=x34x.

Examples of first-order nonlinear differential equations include

(y)4(y)3=(3x2)(y+4)4y+3y3=4x5(y)2=siny+cosx.(y)4(y)3=(3x2)(y+4)4y+3y3=4x5(y)2=siny+cosx.

These equations are nonlinear because of terms like (y)4,y3,(y)4,y3, etc. Due to these terms, it is impossible to put these equations into the same form as Equation 4.14.

Standard Form

Consider the differential equation

(3x24)y+(x3)y=sinx.(3x24)y+(x3)y=sinx.

Our main goal in this section is to derive a solution method for equations of this form. It is useful to have the coefficient of yy be equal to 1.1. To make this happen, we divide both sides by 3x24.3x24.

y+(x33x24)y=sinx3x24y+(x33x24)y=sinx3x24

This is called the standard form of the differential equation. We will use it later when finding the solution to a general first-order linear differential equation. Returning to Equation 4.14, we can divide both sides of the equation by a(x).a(x). This leads to the equation

y+b(x)a(x)y=c(x)a(x).y+b(x)a(x)y=c(x)a(x).
4.15

Now define p(x)=b(x)a(x)p(x)=b(x)a(x) and q(x)=c(x)a(x).q(x)=c(x)a(x). Then Equation 4.14 becomes

y+p(x)y=q(x).y+p(x)y=q(x).
4.16

We can write any first-order linear differential equation in this form, and this is referred to as the standard form for a first-order linear differential equation.

Example 4.15

Writing First-Order Linear Equations in Standard Form

Put each of the following first-order linear differential equations into standard form. Identify p(x)p(x) and q(x)q(x) for each equation.

  1. y=3x4yy=3x4y
  2. 3xy4y3=23xy4y3=2 (here x>0)x>0)
  3. y=3y4x2+5y=3y4x2+5

Solution

  1. Add 4y4y to both sides:
    y+4y=3x.y+4y=3x.

    In this equation, p(x)=4p(x)=4 and q(x)=3x.q(x)=3x.
  2. Multiply both sides by 4y3,4y3, then subtract 8y8y from each side:
    3xy4y3=23xy=2(4y3)3xy=8y63xy8y=−6.3xy4y3=23xy=2(4y3)3xy=8y63xy8y=−6.

    Finally, divide both sides by 3x3x to make the coefficient of yy equal to 1:1:
    y83xy=23x.y83xy=23x.
    4.17
    This is allowable because in the original statement of this problem we assumed that x>0.x>0. (If x=0x=0 then the original equation becomes 0=2,0=2, which is clearly a false statement.)
    In this equation, p(x)=83xp(x)=83x and q(x)=23x.q(x)=23x.
  3. Subtract yy from each side and add 4x25:4x25:
    3yy=4x25.3yy=4x25.

    Next divide both sides by 3:3:
    y13y=43x253.y13y=43x253.

    In this equation, p(x)=13p(x)=13 and q(x)=43x253.q(x)=43x253.
Checkpoint 4.15

Put the equation (x+3)y2x3y4=5(x+3)y2x3y4=5 into standard form and identify p(x)p(x) and q(x).q(x).

Integrating Factors

We now develop a solution technique for any first-order linear differential equation. We start with the standard form of a first-order linear differential equation:

y+p(x)y=q(x).y+p(x)y=q(x).
4.18

The first term on the left-hand side of Equation 4.15 is the derivative of the unknown function, and the second term is the product of a known function with the unknown function. This is somewhat reminiscent of the power rule from the Differentiation Rules section. If we multiply Equation 4.16 by a yet-to-be-determined function μ(x),μ(x), then the equation becomes

μ(x)y+μ(x)p(x)y=μ(x)q(x).μ(x)y+μ(x)p(x)y=μ(x)q(x).
4.19

The left-hand side Equation 4.18 can be matched perfectly to the product rule:

ddx[f(x)g(x)]=f(x)g(x)+f(x)g(x).ddx[f(x)g(x)]=f(x)g(x)+f(x)g(x).

Matching term by term gives y=f(x),g(x)=μ(x),y=f(x),g(x)=μ(x), and g(x)=μ(x)p(x).g(x)=μ(x)p(x). Taking the derivative of g(x)=μ(x)g(x)=μ(x) and setting it equal to the right-hand side of g(x)=μ(x)p(x)g(x)=μ(x)p(x) leads to

μ(x)=μ(x)p(x).μ(x)=μ(x)p(x).

This is a first-order, separable differential equation for μ(x).μ(x). We know p(x)p(x) because it appears in the differential equation we are solving. Separating variables and integrating yields

μ(x)μ(x)=p(x)μ(x)μ(x)dx=p(x)dxln|μ(x)|=p(x)dx+Celn|μ(x)|=ep(x)dx+C|μ(x)|=C1ep(x)dxμ(x)=C2ep(x)dx.μ(x)μ(x)=p(x)μ(x)μ(x)dx=p(x)dxln|μ(x)|=p(x)dx+Celn|μ(x)|=ep(x)dx+C|μ(x)|=C1ep(x)dxμ(x)=C2ep(x)dx.

Here C2C2 can be an arbitrary (positive or negative) constant. This leads to a general method for solving a first-order linear differential equation. We first multiply both sides of Equation 4.16 by the integrating factor μ(x).μ(x). This gives

μ(x)y+μ(x)p(x)y=μ(x)q(x).μ(x)y+μ(x)p(x)y=μ(x)q(x).
4.20

The left-hand side of Equation 4.19 can be rewritten as ddx(μ(x)y).ddx(μ(x)y).

ddx(μ(x)y)=μ(x)q(x).ddx(μ(x)y)=μ(x)q(x).
4.21

Next integrate both sides of Equation 4.20 with respect to x.x.

ddx(μ(x)y)dx=μ(x)q(x)dxμ(x)y=μ(x)q(x)dx.ddx(μ(x)y)dx=μ(x)q(x)dxμ(x)y=μ(x)q(x)dx.
4.22

Divide both sides of Equation 4.21 by μ(x):μ(x):

y=1μ(x)[μ(x)q(x)dx+C].y=1μ(x)[μ(x)q(x)dx+C].
4.23

Since μ(x)μ(x) was previously calculated, we are now finished. An important note about the integrating constant C:C: It may seem that we are inconsistent in the usage of the integrating constant. However, the integral involving p(x)p(x) is necessary in order to find an integrating factor for Equation 4.15. Only one integrating factor is needed in order to solve the equation; therefore, it is safe to assign a value for CC for this integral. We chose C=0.C=0. When calculating the integral inside the brackets in Equation 4.21, it is necessary to keep our options open for the value of the integrating constant, because our goal is to find a general family of solutions to Equation 4.15. This integrating factor guarantees just that.

Problem-Solving Strategy: Solving a First-order Linear Differential Equation
  1. Put the equation into standard form and identify p(x)p(x) and q(x).q(x).
  2. Calculate the integrating factor μ(x)=ep(x)dx.μ(x)=ep(x)dx.
  3. Multiply both sides of the differential equation by μ(x).μ(x).
  4. Integrate both sides of the equation obtained in step 3,3, and divide both sides by μ(x).μ(x).
  5. If there is an initial condition, determine the value of C.C.

Example 4.16

Solving a First-order Linear Equation

Find a general solution for the differential equation xy+3y=4x23x.xy+3y=4x23x. Assume x>0.x>0.

Solution

  1. To put this differential equation into standard form, divide both sides by x:x:
    y+3xy=4x3.y+3xy=4x3.

    Therefore p(x)=3xp(x)=3x and q(x)=4x3.q(x)=4x3.
  2. The integrating factor is μ(x)=e(3/x)dx=e3lnx=x3.μ(x)=e(3/x)dx=e3lnx=x3.
  3. Multiplying both sides of the differential equation by μ(x)μ(x) gives us
    x3y+x3(3x)y=x3(4x3)x3y+3x2y=4x43x3ddx(x3y)=4x43x3.x3y+x3(3x)y=x3(4x3)x3y+3x2y=4x43x3ddx(x3y)=4x43x3.
  4. Integrate both sides of the equation.
    ddx(x3y)dx=4x43x3dxx3y=4x553x44+Cy=4x253x4+Cx−3.ddx(x3y)dx=4x43x3dxx3y=4x553x44+Cy=4x253x4+Cx−3.
  5. There is no initial value, so the problem is complete.

Analysis

You may have noticed the condition that was imposed on the differential equation; namely, x>0.x>0. For any nonzero value of C,C, the general solution is not defined at x=0.x=0. Furthermore, when x<0,x<0, the integrating factor changes. The integrating factor is given by Equation 4.19 as f(x)=ep(x)dx.f(x)=ep(x)dx. For this p(x)p(x) we get

ep(x)dx=e(3/x)dx=e3ln|x|=|x|3,ep(x)dx=e(3/x)dx=e3ln|x|=|x|3,

since x<0.x<0. The behavior of the general solution changes at x=0x=0 largely due to the fact that p(x)p(x) is not defined there.

Checkpoint 4.16

Find the general solution to the differential equation (x2)y+y=3x2+2x.(x2)y+y=3x2+2x. Assume x>2.x>2.

Now we use the same strategy to find the solution to an initial-value problem.

Example 4.17

A First-order Linear Initial-Value Problem

Solve the initial-value problem

y+3y=2x1,y(0)=3.y+3y=2x1,y(0)=3.

Solution

  1. This differential equation is already in standard form with p(x)=3p(x)=3 and q(x)=2x1.q(x)=2x1.
  2. The integrating factor is μ(x)=e3dx=e3x.μ(x)=e3dx=e3x.
  3. Multiplying both sides of the differential equation by μ(x)μ(x) gives
    e3xy+3e3xy=(2x1)e3xddx[ye3x]=(2x1)e3x.e3xy+3e3xy=(2x1)e3xddx[ye3x]=(2x1)e3x.

    Integrate both sides of the equation:
    ddx[ye3x]dx=(2x1)e3xdxye3x=e3x3(2x1)23e3xdxye3x=e3x(2x1)32e3x9+Cy=2x1329+Ce−3xy=2x359+Ce−3x.ddx[ye3x]dx=(2x1)e3xdxye3x=e3x3(2x1)23e3xdxye3x=e3x(2x1)32e3x9+Cy=2x1329+Ce−3xy=2x359+Ce−3x.
  4. Now substitute x=0x=0 and y=3y=3 into the general solution and solve for C:C:
    y=23x59+Ce−3x3=23(0)59+Ce−3(0)3=59+CC=329.y=23x59+Ce−3x3=23(0)59+Ce−3(0)3=59+CC=329.

    Therefore the solution to the initial-value problem is
    y=23x59+329e−3x.y=23x59+329e−3x.
Checkpoint 4.17

Solve the initial-value problem y2y=4x+3y(0)=−2.y2y=4x+3y(0)=−2.

Applications of First-order Linear Differential Equations

We look at two different applications of first-order linear differential equations. The first involves air resistance as it relates to objects that are rising or falling; the second involves an electrical circuit. Other applications are numerous, but most are solved in a similar fashion.

Free fall with air resistance

We discussed air resistance at the beginning of this section. The next example shows how to apply this concept for a ball in vertical motion. Other factors can affect the force of air resistance, such as the size and shape of the object, but we ignore them here.

Example 4.18

A Ball with Air Resistance

A racquetball is hit straight upward with an initial velocity of 22 m/s. The mass of a racquetball is approximately 0.04270.0427 kg. Air resistance acts on the ball with a force numerically equal to 0.5v,0.5v, where vv represents the velocity of the ball at time t.t.

  1. Find the velocity of the ball as a function of time.
  2. How long does it take for the ball to reach its maximum height?
  3. If the ball is hit from an initial height of 11 meter, how high will it reach?

Solution

  1. The mass m=0.0427kg,k=0.5,m=0.0427kg,k=0.5, and g=9.8m/s2.g=9.8m/s2. The initial velocity is v0=2v0=2 m/s. Therefore the initial-value problem is
    0.0427dvdt=−0.5v0.0427(9.8),v0=2.0.0427dvdt=−0.5v0.0427(9.8),v0=2.

    Dividing the differential equation by 0.04270.0427 gives
    dvdt=−11.7096v9.8,v0=2.dvdt=−11.7096v9.8,v0=2.

    The differential equation is linear. Using the problem-solving strategy for linear differential equations:
    Step 1. Rewrite the differential equation as dvdt+11.7096v=−9.8.dvdt+11.7096v=−9.8. This gives p(t)=11.7096p(t)=11.7096 and q(t)=−9.8q(t)=−9.8
    Step 2. The integrating factor is μ(t)=e11.7096dt=e11.7096t.μ(t)=e11.7096dt=e11.7096t.
    Step 3. Multiply the differential equation by μ(t):μ(t):
    e11.7096tdvdt+11.7096ve11.7096t=−9.8e11.7096tddt[ve11.7096t]=−9.8e11.7096t.e11.7096tdvdt+11.7096ve11.7096t=−9.8e11.7096tddt[ve11.7096t]=−9.8e11.7096t.

    Step 4. Integrate both sides:
    ddt[ve11.7096t]dt=−9.8e11.7096tdtve11.7096t=−9.811.7096e11.7096t+Cv(t)=−0.8369+Ce−11.7096t.ddt[ve11.7096t]dt=−9.8e11.7096tdtve11.7096t=−9.811.7096e11.7096t+Cv(t)=−0.8369+Ce−11.7096t.

    Step 5. Solve for CC using the initial condition v0=v(0)=2:v0=v(0)=2:
    v(t)=−0.8369+Ce−11.7096tv(0)=−0.8369+Ce−11.7096(0)2=−0.8369+CC=2.8369.v(t)=−0.8369+Ce−11.7096tv(0)=−0.8369+Ce−11.7096(0)2=−0.8369+CC=2.8369.

    Therefore the solution to the initial-value problem is v(t)=2.8369e−11.7096t0.8369.v(t)=2.8369e−11.7096t0.8369.
  2. The ball reaches its maximum height when the velocity is equal to zero. The reason is that when the velocity is positive, it is rising, and when it is negative, it is falling. Therefore when it is zero, it is neither rising nor falling, and is at its maximum height:
    2.8369e−11.7096t0.8369=02.8369e−11.7096t=0.8369e−11.7096t=0.83692.83690.295lne−11.7096t=ln0.2951.221−11.7096t=−1.221t0.104.2.8369e−11.7096t0.8369=02.8369e−11.7096t=0.8369e−11.7096t=0.83692.83690.295lne−11.7096t=ln0.2951.221−11.7096t=−1.221t0.104.

    Therefore it takes approximately 0.1040.104 second to reach maximum height.
  3. To find the height of the ball as a function of time, use the fact that the derivative of position is velocity, i.e., if h(t)h(t) represents the height at time t,t, then h(t)=v(t).h(t)=v(t). Because we know v(t)v(t) and the initial height, we can form an initial-value problem:
    h(t)=2.8369e−11.7096t0.8369,h(0)=1.h(t)=2.8369e−11.7096t0.8369,h(0)=1.

    Integrating both sides of the differential equation with respect to tt gives
    h(t)dt=2.8369e−11.7096t0.8369dth(t)=2.836911.7096e−11.7096t0.8369t+Ch(t)=−0.2423e−11.7096t0.8369t+C.h(t)dt=2.8369e−11.7096t0.8369dth(t)=2.836911.7096e−11.7096t0.8369t+Ch(t)=−0.2423e−11.7096t0.8369t+C.

    Solve for CC by using the initial condition:
    h(t)=−0.2423e−11.7096t0.8369t+Ch(0)=−0.2423e−11.7096(0)0.8369(0)+C1=−0.2423+CC=1.2423.h(t)=−0.2423e−11.7096t0.8369t+Ch(0)=−0.2423e−11.7096(0)0.8369(0)+C1=−0.2423+CC=1.2423.

    Therefore
    h(t)=−0.2423e−11.7096t0.8369t+1.2423.h(t)=−0.2423e−11.7096t0.8369t+1.2423.

    After 0.1040.104 second, the height is given by
    h(0.2)=−0.2423e−11.7096t0.8369t+1.24231.0836h(0.2)=−0.2423e−11.7096t0.8369t+1.24231.0836 meter.

Checkpoint 4.18

The weight of a penny is 2.52.5 grams (United States Mint, “Coin Specifications,” accessed April 9, 2015, http://www.usmint.gov/about_the_mint/?action=coin_specifications), and the upper observation deck of the Empire State Building is 369369 meters above the street. Since the penny is a small and relatively smooth object, air resistance acting on the penny is actually quite small. We assume the air resistance is numerically equal to 0.0025v.0.0025v. Furthermore, the penny is dropped with no initial velocity imparted to it.

  1. Set up an initial-value problem that represents the falling penny.
  2. Solve the problem for v(t).v(t).
  3. What is the terminal velocity of the penny (i.e., calculate the limit of the velocity as tt approaches infinity)?

Electrical Circuits

A source of electromotive force (e.g., a battery or generator) produces a flow of current in a closed circuit, and this current produces a voltage drop across each resistor, inductor, and capacitor in the circuit. Kirchhoff’s Loop Rule states that the sum of the voltage drops across resistors, inductors, and capacitors is equal to the total electromotive force in a closed circuit. We have the following three results:

  1. The voltage drop across a resistor is given by
    ER=Ri,ER=Ri,

    where RR is a constant of proportionality called the resistance, and ii is the current.
  2. The voltage drop across an inductor is given by
    EL=Li,EL=Li,

    where LL is a constant of proportionality called the inductance, and ii again denotes the current.
  3. The voltage drop across a capacitor is given by
    EC=1Cq,EC=1Cq,

where CC is a constant of proportionality called the capacitance, and qq is the instantaneous charge on the capacitor. The relationship between ii and qq is i=q.i=q.

We use units of volts (V)(V) to measure voltage E,E, amperes (A)(A) to measure current i,i, coulombs (C)(C) to measure charge q,q, ohms (Ω)(Ω) to measure resistance R,R, henrys (H)(H) to measure inductance L,L, and farads (F)(F) to measure capacitance C.C. Consider the circuit in Figure 4.25.

A diagram of an electric circuit in a rectangle. The top has a capacitor C, the left has a voltage generator Vs, the bottom was a resistor R, and the right has an inductor L.
Figure 4.25 A typical electric circuit, containing a voltage generator (VS),(VS), capacitor (C),(C), inductor (L),(L), and resistor (R).(R).

Applying Kirchhoff’s Loop Rule to this circuit, we let EE denote the electromotive force supplied by the voltage generator. Then

EL+ER+EC=E.EL+ER+EC=E.

Substituting the expressions for EL,ER,EL,ER, and ECEC into this equation, we obtain

Li+Ri+1Cq=E.Li+Ri+1Cq=E.
4.24

If there is no capacitor in the circuit, then the equation becomes

Li+Ri=E.Li+Ri=E.
4.25

This is a first-order differential equation in i.i. The circuit is referred to as an LRLR circuit.

Next, suppose there is no inductor in the circuit, but there is a capacitor and a resistor, so L=0,R0,L=0,R0, and C0.C0. Then Equation 4.23 can be rewritten as

Rq+1Cq=E,Rq+1Cq=E,
4.26

which is a first-order linear differential equation. This is referred to as an RC circuit. In either case, we can set up and solve an initial-value problem.

Example 4.19

Finding Current in an RL Electric Circuit

A circuit has in series an electromotive force given by E=50sin20tV,E=50sin20tV, a resistor of 5Ω,5Ω, and an inductor of 0.4H.0.4H. If the initial current is 0,0, find the current at time t>0.t>0.

Solution

We have a resistor and an inductor in the circuit, so we use Equation 4.24. The voltage drop across the resistor is given by ER=Ri=5i.ER=Ri=5i. The voltage drop across the inductor is given by EL=Li=0.4i.EL=Li=0.4i. The electromotive force becomes the right-hand side of Equation 4.24. Therefore Equation 4.24 becomes

0.4i+5i=50sin20t.0.4i+5i=50sin20t.

Dividing both sides by 0.40.4 gives the equation

i+12.5i=125sin20t.i+12.5i=125sin20t.

Since the initial current is 0, this result gives an initial condition of i(0)=0.i(0)=0. We can solve this initial-value problem using the five-step strategy for solving first-order differential equations.

Step 1. Rewrite the differential equation as i+12.5i=125sin20t.i+12.5i=125sin20t. This gives p(t)=12.5p(t)=12.5 and q(t)=125sin20t.q(t)=125sin20t.

Step 2. The integrating factor is μ(t)=e12.5dt=e12.5t.μ(t)=e12.5dt=e12.5t.

Step 3. Multiply the differential equation by μ(t):μ(t):

e12.5ti+12.5e12.5ti=125e12.5tsin20tddt[ie12.5t]=125e12.5tsin20t.e12.5ti+12.5e12.5ti=125e12.5tsin20tddt[ie12.5t]=125e12.5tsin20t.

Step 4. Integrate both sides:

ddt[ie12.5t]dt=125e12.5tsin20tdtie12.5t=(250sin20t400cos20t89)e12.5t+Ci(t)=250sin20t400cos20t89+Ce−12.5t.ddt[ie12.5t]dt=125e12.5tsin20tdtie12.5t=(250sin20t400cos20t89)e12.5t+Ci(t)=250sin20t400cos20t89+Ce−12.5t.

Step 5. Solve for CC using the initial condition v(0)=2:v(0)=2:

i(t)=250sin20t400cos20t89+Ce−12.5ti(0)=250sin20(0)400cos20(0)89+Ce−12.5(0)0=40089+CC=40089.i(t)=250sin20t400cos20t89+Ce−12.5ti(0)=250sin20(0)400cos20(0)89+Ce−12.5(0)0=40089+CC=40089.

Therefore the solution to the initial-value problem is i(t)=250sin20t400cos20t+400e−12.5t89=250sin20t400cos20t89+400e−12.5t89.i(t)=250sin20t400cos20t+400e−12.5t89=250sin20t400cos20t89+400e−12.5t89.

The first term can be rewritten as a single cosine function. First, multiply and divide by 2502+4002=5089:2502+4002=5089:

250sin20t400cos20t89=508989(250sin20t400cos20t5089)=508989(8cos20t895sin20t89).250sin20t400cos20t89=508989(250sin20t400cos20t5089)=508989(8cos20t895sin20t89).

Next, define φφ to be an acute angle such that cosφ=889.cosφ=889. Then sinφ=589sinφ=589 and

508989(8cos20t895sin20t89)=508989(cosφcos20tsinφsin20t)=508989cos(20t+φ).508989(8cos20t895sin20t89)=508989(cosφcos20tsinφsin20t)=508989cos(20t+φ).

Therefore the solution can be written as

i(t)=508989cos(20t+φ)+400e−12.5t89.i(t)=508989cos(20t+φ)+400e−12.5t89.

The second term is called the attenuation term, because it disappears rapidly as t grows larger. The phase shift is given by φ,φ, and the amplitude of the steady-state current is given by 508989.508989. The graph of this solution appears in Figure 4.26:

A graph of the given solution over [0, 6] on the x axis. It is an oscillating function, rapidly going from just below -5 to just above 5.
Figure 4.26

Checkpoint 4.19

A circuit has in series an electromotive force given by E=20sin5tE=20sin5t V, a capacitor with capacitance 0.02F,0.02F, and a resistor of 8Ω.8Ω. If the initial charge is 4C,4C, find the charge at time t>0.t>0.

Section 4.5 Exercises

Are the following differential equations linear? Explain your reasoning.

208.

dydx=x2y+sinxdydx=x2y+sinx

209.

dydt=tydydt=ty

210.

dydt+y2=xdydt+y2=x

211.

y=x3+exy=x3+ex

212.

y=y+eyy=y+ey

Write the following first-order differential equations in standard form.

213.

y=x3y+sinxy=x3y+sinx

214.

y+3ylnx=0y+3ylnx=0

215.

xy=(3x+2)y+xexxy=(3x+2)y+xex

216.

dydt=4y+ty+tantdydt=4y+ty+tant

217.

dydt=yx(x+1)dydt=yx(x+1)

What are the integrating factors for the following differential equations?

218.

y=xy+3y=xy+3

219.

y+exy=sinxy+exy=sinx

220.

y=xln(x)y+3xy=xln(x)y+3x

221.

dydx=tanh(x)y+1dydx=tanh(x)y+1

222.

dydt+3ty=etydydt+3ty=ety

Solve the following differential equations by using integrating factors.

223.

y=3y+2y=3y+2

224.

y=2yx2y=2yx2

225.

xy=3y6x2xy=3y6x2

226.

(x+2)y=3x+y(x+2)y=3x+y

227.

y=3x+xyy=3x+xy

228.

xy=x+yxy=x+y

229.

sin(x)y=y+2xsin(x)y=y+2x

230.

y=y+exy=y+ex

231.

xy=3y+x2xy=3y+x2

232.

y+lnx=yxy+lnx=yx

Solve the following differential equations. Use your calculator to draw a family of solutions. Are there certain initial conditions that change the behavior of the solution?

233.

[T] (x+2)y=2y1(x+2)y=2y1

234.

[T] y=3et/32yy=3et/32y

235.

[T] xy+y2=sin(3t)xy+y2=sin(3t)

236.

[T] xy=2cosxx3yxy=2cosxx3y

237.

[T] (x+1)y=3y+x2+2x+1(x+1)y=3y+x2+2x+1

238.

[T] sin(x)y+cos(x)y=2xsin(x)y+cos(x)y=2x

239.

[T] x2+1y=y+2x2+1y=y+2

240.

[T] x3y+2x2y=x+1x3y+2x2y=x+1

Solve the following initial-value problems by using integrating factors.

241.

y+y=x,y(0)=3y+y=x,y(0)=3

242.

y=y+2x2,y(0)=0y=y+2x2,y(0)=0

243.

xy=y3x3,y(1)=0xy=y3x3,y(1)=0

244.

x2y=xylnx,y(1)=1x2y=xylnx,y(1)=1

245.

(1+x2)y=y1,y(0)=0(1+x2)y=y1,y(0)=0

246.

xy=y+2xlnx,y(1)=5xy=y+2xlnx,y(1)=5

247.

(2+x)y=y+2+x,y(0)=0(2+x)y=y+2+x,y(0)=0

248.

y=xy+2xex,y(0)=2y=xy+2xex,y(0)=2

249.

xy=y+2x,y(0)=1xy=y+2x,y(0)=1

250.

y=2y+xex,y(0)=−1y=2y+xex,y(0)=−1

251.

A falling object of mass mm can reach terminal velocity when the drag force is proportional to its velocity, with proportionality constant k.k. Set up the differential equation and solve for the velocity given an initial velocity of 0.0.

252.

Using your expression from the preceding problem, what is the terminal velocity? (Hint: Examine the limiting behavior; does the velocity approach a value?)

253.

[T] Using your equation for terminal velocity, solve for the distance fallen. How long does it take to fall 50005000 meters if the mass is 100100 kilograms, the acceleration due to gravity is 9.89.8 m/s2 and the proportionality constant is 4?4?

254.

A more accurate way to describe terminal velocity is that the drag force is proportional to the square of velocity, with a proportionality constant k.k. Set up the differential equation and solve for the velocity.

255.

Using your expression from the preceding problem, what is the terminal velocity? (Hint: Examine the limiting behavior: Does the velocity approach a value?)

256.

[T] Using your equation for terminal velocity, solve for the distance fallen. How long does it take to fall 50005000 meters if the mass is 100100 kilograms, the acceleration due to gravity is 9.8m/s29.8m/s2 and the proportionality constant is 4?4? Does it take more or less time than your initial estimate?

For the following problems, determine how parameter aa affects the solution.

257.

Solve the generic equation y=ax+y.y=ax+y. How does varying aa change the behavior?

258.

Solve the generic equation y=ax+y.y=ax+y. How does varying aa change the behavior?

259.

Solve the generic equation y=ax+xy.y=ax+xy. How does varying aa change the behavior?

260.

Solve the generic equation y=x+axy.y=x+axy. How does varying aa change the behavior?

261.

Solve yy=ektyy=ekt with the initial condition y(0)=0.y(0)=0. As kk approaches 1,1, what happens to your formula?

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