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Calculus Volume 2

4.3 Separable Equations

Calculus Volume 24.3 Separable Equations

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-2)(y+3),(x-2)(y+3), 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

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.

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)=-1.y=(2x+3)(y24),y(0)=-1.

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?

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 pizza when it comes out of the oven is 200°F.200°F. The temperature of the kitchen is 75°F,75°F, and after 11 minute the temperature of the pizza is 190°F.190°F. We would like to wait until the temperature of the pizza reaches 150°F150°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 200 degrees.
Figure 4.17 From Newton’s law of cooling, if the pizza cools 10°F10°F in 11 minute, how long before it cools to 150°F?150°F?

Checkpoint 4.13

A cake is removed from the oven after baking thoroughly, and the temperature of the cake when it comes out 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 330°F.330°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.

d y d t = y + 1 d y d t = y + 1

120.

d y d t = y 1 d y d t = y 1

121.

d y d t = –y + 1 d y d t = –y + 1

122.

d y d t = y 1 d y d t = y 1

Find the general solution to the differential equation.

123.

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

124.

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

125.

y = 2 x y 2 y = 2 x y 2

126.

d y d t = y cos ( 3 t + 2 ) d y d t = y cos ( 3 t + 2 )

127.

2 x d y d x = y 2 2 x d y d x = y 2

128.

y = e y x 2 y = e y x 2

129.

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

130.

d x d t = 3 t 2 ( x 2 + 4 ) d x d t = 3 t 2 ( x 2 + 4 )

131.

t d y d t = 1 y 2 t d y d t = 1 y 2

132.

y = e x e y y = e x e y

Find the solution to the initial-value problem.

133.

y = e y x , y ( 0 ) = 0 y = e y x , y ( 0 ) = 0

134.

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

135.

d y d x = y 3 x e x 2 , y ( 0 ) = 1 d y d x = y 3 x e x 2 , y ( 0 ) = 1

136.

d y d t = y 2 e x sin ( 3 x ) , y ( 0 ) = 1 d y d t = y 2 e x sin ( 3 x ) , y ( 0 ) = 1

137.

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

138.

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

139.

d x d t = ln ( t ) 1 x 2 , x ( 1 ) = 0 d x d t = ln ( t ) 1 x 2 , x ( 1 ) = 0

140.

y = 3 x 2 ( y 2 + 4 ) , y ( 0 ) = 0 y = 3 x 2 ( y 2 + 4 ) , y ( 0 ) = 0

141.

y = e y 5 x , y ( 0 ) = ln ( ln ( 5 ) ) y = e y 5 x , y ( 0 ) = ln ( ln ( 5 ) )

142.

y = −2 x tan ( y ) , y ( 0 ) = π 2 y = −2 x tan ( y ) , y ( 0 ) = π 2

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-sectional area of ATAT 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=AT2gh,dVdt=AT2gh, where gg is the acceleration due to gravity. Note that dVdt=Adhdt,dVdt=Adhdt, where ATAT is the cross-sectional area of the tank. Solve the resulting initial-value problem for the height of water, assuming a tankof radius 242242 with a circular hole of radius 22 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?

For Exercises 159—162, assume a cooling constant of k=-0.125k=-0.125 and assume time tt is in minutes.

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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