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Calculus Volume 2

4.3 Separable Equations

Calculus Volume 24.3 Separable 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

  • 4.3.1. Use separation of variables to solve a differential equation.
  • 4.3.2. Solve applications using separation of variables.

We now examine a solution technique for finding exact solutions to a class of differential equations known as separable differential equations. These equations are common in a wide variety of disciplines, including physics, chemistry, and engineering. We illustrate a few applications at the end of the section.

Separation of Variables

We start with a definition and some examples.

Definition

A separable differential equation is any equation that can be written in the form

y=f(x)g(y).y=f(x)g(y).
4.3

The term ‘separable’ refers to the fact that the right-hand side of the equation can be separated into a function of xx times a function of y.y. Examples of separable differential equations include

y=(x24)(3y+2)y=6x2+4xy=secy+tanyy=xy+3x2y6.y=(x24)(3y+2)y=6x2+4xy=secy+tanyy=xy+3x2y6.

The second equation is separable with f(x)=6x2+4xf(x)=6x2+4x and g(y)=1,g(y)=1, the third equation is separable with f(x)=1f(x)=1 and g(y)=secy+tany,g(y)=secy+tany, and the right-hand side of the fourth equation can be factored as (x+3)(y2),(x+3)(y2), so it is separable as well. The third equation is also called an autonomous differential equation because the right-hand side of the equation is a function of yy alone. If a differential equation is separable, then it is possible to solve the equation using the method of separation of variables.

Problem-Solving Strategy: Separation of Variables
  1. Check for any values of yy that make g(y)=0.g(y)=0. These correspond to constant solutions.
  2. Rewrite the differential equation in the form dyg(y)=f(x)dx.dyg(y)=f(x)dx.
  3. Integrate both sides of the equation.
  4. Solve the resulting equation for yy if possible.
  5. If an initial condition exists, substitute the appropriate values for xx and yy into the equation and solve for the constant.

Note that Step 4. states “Solve the resulting equation for yy if possible.” It is not always possible to obtain yy as an explicit function of x.x. Quite often we have to be satisfied with finding yy as an implicit function of x.x.

Example 4.10

Using Separation of Variables

Find a general solution to the differential equation y=(x24)(3y+2)y=(x24)(3y+2) using the method of separation of variables.

Solution

Follow the five-step method of separation of variables.

  1. In this example, f(x)=x24f(x)=x24 and g(y)=3y+2.g(y)=3y+2. Setting g(y)=0g(y)=0 gives y=23y=23 as a constant solution.
  2. Rewrite the differential equation in the form
    dy3y+2=(x24)dx.dy3y+2=(x24)dx.
  3. Integrate both sides of the equation:
    dy3y+2=(x24)dx.dy3y+2=(x24)dx.

    Let u=3y+2.u=3y+2. Then du=3dydxdx,du=3dydxdx, so the equation becomes
    131udu=13x34x+C13ln|u|=13x34x+C13ln|3y+2|=13x34x+C.131udu=13x34x+C13ln|u|=13x34x+C13ln|3y+2|=13x34x+C.
  4. To solve this equation for y,y, first multiply both sides of the equation by 3.3.
    ln|3y+2|=x312x+3Cln|3y+2|=x312x+3C

    Now we use some logic in dealing with the constant C.C. Since CC represents an arbitrary constant, 3C3C also represents an arbitrary constant. If we call the second arbitrary constant C1,C1, the equation becomes
    ln|3y+2|=x312x+C1.ln|3y+2|=x312x+C1.

    Now exponentiate both sides of the equation (i.e., make each side of the equation the exponent for the base e).e).
    eln|3y+2|=ex312x+C1|3y+2|=eC1ex312xeln|3y+2|=ex312x+C1|3y+2|=eC1ex312x

    Again define a new constant C2=ec1C2=ec1 (note that C2>0):C2>0):
    |3y+2|=C2ex312x.|3y+2|=C2ex312x.

    This corresponds to two separate equations: 3y+2=C2ex312x3y+2=C2ex312x and 3y+2=C2ex312x.3y+2=C2ex312x.
    The solution to either equation can be written in the form y=−2±C2ex312x3.y=−2±C2ex312x3.
    Since C2>0,C2>0, it does not matter whether we use plus or minus, so the constant can actually have either sign. Furthermore, the subscript on the constant CC is entirely arbitrary, and can be dropped. Therefore the solution can be written as
    y=−2+Cex312x3.y=−2+Cex312x3.
  5. No initial condition is imposed, so we are finished.
Checkpoint 4.10

Use the method of separation of variables to find a general solution to the differential equation y=2xy+3y4x6.y=2xy+3y4x6.

Example 4.11

Solving an Initial-Value Problem

Using the method of separation of variables, solve the initial-value problem

y=(2x+3)(y24),y(0)=−3.y=(2x+3)(y24),y(0)=−3.

Solution

Follow the five-step method of separation of variables.

  1. In this example, f(x)=2x+3f(x)=2x+3 and g(y)=y24.g(y)=y24. Setting g(y)=0g(y)=0 gives y=±2y=±2 as constant solutions.
  2. Divide both sides of the equation by y24y24 and multiply by dx.dx. This gives the equation
    dyy24=(2x+3)dx.dyy24=(2x+3)dx.
  3. Next integrate both sides:
    1y24dy=(2x+3)dx.1y24dy=(2x+3)dx.
    4.4

    To evaluate the left-hand side, use the method of partial fraction decomposition. This leads to the identity
    1y24=14(1y21y+2).1y24=14(1y21y+2).

    Then Equation 4.4 becomes
    14(1y21y+2)dy=(2x+3)dx14(ln|y2|ln|y+2|)=x2+3x+C.14(1y21y+2)dy=(2x+3)dx14(ln|y2|ln|y+2|)=x2+3x+C.

    Multiplying both sides of this equation by 44 and replacing 4C4C with C1C1 gives
    ln|y2|ln|y+2|=4x2+12x+C1ln|y2y+2|=4x2+12x+C1.ln|y2|ln|y+2|=4x2+12x+C1ln|y2y+2|=4x2+12x+C1.
  4. It is possible to solve this equation for y. First exponentiate both sides of the equation and define C2=eC1:C2=eC1:
    |y2y+2|=C2e4x2+12x.|y2y+2|=C2e4x2+12x.

    Next we can remove the absolute value and let C2C2 be either positive or negative. Then multiply both sides by y+2.y+2.
    y2=C2(y+2)e4x2+12xy2=C2ye4x2+12x+2C2e4x2+12x.y2=C2(y+2)e4x2+12xy2=C2ye4x2+12x+2C2e4x2+12x.

    Now collect all terms involving y on one side of the equation, and solve for y:y:
    yC2ye4x2+12x=2+2C2e4x2+12xy(1C2e4x2+12x)=2+2C2e4x2+12xy=2+2C2e4x2+12x1C2e4x2+12x.yC2ye4x2+12x=2+2C2e4x2+12xy(1C2e4x2+12x)=2+2C2e4x2+12xy=2+2C2e4x2+12x1C2e4x2+12x.
  5. To determine the value of C2,C2, substitute x=0x=0 and y=−1y=−1 into the general solution. Alternatively, we can put the same values into an earlier equation, namely the equation y2y+2=C2e4x2+12.y2y+2=C2e4x2+12. This is much easier to solve for C2:C2:
    y2y+2=C2e4x2+12x−12−1+2=C2e4(0)2+12(0)C2=−3.y2y+2=C2e4x2+12x−12−1+2=C2e4(0)2+12(0)C2=−3.

    Therefore the solution to the initial-value problem is
    y=26e4x2+12x1+3e4x2+12x.y=26e4x2+12x1+3e4x2+12x.

    A graph of this solution appears in Figure 4.15.
    A graph of the solution over [-5, 3] for x and [-3, 2] for y. It begins as a horizontal line at y = -2 from x = -5 to just before -3, almost immediately steps up to y = 2 from just after x = -3 to just before x = 0, and almost immediately steps back down to y = -2 just after x = 0 to x = 3.
    Figure 4.15 Graph of the solution to the initial-value problem y=(2x+3)(y24),y(0)=−3.y=(2x+3)(y24),y(0)=−3.
Checkpoint 4.11

Find the solution to the initial-value problem

6y=(2x+1)(y22y8),y(0)=−36y=(2x+1)(y22y8),y(0)=−3

using the method of separation of variables.

Applications of Separation of Variables

Many interesting problems can be described by separable equations. We illustrate two types of problems: solution concentrations and Newton’s law of cooling.

Solution concentrations

Consider a tank being filled with a salt solution. We would like to determine the amount of salt present in the tank as a function of time. We can apply the process of separation of variables to solve this problem and similar problems involving solution concentrations.

Example 4.12

Determining Salt Concentration over Time

A tank containing 100L100L of a brine solution initially has 4kg4kg of salt dissolved in the solution. At time t=0,t=0, another brine solution flows into the tank at a rate of 2L/min.2L/min. This brine solution contains a concentration of 0.5kg/L0.5kg/L of salt. At the same time, a stopcock is opened at the bottom of the tank, allowing the combined solution to flow out at a rate of 2L/min,2L/min, so that the level of liquid in the tank remains constant (Figure 4.16). Find the amount of salt in the tank as a function of time (measured in minutes), and find the limiting amount of salt in the tank, assuming that the solution in the tank is well mixed at all times.

A diagram of a cylinder filled with water with input and output. It is a 100 liter tank which initially contains 4 kg of salt. The input is 0.5 kg salt / liter and 2 liters / minute. The output is 2 liters / minute.
Figure 4.16 A brine tank with an initial amount of salt solution accepts an input flow and delivers an output flow. How does the amount of salt change with time?

Solution

First we define a function u(t)u(t) that represents the amount of salt in kilograms in the tank as a function of time. Then dudtdudt represents the rate at which the amount of salt in the tank changes as a function of time. Also, u(0)u(0) represents the amount of salt in the tank at time t=0,t=0, which is 44 kilograms.

The general setup for the differential equation we will solve is of the form

dudt=INFLOW RATEOUTFLOW RATE.dudt=INFLOW RATEOUTFLOW RATE.
4.5

INFLOW RATE represents the rate at which salt enters the tank, and OUTFLOW RATE represents the rate at which salt leaves the tank. Because solution enters the tank at a rate of 22 L/min, and each liter of solution contains 0.50.5 kilogram of salt, every minute 2(0.5)=1kilogram2(0.5)=1kilogram of salt enters the tank. Therefore INFLOW RATE = 1.1.

To calculate the rate at which salt leaves the tank, we need the concentration of salt in the tank at any point in time. Since the actual amount of salt varies over time, so does the concentration of salt. However, the volume of the solution remains fixed at 100 liters. The number of kilograms of salt in the tank at time tt is equal to u(t).u(t). Thus, the concentration of salt is u(t)100u(t)100 kg/L, and the solution leaves the tank at a rate of 22 L/min. Therefore salt leaves the tank at a rate of u(t)100·2=u(t)50u(t)100·2=u(t)50 kg/min, and OUTFLOW RATE is equal to u(t)50.u(t)50. Therefore the differential equation becomes dudt=1u50,dudt=1u50, and the initial condition is u(0)=4.u(0)=4. The initial-value problem to be solved is

dudt=1u50,u(0)=4.dudt=1u50,u(0)=4.

The differential equation is a separable equation, so we can apply the five-step strategy for solution.

Step 1. Setting 1u50=01u50=0 gives u=50u=50 as a constant solution. Since the initial amount of salt in the tank is 44 kilograms, this solution does not apply.

Step 2. Rewrite the equation as

dudt=50u50.dudt=50u50.

Then multiply both sides by dtdt and divide both sides by 50u:50u:

du50u=dt50.du50u=dt50.

Step 3. Integrate both sides:

du50u=dt50ln|50u|=t50+C.du50u=dt50ln|50u|=t50+C.

Step 4. Solve for u(t):u(t):

ln|50u|=t50Celn|50u|=e(t/50)C|50u|=C1et/50.ln|50u|=t50Celn|50u|=e(t/50)C|50u|=C1et/50.

Eliminate the absolute value by allowing the constant to be either positive or negative:

50u=C1et/50.50u=C1et/50.

Finally, solve for u(t):u(t):

u(t)=50C1et/50.u(t)=50C1et/50.

Step 5. Solve for C1:C1:

u(0)=50C1e−0/504=50C1C1=46.u(0)=50C1e−0/504=50C1C1=46.

The solution to the initial value problem is u(t)=5046et/50.u(t)=5046et/50. To find the limiting amount of salt in the tank, take the limit as tt approaches infinity:

limtu(t)=5046et/50=5046(0)=50.limtu(t)=5046et/50=5046(0)=50.

Note that this was the constant solution to the differential equation. If the initial amount of salt in the tank is 5050 kilograms, then it remains constant. If it starts at less than 50 kilograms, then it approaches 50 kilograms over time.

Checkpoint 4.12

A tank contains 33 kilograms of salt dissolved in 7575 liters of water. A salt solution of 0.4kg salt/L0.4kg salt/L is pumped into the tank at a rate of 6L/min6L/min and is drained at the same rate. Solve for the salt concentration at time t.t. Assume the tank is well mixed at all times.

Newton’s law of cooling

Newton’s law of cooling states that the rate of change of an object’s temperature is proportional to the difference between its own temperature and the ambient temperature (i.e., the temperature of its surroundings). If we let T(t)T(t) represent the temperature of an object as a function of time, then dTdtdTdt represents the rate at which that temperature changes. The temperature of the object’s surroundings can be represented by Ts.Ts. Then Newton’s law of cooling can be written in the form

dTdt=k(T(t)Ts)dTdt=k(T(t)Ts)

or simply

dTdt=k(TTs).dTdt=k(TTs).
4.6

The temperature of the object at the beginning of any experiment is the initial value for the initial-value problem. We call this temperature T0.T0. Therefore the initial-value problem that needs to be solved takes the form

dTdt=k(TTs),T(0)=T0,dTdt=k(TTs),T(0)=T0,
4.7

where kk is a constant that needs to be either given or determined in the context of the problem. We use these equations in Example 4.13.

Example 4.13

Waiting for a Pizza to Cool

A pizza is removed from the oven after baking thoroughly, and the temperature of the oven is 350°F.350°F. The temperature of the kitchen is 75°F,75°F, and after 55 minutes the temperature of the pizza is 340°F.340°F. We would like to wait until the temperature of the pizza reaches 300°F300°F before cutting and serving it (Figure 4.17). How much longer will we have to wait?

A diagram of a pizza pie. The room temperature is 75 degrees, and the pizza temperature is 350 degrees.
Figure 4.17 From Newton’s law of cooling, if the pizza cools 10°F10°F in 55 minutes, how long before it cools to 300°F?300°F?

Solution

The ambient temperature (surrounding temperature) is 75°F,75°F, so Ts=75.Ts=75. The temperature of the pizza when it comes out of the oven is 350°F,350°F, which is the initial temperature (i.e., initial value), so T0=350.T0=350. Therefore Equation 4.4 becomes

dTdt=k(T75),T(0)=350.dTdt=k(T75),T(0)=350.

To solve the differential equation, we use the five-step technique for solving separable equations.

  1. Setting the right-hand side equal to zero gives T=75T=75 as a constant solution. Since the pizza starts at 350°F,350°F, this is not the solution we are seeking.
  2. Rewrite the differential equation by multiplying both sides by dtdt and dividing both sides by T75:T75:
    dTT75=kdt.dTT75=kdt.
  3. Integrate both sides:
    dTT75=kdtln|T75|=kt+C.dTT75=kdtln|T75|=kt+C.
  4. Solve for TT by first exponentiating both sides:
    eln|T75|=ekt+C|T75|=C1ektT75=C1ektT(t)=75+C1ekt.eln|T75|=ekt+C|T75|=C1ektT75=C1ektT(t)=75+C1ekt.
  5. Solve for C1C1 by using the initial condition T(0)=350:T(0)=350:
    T(t)=75+C1ektT(0)=75+C1ek(0)350=75+C1C1=275.T(t)=75+C1ektT(0)=75+C1ek(0)350=75+C1C1=275.

    Therefore the solution to the initial-value problem is
    T(t)=75+275ekt.T(t)=75+275ekt.

    To determine the value of k,k, we need to use the fact that after 55 minutes the temperature of the pizza is 340°F.340°F. Therefore T(5)=340.T(5)=340. Substituting this information into the solution to the initial-value problem, we have
    T(t)=75+275ektT(5)=340=75+275e5k265=275e5ke5k=5355lne5k=ln(5355)5k=ln(5355)k=15ln(5355)0.007408.T(t)=75+275ektT(5)=340=75+275e5k265=275e5ke5k=5355lne5k=ln(5355)5k=ln(5355)k=15ln(5355)0.007408.

    So now we have T(t)=75+275e−0.007048t.T(t)=75+275e−0.007048t. When is the temperature 300°F?300°F? Solving for t,t, we find
    T(t)=75+275e−0.007048t300=75+275e−0.007048t225=275e−0.007048te−0.007048t=911lne−0.007048t=ln911−0.007048t=ln911t=10.007048ln91128.5.T(t)=75+275e−0.007048t300=75+275e−0.007048t225=275e−0.007048te−0.007048t=911lne−0.007048t=ln911−0.007048t=ln911t=10.007048ln91128.5.

    Therefore we need to wait an additional 23.523.5 minutes (after the temperature of the pizza reached 340°F).340°F). That should be just enough time to finish this calculation.

Checkpoint 4.13

A cake is removed from the oven after baking thoroughly, and the temperature of the oven is 450°F.450°F. The temperature of the kitchen is 70°F,70°F, and after 1010 minutes the temperature of the cake is 430°F.430°F.

  1. Write the appropriate initial-value problem to describe this situation.
  2. Solve the initial-value problem for T(t).T(t).
  3. How long will it take until the temperature of the cake is within 5°F5°F of room temperature?

Section 4.3 Exercises

Solve the following initial-value problems with the initial condition y0=0y0=0 and graph the solution.

119.

dydt=y+1dydt=y+1

120.

dydt=y1dydt=y1

121.

dydt=y+1dydt=y+1

122.

dydt=y1dydt=y1

Find the general solution to the differential equation.

123.

x2y=(x+1)yx2y=(x+1)y

124.

y=tan(y)xy=tan(y)x

125.

y=2xy2y=2xy2

126.

dydt=ycos(3t+2)dydt=ycos(3t+2)

127.

2xdydx=y22xdydx=y2

128.

y=eyx2y=eyx2

129.

(1+x)y=(x+2)(y1)(1+x)y=(x+2)(y1)

130.

dxdt=3t2(x2+4)dxdt=3t2(x2+4)

131.

tdydt=1y2tdydt=1y2

132.

y=exeyy=exey

Find the solution to the initial-value problem.

133.

y=eyx,y(0)=0y=eyx,y(0)=0

134.

y=y2(x+1),y(0)=2y=y2(x+1),y(0)=2

135.

dydx=y3xex2,y(0)=1dydx=y3xex2,y(0)=1

136.

dydt=y2exsin(3x),y(0)=1dydt=y2exsin(3x),y(0)=1

137.

y=xsech2y,y(0)=0y=xsech2y,y(0)=0

138.

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

139.

dxdt=ln(t)1x2,x(0)=0dxdt=ln(t)1x2,x(0)=0

140.

y=3x2(y2+4),y(0)=0y=3x2(y2+4),y(0)=0

141.

y=ey5x,y(0)=ln(ln(5))y=ey5x,y(0)=ln(ln(5))

142.

y=−2xtan(y),y(0)=π2y=−2xtan(y),y(0)=π2

For the following problems, use a software program or your calculator to generate the directional fields. Solve explicitly and draw solution curves for several initial conditions. Are there some critical initial conditions that change the behavior of the solution?

143.

[T] y=12yy=12y

144.

[T] y=y2x3y=y2x3

145.

[T] y=y3exy=y3ex

146.

[T] y=eyy=ey

147.

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

148.

Most drugs in the bloodstream decay according to the equation y=cy,y=cy, where yy is the concentration of the drug in the bloodstream. If the half-life of a drug is 22 hours, what fraction of the initial dose remains after 66 hours?

149.

A drug is administered intravenously to a patient at a rate rr mg/h and is cleared from the body at a rate proportional to the amount of drug still present in the body, dd Set up and solve the differential equation, assuming there is no drug initially present in the body.

150.

[T] How often should a drug be taken if its dose is 33 mg, it is cleared at a rate c=0.1c=0.1 mg/h, and 11 mg is required to be in the bloodstream at all times?

151.

A tank contains 11 kilogram of salt dissolved in 100100 liters of water. A salt solution of 0.10.1 kg salt/L is pumped into the tank at a rate of 22 L/min and is drained at the same rate. Solve for the salt concentration at time t.t. Assume the tank is well mixed.

152.

A tank containing 1010 kilograms of salt dissolved in 10001000 liters of water has two salt solutions pumped in. The first solution of 0.20.2 kg salt/L is pumped in at a rate of 2020 L/min and the second solution of 0.050.05 kg salt/L is pumped in at a rate of 55 L/min. The tank drains at 2525 L/min. Assume the tank is well mixed. Solve for the salt concentration at time t.t.

153.

[T] For the preceding problem, find how much salt is in the tank 11 hour after the process begins.

154.

Torricelli’s law states that for a water tank with a hole in the bottom that has a cross-section of AA and with a height of water hh above the bottom of the tank, the rate of change of volume of water flowing from the tank is proportional to the square root of the height of water, according to dVdt=A2gh,dVdt=A2gh, where gg is the acceleration due to gravity. Note that dVdt=Adhdt.dVdt=Adhdt. Solve the resulting initial-value problem for the height of the water, assuming a tank with a hole of radius 22 ft. The initial height of water is 100100 ft.

155.

For the preceding problem, determine how long it takes the tank to drain.

For the following problems, use Newton’s law of cooling.

156.

The liquid base of an ice cream has an initial temperature of 200°F200°F before it is placed in a freezer with a constant temperature of 0°F.0°F. After 11 hour, the temperature of the ice-cream base has decreased to 140°F.140°F. Formulate and solve the initial-value problem to determine the temperature of the ice cream.

157.

[T] The liquid base of an ice cream has an initial temperature of 210°F210°F before it is placed in a freezer with a constant temperature of 20°F.20°F. After 22 hours, the temperature of the ice-cream base has decreased to 170°F.170°F. At what time will the ice cream be ready to eat? (Assume 30°F30°F is the optimal eating temperature.)

158.

[T] You are organizing an ice cream social. The outside temperature is 80°F80°F and the ice cream is at 10°F.10°F. After 1010 minutes, the ice cream temperature has risen by 10°F.10°F. How much longer can you wait before the ice cream melts at 40°F?40°F?

159.

You have a cup of coffee at temperature 70°C70°C and the ambient temperature in the room is 20°C.20°C. Assuming a cooling rate kof0.125,kof0.125, write and solve the differential equation to describe the temperature of the coffee with respect to time.

160.

[T] You have a cup of coffee at temperature 70°C70°C that you put outside, where the ambient temperature is 0°C.0°C. After 55 minutes, how much colder is the coffee?

161.

You have a cup of coffee at temperature 70°C70°C and you immediately pour in 11 part milk to 55 parts coffee. The milk is initially at temperature 1°C.1°C. Write and solve the differential equation that governs the temperature of this coffee.

162.

You have a cup of coffee at temperature 70°C,70°C, which you let cool 1010 minutes before you pour in the same amount of milk at 1°C1°C as in the preceding problem. How does the temperature compare to the previous cup after 1010 minutes?

163.

Solve the generic problem y=ay+by=ay+b with initial condition y(0)=c.y(0)=c.

164.

Prove the basic continual compounded interest equation. Assuming an initial deposit of P0P0 and an interest rate of r,r, set up and solve an equation for continually compounded interest.

165.

Assume an initial nutrient amount of II kilograms in a tank with LL liters. Assume a concentration of cc kg/L being pumped in at a rate of rr L/min. The tank is well mixed and is drained at a rate of rr L/min. Find the equation describing the amount of nutrient in the tank.

166.

Leaves accumulate on the forest floor at a rate of 22 g/cm2/yr and also decompose at a rate of 90%90% per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time 00 there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?

167.

Leaves accumulate on the forest floor at a rate of 44 g/cm2/yr. These leaves decompose at a rate of 10%10% per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor. Does this amount approach a steady value? What is that value?

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