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Calculus Volume 2

4.1 Basics of Differential Equations

Calculus Volume 24.1 Basics of Differential 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

  • 4.1.1. Identify the order of a differential equation.
  • 4.1.2. Explain what is meant by a solution to a differential equation.
  • 4.1.3. Distinguish between the general solution and a particular solution of a differential equation.
  • 4.1.4. Identify an initial-value problem.
  • 4.1.5. Identify whether a given function is a solution to a differential equation or an initial-value problem.

Calculus is the mathematics of change, and rates of change are expressed by derivatives. Thus, one of the most common ways to use calculus is to set up an equation containing an unknown function y=f(x)y=f(x) and its derivative, known as a differential equation. Solving such equations often provides information about how quantities change and frequently provides insight into how and why the changes occur.

Techniques for solving differential equations can take many different forms, including direct solution, use of graphs, or computer calculations. We introduce the main ideas in this chapter and describe them in a little more detail later in the course. In this section we study what differential equations are, how to verify their solutions, some methods that are used for solving them, and some examples of common and useful equations.

General Differential Equations

Consider the equation y=3x2,y=3x2, which is an example of a differential equation because it includes a derivative. There is a relationship between the variables xx and y:yy:y is an unknown function of x.x. Furthermore, the left-hand side of the equation is the derivative of y.y. Therefore we can interpret this equation as follows: Start with some function y=f(x)y=f(x) and take its derivative. The answer must be equal to 3x2.3x2. What function has a derivative that is equal to 3x2?3x2? One such function is y=x3,y=x3, so this function is considered a solution to a differential equation.

Definition

A differential equation is an equation involving an unknown function y=f(x)y=f(x) and one or more of its derivatives. A solution to a differential equation is a function y=f(x)y=f(x) that satisfies the differential equation when ff and its derivatives are substituted into the equation.

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Some examples of differential equations and their solutions appear in Table 4.1.

Equation Solution
y=2xy=2x y=x2y=x2
y+3y=6x+11y+3y=6x+11 y=e−3x+2x+3y=e−3x+2x+3
y3y+2y=24e−2xy3y+2y=24e−2x y=3ex4e2x+2e−2xy=3ex4e2x+2e−2x
Table 4.1 Examples of Differential Equations and Their Solutions

Note that a solution to a differential equation is not necessarily unique, primarily because the derivative of a constant is zero. For example, y=x2+4y=x2+4 is also a solution to the first differential equation in Table 4.1. We will return to this idea a little bit later in this section. For now, let’s focus on what it means for a function to be a solution to a differential equation.

Example 4.1

Verifying Solutions of Differential Equations

Verify that the function y=e−3x+2x+3y=e−3x+2x+3 is a solution to the differential equation y+3y=6x+11.y+3y=6x+11.

Solution

To verify the solution, we first calculate yy using the chain rule for derivatives. This gives y=−3e−3x+2.y=−3e−3x+2. Next we substitute yy and yy into the left-hand side of the differential equation:

(−3e−2x+2)+3(e−2x+2x+3).(−3e−2x+2)+3(e−2x+2x+3).

The resulting expression can be simplified by first distributing to eliminate the parentheses, giving

−3e−2x+2+3e−2x+6x+9.−3e−2x+2+3e−2x+6x+9.

Combining like terms leads to the expression 6x+11,6x+11, which is equal to the right-hand side of the differential equation. This result verifies that y=e−3x+2x+3y=e−3x+2x+3 is a solution of the differential equation.

Checkpoint 4.1

Verify that y=2e3x2x2y=2e3x2x2 is a solution to the differential equation y3y=6x+4.y3y=6x+4.

It is convenient to define characteristics of differential equations that make it easier to talk about them and categorize them. The most basic characteristic of a differential equation is its order.

Definition

The order of a differential equation is the highest order of any derivative of the unknown function that appears in the equation.

Example 4.2

Identifying the Order of a Differential Equation

What is the order of each of the following differential equations?

  1. y4y=x23x+4y4y=x23x+4
  2. x2y3xy+xy3y=sinxx2y3xy+xy3y=sinx
  3. 4xy(4)6x2y+12x4y=x33x2+4x124xy(4)6x2y+12x4y=x33x2+4x12

Solution

  1. The highest derivative in the equation is y,y, so the order is 1.1.
  2. The highest derivative in the equation is y,y, so the order is 3.3.
  3. The highest derivative in the equation is y(4),y(4), so the order is 4.4.
Checkpoint 4.2

What is the order of the following differential equation?

(x43x)y(5)(3x2+1)y+3y=sinxcosx(x43x)y(5)(3x2+1)y+3y=sinxcosx

General and Particular Solutions

We already noted that the differential equation y=2xy=2x has at least two solutions: y=x2y=x2 and y=x2+4.y=x2+4. The only difference between these two solutions is the last term, which is a constant. What if the last term is a different constant? Will this expression still be a solution to the differential equation? In fact, any function of the form y=x2+C,y=x2+C, where CC represents any constant, is a solution as well. The reason is that the derivative of x2+Cx2+C is 2x,2x, regardless of the value of C.C. It can be shown that any solution of this differential equation must be of the form y=x2+C.y=x2+C. This is an example of a general solution to a differential equation. A graph of some of these solutions is given in Figure 4.2. (Note: in this graph we used even integer values for CC ranging between −4−4 and 4.4. In fact, there is no restriction on the value of C;C; it can be an integer or not.)

A graph of a family of solutions to the differential equation y’ = 2 x, which are of the form y = x ^ 2 + C. Parabolas are drawn for values of C: -4, -2, 0, 2, and 4.
Figure 4.2 Family of solutions to the differential equation y=2x.y=2x.

In this example, we are free to choose any solution we wish; for example, y=x23y=x23 is a member of the family of solutions to this differential equation. This is called a particular solution to the differential equation. A particular solution can often be uniquely identified if we are given additional information about the problem.

Example 4.3

Finding a Particular Solution

Find the particular solution to the differential equation y=2xy=2x passing through the point (2,7).(2,7).

Solution

Any function of the form y=x2+Cy=x2+C is a solution to this differential equation. To determine the value of C,C, we substitute the values x=2x=2 and y=7y=7 into this equation and solve for C:C:

y=x2+C7=22+C=4+CC=3.y=x2+C7=22+C=4+CC=3.

Therefore the particular solution passing through the point (2,7)(2,7) is y=x2+3.y=x2+3.

Checkpoint 4.3

Find the particular solution to the differential equation

y=4x+3y=4x+3

passing through the point (1,7),(1,7), given that y=2x2+3x+Cy=2x2+3x+C is a general solution to the differential equation.

Initial-Value Problems

Usually a given differential equation has an infinite number of solutions, so it is natural to ask which one we want to use. To choose one solution, more information is needed. Some specific information that can be useful is an initial value, which is an ordered pair that is used to find a particular solution.

A differential equation together with one or more initial values is called an initial-value problem. The general rule is that the number of initial values needed for an initial-value problem is equal to the order of the differential equation. For example, if we have the differential equation y=2x,y=2x, then y(3)=7y(3)=7 is an initial value, and when taken together, these equations form an initial-value problem. The differential equation y3y+2y=4exy3y+2y=4ex is second order, so we need two initial values. With initial-value problems of order greater than one, the same value should be used for the independent variable. An example of initial values for this second-order equation would be y(0)=2y(0)=2 and y(0)=−1.y(0)=−1. These two initial values together with the differential equation form an initial-value problem. These problems are so named because often the independent variable in the unknown function is t,t, which represents time. Thus, a value of t=0t=0 represents the beginning of the problem.

Example 4.4

Verifying a Solution to an Initial-Value Problem

Verify that the function y=2e−2t+ety=2e−2t+et is a solution to the initial-value problem

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

Solution

For a function to satisfy an initial-value problem, it must satisfy both the differential equation and the initial condition. To show that yy satisfies the differential equation, we start by calculating y.y. This gives y=−4e−2t+et.y=−4e−2t+et. Next we substitute both yy and yy into the left-hand side of the differential equation and simplify:

y+2y=(−4e−2t+et)+2(2e−2t+et)=−4e−2t+et+4e−2t+2et=3et.y+2y=(−4e−2t+et)+2(2e−2t+et)=−4e−2t+et+4e−2t+2et=3et.

This is equal to the right-hand side of the differential equation, so y=2e−2t+ety=2e−2t+et solves the differential equation. Next we calculate y(0):y(0):

y(0)=2e−2(0)+e0=2+1=3.y(0)=2e−2(0)+e0=2+1=3.

This result verifies the initial value. Therefore the given function satisfies the initial-value problem.

Checkpoint 4.4

Verify that y=3e2t+4sinty=3e2t+4sint is a solution to the initial-value problem

y2y=4cost8sint,y(0)=3.y2y=4cost8sint,y(0)=3.

In Example 4.4, the initial-value problem consisted of two parts. The first part was the differential equation y+2y=3ex,y+2y=3ex, and the second part was the initial value y(0)=3.y(0)=3. These two equations together formed the initial-value problem.

The same is true in general. An initial-value problem will consists of two parts: the differential equation and the initial condition. The differential equation has a family of solutions, and the initial condition determines the value of C.C. The family of solutions to the differential equation in Example 4.4 is given by y=2e−2t+Cet.y=2e−2t+Cet. This family of solutions is shown in Figure 4.3, with the particular solution y=2e−2t+ety=2e−2t+et labeled.

A graph of a family of solutions to the differential equation y’ + 2 y = 3 e ^ t, which are of the form y = 2 e ^ (-2 t) + C e ^ t. The versions with C = 1, 0.5, and -0.2 are shown, among others not labeled. For all values of C, the function increases rapidly for t < 0 as t goes to negative infinity. For C > 0, the function changes direction and increases in a gentle curve as t goes to infinity. Larger values of C have a tighter curve closer to the y axis and at a higher y value. For C = 0, the function goes to 0 as t goes to infinity. For C < 0, the function continues to decrease as t goes to infinity.
Figure 4.3 A family of solutions to the differential equation y+2y=3et.y+2y=3et. The particular solution y=2e−2t+ety=2e−2t+et is labeled.

Example 4.5

Solving an Initial-value Problem

Solve the following initial-value problem:

y=3ex+x24,y(0)=5.y=3ex+x24,y(0)=5.

Solution

The first step in solving this initial-value problem is to find a general family of solutions. To do this, we find an antiderivative of both sides of the differential equation

ydx=(3ex+x24)dx,ydx=(3ex+x24)dx,

namely,

y+C1=3ex+13x34x+C2.y+C1=3ex+13x34x+C2.
4.1

We are able to integrate both sides because the y term appears by itself. Notice that there are two integration constants: C1C1 and C2.C2. Solving Equation 4.1 for yy gives

y=3ex+13x34x+C2C1.y=3ex+13x34x+C2C1.

Because C1C1 and C2C2 are both constants, C2C1C2C1 is also a constant. We can therefore define C=C2C1,C=C2C1, which leads to the equation

y=3ex+13x34x+C.y=3ex+13x34x+C.

Next we determine the value of C.C. To do this, we substitute x=0x=0 and y=5y=5 into Equation 4.1 and solve for C:C:

5=3e0+13034(0)+C5=3+CC=2.5=3e0+13034(0)+C5=3+CC=2.

Now we substitute the value C=2C=2 into Equation 4.1. The solution to the initial-value problem is y=3ex+13x34x+2.y=3ex+13x34x+2.

Analysis

The difference between a general solution and a particular solution is that a general solution involves a family of functions, either explicitly or implicitly defined, of the independent variable. The initial value or values determine which particular solution in the family of solutions satisfies the desired conditions.

Checkpoint 4.5

Solve the initial-value problem

y=x24x+36ex,y(0)=8.y=x24x+36ex,y(0)=8.

In physics and engineering applications, we often consider the forces acting upon an object, and use this information to understand the resulting motion that may occur. For example, if we start with an object at Earth’s surface, the primary force acting upon that object is gravity. Physicists and engineers can use this information, along with Newton’s second law of motion (in equation form F=ma,F=ma, where FF represents force, mm represents mass, and aa represents acceleration), to derive an equation that can be solved.

A picture of a baseball with an arrow underneath it pointing down. The arrow is labeled g = -9.8 m/sec ^ 2.
Figure 4.4 For a baseball falling in air, the only force acting on it is gravity (neglecting air resistance).

In Figure 4.4 we assume that the only force acting on a baseball is the force of gravity. This assumption ignores air resistance. (The force due to air resistance is considered in a later discussion.) The acceleration due to gravity at Earth’s surface, g,g, is approximately 9.8m/s2.9.8m/s2. We introduce a frame of reference, where Earth’s surface is at a height of 0 meters. Let v(t)v(t) represent the velocity of the object in meters per second. If v(t)>0,v(t)>0, the ball is rising, and if v(t)<0,v(t)<0, the ball is falling (Figure 4.5).

A picture of a baseball with an arrow above it pointing up and an arrow below it pointing down. The arrow pointing up is labeled v(t) > 0, and the arrow pointing down is labeled v(t) < 0.
Figure 4.5 Possible velocities for the rising/falling baseball.

Our goal is to solve for the velocity v(t)v(t) at any time t.t. To do this, we set up an initial-value problem. Suppose the mass of the ball is m,m, where mm is measured in kilograms. We use Newton’s second law, which states that the force acting on an object is equal to its mass times its acceleration (F=ma).(F=ma). Acceleration is the derivative of velocity, so a(t)=v(t).a(t)=v(t). Therefore the force acting on the baseball is given by F=mv(t).F=mv(t). However, this force must be equal to the force of gravity acting on the object, which (again using Newton’s second law) is given by Fg=mg,Fg=mg, since this force acts in a downward direction. Therefore we obtain the equation F=Fg,F=Fg, which becomes mv(t)=mg.mv(t)=mg. Dividing both sides of the equation by mm gives the equation

v(t)=g.v(t)=g.

Notice that this differential equation remains the same regardless of the mass of the object.

We now need an initial value. Because we are solving for velocity, it makes sense in the context of the problem to assume that we know the initial velocity, or the velocity at time t=0.t=0. This is denoted by v(0)=v0.v(0)=v0.

Example 4.6

Velocity of a Moving Baseball

A baseball is thrown upward from a height of 33 meters above Earth’s surface with an initial velocity of 10m/s,10m/s, and the only force acting on it is gravity. The ball has a mass of 0.15kg0.15kg at Earth’s surface.

  1. Find the velocity v(t)v(t) of the baseball at time t.t.
  2. What is its velocity after 22 seconds?

Solution

  1. From the preceding discussion, the differential equation that applies in this situation is
    v(t)=g,v(t)=g,

    where g=9.8m/s2.g=9.8m/s2. The initial condition is v(0)=v0,v(0)=v0, where v0=10m/s.v0=10m/s. Therefore the initial-value problem is v(t)=−9.8m/s2,v(0)=10m/s.v(t)=−9.8m/s2,v(0)=10m/s.
    The first step in solving this initial-value problem is to take the antiderivative of both sides of the differential equation. This gives
    v(t)dt=−9.8dtv(t)=−9.8t+C.v(t)dt=−9.8dtv(t)=−9.8t+C.

    The next step is to solve for C.C. To do this, substitute t=0t=0 and v(0)=10:v(0)=10:
    v(t)=−9.8t+Cv(0)=−9.8(0)+C10=C.v(t)=−9.8t+Cv(0)=−9.8(0)+C10=C.

    Therefore C=10C=10 and the velocity function is given by v(t)=−9.8t+10.v(t)=−9.8t+10.
  2. To find the velocity after 22 seconds, substitute t=2t=2 into v(t).v(t).
    v(t)=−9.8t+10v(2)=−9.8(2)+10v(2)=−9.6.v(t)=−9.8t+10v(2)=−9.8(2)+10v(2)=−9.6.

    The units of velocity are meters per second. Since the answer is negative, the object is falling at a speed of 9.6m/s.9.6m/s.
Checkpoint 4.6

Suppose a rock falls from rest from a height of 100100 meters and the only force acting on it is gravity. Find an equation for the velocity v(t)v(t) as a function of time, measured in meters per second.

A natural question to ask after solving this type of problem is how high the object will be above Earth’s surface at a given point in time. Let s(t)s(t) denote the height above Earth’s surface of the object, measured in meters. Because velocity is the derivative of position (in this case height), this assumption gives the equation s(t)=v(t).s(t)=v(t). An initial value is necessary; in this case the initial height of the object works well. Let the initial height be given by the equation s(0)=s0.s(0)=s0. Together these assumptions give the initial-value problem

s(t)=v(t),s(0)=s0.s(t)=v(t),s(0)=s0.

If the velocity function is known, then it is possible to solve for the position function as well.

Example 4.7

Height of a Moving Baseball

A baseball is thrown upward from a height of 33 meters above Earth’s surface with an initial velocity of 10m/s,10m/s, and the only force acting on it is gravity. The ball has a mass of 0.150.15 kilogram at Earth’s surface.

  1. Find the position s(t)s(t) of the baseball at time t.t.
  2. What is its height after 22 seconds?

Solution

  1. We already know the velocity function for this problem is v(t)=−9.8t+10.v(t)=−9.8t+10. The initial height of the baseball is 33 meters, so s0=3.s0=3. Therefore the initial-value problem for this example is
    To solve the initial-value problem, we first find the antiderivatives:
    s(t)dt=−9.8t+10dts(t)=−4.9t2+10t+C.s(t)dt=−9.8t+10dts(t)=−4.9t2+10t+C.

    Next we substitute t=0t=0 and solve for C:C:
    s(t)=−4.9t2+10t+Cs(0)=−4.9(0)2+10(0)+C3=C.s(t)=−4.9t2+10t+Cs(0)=−4.9(0)2+10(0)+C3=C.

    Therefore the position function is s(t)=−4.9t2+10t+3.s(t)=−4.9t2+10t+3.
  2. The height of the baseball after 2s2s is given by s(2):s(2):
    s(2)=−4.9(2)2+10(2)+3=−4.9(4)+23=3.4.s(2)=−4.9(2)2+10(2)+3=−4.9(4)+23=3.4.

    Therefore the baseball is 3.43.4 meters above Earth’s surface after 22 seconds. It is worth noting that the mass of the ball cancelled out completely in the process of solving the problem.

Section 4.1 Exercises

Determine the order of the following differential equations.

1.

y+y=3y2y+y=3y2

2.

(y)2=y+2y(y)2=y+2y

3.

y+yy=3x2y+yy=3x2

4.

y=y+3t2y=y+3t2

5.

dydt=tdydt=t

6.

dydx+d2ydx2=3x4dydx+d2ydx2=3x4

7.

(dydt)2+8dydt+3y=4t(dydt)2+8dydt+3y=4t

Verify that the following functions are solutions to the given differential equation.

8.

y=x33y=x33 solves y=x2y=x2

9.

y=2ex+x1y=2ex+x1 solves y=xyy=xy

10.

y=e3xex2y=e3xex2 solves y=3y+exy=3y+ex

11.

y=11xy=11x solves y=y2y=y2

12.

y=ex2/2y=ex2/2 solves y=xyy=xy

13.

y=4+lnxy=4+lnx solves xy=1xy=1

14.

y=3x+xlnxy=3x+xlnx solves y=lnxy=lnx

15.

y=2exx1y=2exx1 solves y=y+xy=y+x

16.

y=ex+sinx2cosx2y=ex+sinx2cosx2 solves y=cosx+yy=cosx+y

17.

y=πecosxy=πecosx solves y=ysinxy=ysinx

Verify the following general solutions and find the particular solution.

18.

Find the particular solution to the differential equation y=4x2y=4x2 that passes through (−3,−30),(−3,−30), given that y=C+4x33y=C+4x33 is a general solution.

19.

Find the particular solution to the differential equation y=3x3y=3x3 that passes through (1,4.75),(1,4.75), given that y=C+3x44y=C+3x44 is a general solution.

20.

Find the particular solution to the differential equation y=3x2yy=3x2y that passes through (0,12),(0,12), given that y=Cex3y=Cex3 is a general solution.

21.

Find the particular solution to the differential equation y=2xyy=2xy that passes through (0,12),(0,12), given that y=Cex2y=Cex2 is a general solution.

22.

Find the particular solution to the differential equation y=(2xy)2y=(2xy)2 that passes through (1,12),(1,12), given that y=3C+4x3y=3C+4x3 is a general solution.

23.

Find the particular solution to the differential equation yx2=yyx2=y that passes through (1,2e),(1,2e), given that y=Ce1/xy=Ce1/x is a general solution.

24.

Find the particular solution to the differential equation 8dxdt=−2cos(2t)cos(4t)8dxdt=−2cos(2t)cos(4t) that passes through (π,π),(π,π), given that x=C18sin(2t)132sin(4t)x=C18sin(2t)132sin(4t) is a general solution.

25.

Find the particular solution to the differential equation dudt=tanududt=tanu that passes through (1,π2),(1,π2), given that u=sin−1(eC+t)u=sin−1(eC+t) is a general solution.

26.

Find the particular solution to the differential equation dydt=e(t+y)dydt=e(t+y) that passes through (1,0),(1,0), given that y=ln(Cet)y=ln(Cet) is a general solution.

27.

Find the particular solution to the differential equation y(1x2)=1+yy(1x2)=1+y that passes through (0,−2),(0,−2), given that y=Cx+11x1y=Cx+11x1 is a general solution.

For the following problems, find the general solution to the differential equation.

28.

y=3x+exy=3x+ex

29.

y=lnx+tanxy=lnx+tanx

30.

y=sinxecosxy=sinxecosx

31.

y=4xy=4x

32.

y=sin−1(2x)y=sin−1(2x)

33.

y=2tt2+16y=2tt2+16

34.

x=cotht+lnt+3t2x=cotht+lnt+3t2

35.

x=t4+tx=t4+t

36.

y=yy=y

37.

y=yxy=yx

Solve the following initial-value problems starting from y(t=0)=1y(t=0)=1 and y(t=0)=−1.y(t=0)=−1. Draw both solutions on the same graph.

38.

dydt=2tdydt=2t

39.

dydt=tdydt=t

40.

dydt=2ydydt=2y

41.

dydt=ydydt=y

42.

dydt=2dydt=2

Solve the following initial-value problems starting from y0=10.y0=10. At what time does yy increase to 100100 or drop to 1?1?

43.

dydt=4tdydt=4t

44.

dydt=4ydydt=4y

45.

dydt=−2ydydt=−2y

46.

dydt=e4tdydt=e4t

47.

dydt=e−4tdydt=e−4t

Recall that a family of solutions includes solutions to a differential equation that differ by a constant. For the following problems, use your calculator to graph a family of solutions to the given differential equation. Use initial conditions from y(t=0)=−10y(t=0)=−10 to y(t=0)=10y(t=0)=10 increasing by 2.2. Is there some critical point where the behavior of the solution begins to change?

48.

[T] y=y(x)y=y(x)

49.

[T] xy=yxy=y

50.

[T] y=t3y=t3

51.

[T] y=x+yy=x+y (Hint: y=Cexx1y=Cexx1 is the general solution)

52.

[T] y=xlnx+sinxy=xlnx+sinx

53.

Find the general solution to describe the velocity of a ball of mass 1lb1lb that is thrown upward at a rate aa ft/sec.

54.

In the preceding problem, if the initial velocity of the ball thrown into the air is a=25a=25 ft/s, write the particular solution to the velocity of the ball. Solve to find the time when the ball hits the ground.

55.

You throw two objects with differing masses m1m1 and m2m2 upward into the air with the same initial velocity aa ft/s. What is the difference in their velocity after 11 second?

56.

[T] You throw a ball of mass 11 kilogram upward with a velocity of a=25a=25 m/s on Mars, where the force of gravity is g=−3.711g=−3.711 m/s2. Use your calculator to approximate how much longer the ball is in the air on Mars.

57.

[T] For the previous problem, use your calculator to approximate how much higher the ball went on Mars.

58.

[T] A car on the freeway accelerates according to a=15cos(πt),a=15cos(πt), where tt is measured in hours. Set up and solve the differential equation to determine the velocity of the car if it has an initial speed of 5151 mph. After 4040 minutes of driving, what is the driver’s velocity?

59.

[T] For the car in the preceding problem, find the expression for the distance the car has traveled in time t,t, assuming an initial distance of 0.0. How long does it take the car to travel 100100 miles? Round your answer to hours and minutes.

60.

[T] For the previous problem, find the total distance traveled in the first hour.

61.

Substitute y=Be3ty=Be3t into yy=8e3tyy=8e3t to find a particular solution.

62.

Substitute y=acos(2t)+bsin(2t)y=acos(2t)+bsin(2t) into y+y=4sin(2t)y+y=4sin(2t) to find a particular solution.

63.

Substitute y=a+bt+ct2y=a+bt+ct2 into y+y=1+t2y+y=1+t2 to find a particular solution.

64.

Substitute y=aetcost+betsinty=aetcost+betsint into y=2etcosty=2etcost to find a particular solution.

65.

Solve y=ekty=ekt with the initial condition y(0)=0y(0)=0 and solve y=1y=1 with the same initial condition. As kk approaches 0,0, what do you notice?

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