9: 650
Chapter Further Applications of Integration
9.2
First-Order Linear Differential Equations
The exponential growth> decay equation dy>dx = ky (Section 7.5) is a separable differential equation. It is also a special case of a differential equation having a linear form. Linear differential equations model a number of real-world phenomena, including electrical circuits and chemical mixture problems. A first-order linear differential equation is one that can be written in the form dy (1) + Psxdy = Qsxd, dx where P and Q are continuous functions of x. Equation (1) is the linear equation's standard form. Since the exponential growth> decay equation can be put in the standard form dy - ky = 0, dx
we see it is a linear equation with Psxd = -k and Qsxd = 0. Equation (1) is linear (in y) because y and its derivative dy> dx occur only to the first power, are not multiplied together, nor do they appear as the argument of a function (such as sin y, e y , or 2dy>dx).
EXAMPLE 1
Finding the Standard Form
Put the following equation in standard form: dy x = x 2 + 3y, dx
Solution
x 7 0.
x
dy = x2 + 3y dx dy 3 = x + xy dx
Divide by x Standard form with Psxd = -3>x and Qsxd = x
dy 3 - xy = x dx
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9.2
First-Order Linear Differential Equations
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Notice that P(x) is -3>x, not +3>x. The standard form is y + Psxdy = Qsxd, so the minus sign is part of the formula for P(x).
Solving Linear Equations
We solve the equation dy + Psxdy = Qsxd dx
(2)
by multiplying both sides by a positive function y(x) that transforms the left side into the derivative of the product ysxd # y. We will show how to find y in a moment, but first we want to show how, once found, it provides the solution we seek. Here is why multiplying by y(x) works: dy + Psxdy = Qsxd dx ysxd dy + Psxdysxdy = ysxdQsxd dx d sysxd # yd = ysxdQsxd dx ysxd # y = y = L ysxdQsxd dx
Original equation is in standard form. Multiply by positive y(x). ysxd is chosen to make dy d y sy # yd . + Pyy = dx dx Integrate with respect to x.
1 ysxdQsxd dx ysxd L
(3)
Equation (3) expresses the solution of Equation (2) in terms of the function y(x) and Q(x). We call y(x) an integrating factor for Equation (2) because its presence makes the equation integrable. Why doesn't the formula for P(x) appear in the solution as well? It does, but indirectly, in the construction of the positive function y(x). We have dy d syyd = y + Pyy dx dx dy dy dy y + y = y + Pyy dx dx dx dy y = Pyy dx This last equation will hold if dy = Py dx dy y = P dx L dy y = L L P dx P dx
Condition imposed on y Product Rule for derivatives The terms y dy cancel. dx
Variables separated, y 7 0 Integrate both sides. Since y 7 0 , we do not need absolute value signs in ln y. Exponentiate both sides to solve for y.
ln y = e
ln y
= e 1 P dx
1 P dx
y = e
(4)
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652
Chapter 9: Further Applications of Integration
Thus a formula for the general solution to Equation (1) is given by Equation (3), where y(x) is given by Equation (4). However, rather than memorizing the formula, just remember how to find the integrating factor once you have the standard form so P(x) is correctly identified.
To solve the linear equation y + Psxdy = Qsxd, multiply both sides by the integrating factor ysxd = e 1 Psxd dx and integrate both sides.
When you integrate the left-side product in this procedure, you always obtain the product y(x)y of the integrating factor and solution function y because of the way y is defined.
EXAMPLE 2
Solving a First-Order Linear Differential Equation
dy = x 2 + 3y, dx
Solve the equation x HISTORICAL BIOGRAPHY
Adrien Marie Legendre (17521833) Solution
x 7 0.
First we put the equation in standard form (Example 1): dy 3 - x y = x, dx
so Psxd = -3>x is identified. The integrating factor is ysxd = e 1 Psxd dx = e 1s-3>xd dx = e -3 ln x = e -3 ln x = e ln x
-3
Constant of integration is 0, so y is as simple as possible. x 7 0
=
1 . x3
Next we multiply both sides of the standard form by y(x) and integrate: 3 1 # dy 1 a - x yb = 3 # x x 3 dx x 3 1 dy 1 - 4y = 2 x 3 dx x x d 1 1 a yb = 2 dx x 3 x 1 1 y = dx x3 x2 L 1 1 y = - x + C. x3
Left side is
d dx sy
# yd.
Integrate both sides.
Solving this last equation for y gives the general solution: 1 y = x 3 a- x + Cb = -x 2 + Cx 3, x 7 0.
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9.2
First-Order Linear Differential Equations
653
EXAMPLE 3
Solving a First-Order Linear Initial Value Problem
xy = x 2 + 3y, x 7 0,
Solve the equation
given the initial condition ys1d = 2.
Solution
We first solve the differential equation (Example 2), obtaining y = -x 2 + Cx 3, x 7 0.
We then use the initial condition to find C: y = -x2 + Cx3 2 = -s1d2 + Cs1d3 C = 2 + s1d2 = 3. The solution of the initial value problem is the function y = -x 2 + 3x 3 .
y = 2 when x = 1
EXAMPLE 4
Find the particular solution of 3xy - y = ln x + 1, x 7 0,
satisfying ys1d = -2.
Solution
With x 7 0, we write the equation in standard form: y ln x + 1 1 y = . 3x 3x
Then the integrating factor is given by y = e 1 - dx>3x = e s-1>3dln x = x -1>3 . Thus x -1>3y = 1 sln x + 1dx -4>3 dx. 3L
Left side is yy. x 7 0
Integration by parts of the right side gives x -1>3y = -x -1>3sln x + 1d + Therefore x -1>3y = -x -1>3sln x + 1d - 3x -1>3 + C or, solving for y, y = -sln x + 4d + Cx 1>3 . When x = 1 and y = -2 this last equation becomes -2 = -s0 + 4d + C, so C = 2. L x -4>3 dx + C.
Copyright 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley
654
Chapter 9: Further Applications of Integration
Substitution into the equation for y gives the particular solution y = 2x 1>3 - ln x - 4. In solving the linear equation in Example 2, we integrated both sides of the equation after multiplying each side by the integrating factor. However, we can shorten the amount of work, as in Example 4, by remembering that the left side always integrates into the product ysxd # y of the integrating factor times the solution function. From Equation (3) this means that ysxdy = L ysxdQsxd dx.
We need only integrate the product of the integrating factor y(x) with the right side Q(x) of Equation (1) and then equate the result with y(x)y obtain to the general solution. Nevertheless, to emphasize the role of y(x) in the solution process, we sometimes follow the complete procedure as illustrated in Example 2. Observe that if the function Q(x) is identically zero in the standard form given by Equation (1), the linear equation is separable: dy + Psxdy = Qsxd dx dy + Psxdy = 0 dx dy = -Psxd dx
Qsxd K 0 Separating the variables
We now present two applied problems modeled by a first-order linear differential equation.
RL Circuits
The diagram in Figure 9.5 represents an electrical circuit whose total resistance is a constant R ohms and whose self-inductance, shown as a coil, is L henries, also a constant. There is a switch whose terminals at a and b can be closed to connect a constant electrical source of V volts. Ohm's Law, V = RI, has to be modified for such a circuit. The modified form is L di + Ri = V, dt
(5)
where i is the intensity of the current in amperes and t is the time in seconds. By solving this equation, we can predict how the current will flow after the switch is closed.
V
EXAMPLE 5
Switch a i b
Electric Current Flow
The switch in the RL circuit in Figure 9.5 is closed at time t = 0. How will the current flow as a function of time?
Solution Equation (5) is a first-order linear differential equation for i as a function of t. Its standard form is
R
L
FIGURE 9.5 The RL circuit in Example 5.
V di R + i = , L L dt
(6)
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9.2
First-Order Linear Differential Equations
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i I V R
and the corresponding solution, given that i = 0 when t = 0, is
I e
i =
i V (1 R e
Rt/L
)
V V - e -sR>Ldt R R
(7)
(Exercise 32). Since R and L are positive, -sR>Ld is negative and e -sR>Ldt : 0 as t : q . Thus,
4 L R t
t: q
0
L R
2
L R
3
L R
lim i = lim a
t: q
V V V V V - e -sR>Ldt b = - #0 = . R R R R R
FIGURE 9.6 The growth of the current in the RL circuit in Example 5. I is the current's steady-state value. The number t = L>R is the time constant of the circuit. The current gets to within 5% of its steady-state value in 3 time constants (Exercise 31).
At any given time, the current is theoretically less than V> R, but as time passes, the current approaches the steady-state value V> R. According to the equation L di + Ri = V, dt
I = V>R is the current that will flow in the circuit if either L = 0 (no inductance) or di>dt = 0 (steady current, i = constant) (Figure 9.6). Equation (7) expresses the solution of Equation (6) as the sum of two terms: a steady-state solution V> R and a transient solution -sV>Rde -sR>Ldt that tends to zero as t: q.
Mixture Problems
A chemical in a liquid solution (or dispersed in a gas) runs into a container holding the liquid (or the gas) with, possibly, a specified amount of the chemical dissolved as well. The mixture is kept uniform by stirring and flows out of the container at a known rate. In this process, it is often important to know the concentration of the chemical in the container at any given time. The differential equation describing the process is based on the formula Rate of change rate at which rate at which of amount = chemical - chemical in container arrives departs. ystd # soutflow rated Vstd
(9) (8)
If y(t) is the amount of chemical in the container at time t and V(t) is the total volume of liquid in the container at time t, then the departure rate of the chemical at time t is Departure rate =
concentration in = acontainer at time t b # soutflow rated. Accordingly, Equation (8) becomes dy ystd # soutflow rated. = schemical's arrival rated dt Vstd pounds pounds gallons pounds # = . minutes minutes gallons minutes
(10)
If, say, y is measured in pounds, V in gallons, and t in minutes, the units in Equation (10) are
EXAMPLE 6
Oil Refinery Storage Tank
In an oil refinery, a storage tank contains 2000 gal of gasoline that initially has 100 lb of an additive dissolved in it. In preparation for winter weather, gasoline containing 2 lb of
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656
Chapter 9: Further Applications of Integration
40 gal/min containing 2 lb/gal
45 gal/min containing y lb/gal V
FIGURE 9.7 The storage tank in Example 6 mixes input liquid with stored liquid to produce an output liquid.
additive per gallon is pumped into the tank at a rate of 40 gal> min. The well-mixed solution is pumped out at a rate of 45 gal> min. How much of the additive is in the tank 20 min after the pumping process begins (Figure 9.7)? Let y be the amount (in pounds) of additive in the tank at time t. We know that y = 100 when t = 0. The number of gallons of gasoline and additive in solution in the tank at any time t is
Solution
Vstd = 2000 gal + a40 Therefore, Rate out =
gal gal - 45 b st mind min min
= s2000 - 5td gal.
ystd # outflow rate Vstd y b 45 2000 - 5t
Eq. (9) Outflow rate is 45 gal> min. and y = 2000 - 5t .
= a = Also,
45y lb . 2000 - 5t min
Rate in = a2
gal lb b a40 b gal min lb . min
Eq. (10)
= 80
The differential equation modeling the mixture process is dy 45y = 80 2000 - 5t dt in pounds per minute. To solve this differential equation, we first write it in standard form: dy 45 + y = 80. 2000 - 5t dt Thus, Pstd = 45>s2000 - 5td and Qstd = 80.
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9.2 First-Order Linear Differential Equations
657
The integrating factor is ystd = e 1 P dt = e 1 2000 - 5t dt = e -9 ln s2000 - 5td = s2000 - 5td-9 .
2000 - 5t 7 0
45
Multiplying both sides of the standard equation by y(t) and integrating both sides gives, s2000 - 5td-9 # a s2000 - 5td-9 dy 45 + yb = 80s2000 - 5td-9 2000 - 5t dt
dy + 45s2000 - 5td-10 y = 80s2000 - 5td-9 dt d C s2000 - 5td-9y D = 80s2000 - 5td-9 dt s2000 - 5td-9y = L 80s2000 - 5td-9 dt s2000 - 5td-8 + C. s -8ds -5d
s2000 - 5td-9y = 80 # The general solution is
y = 2s2000 - 5td + Cs2000 - 5td9 . Because y = 100 when t = 0, we can determine the value of C: 100 = 2s2000 - 0d + Cs2000 - 0d9 C = 3900 . s2000d9
The particular solution of the initial value problem is y = 2s2000 - 5td 3900 s2000 - 5td9 . s2000d9
The amount of additive 20 min after the pumping begins is ys20d = 2[2000 - 5s20d] 3900 [2000 - 5s20d]9 L 1342 lb. s2000d9
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