2. Analytic Solutions and Approximations
2.2 First-Order Linear Equations
A differential equation of the form
u =p(t)u+q(t). (2.3)
is called afirst-order linear equation. The given functionspand qare as- sumed to be continuous. Ifq(t) = 0, then the equation is calledhomogeneous;
otherwise it is callednonhomogeneous. Linear equations have a nice struc- ture to their solution set, and we are able to derive the general solution. The homogeneous equation
u=p(t)u, (2.4)
without the nonhomogeneous termq(t), can readily be solved by separation of variables to obtain
uh(t) =CeP(t), where P(t) =
p(t)dt, (2.5)
whereCis an arbitrary constant. We have placed a subscripthon this solution to distinguish it from the solution of the nonhomogeneous equation (2.3). (The solution uh(t) to the homogeneous equation is sometimes called the comple- mentary solution; we just refer to it as the homogeneous solution.) Also, note that we have used the indefinite integral notation for the functionP(t); in some cases we have to representP(t) in the form
P(t) = t
a
p(s)ds,
with limits of integration. We always choose the lower limitato be the value of time where the initial condition is prescribed.
We now describe a standard technique to solve the nonhomogeneous equa- tion (2.3). The idea is to try a solution of the form (2.5) where we let the constantCin the homogeneous solution vary as a function oft; we then substi- tute this form into (2.3) to determine theC=C(t). The method is, for obvious reasons, called variation of parameters.1 Thus, assume a solution to (2.3) of the form
u(t) =C(t)eP(t). Then, plugging in,
C(t)eP(t)+C(t)eP(t)P(t) =p(t)C(t)eP(t)+q(t).
ButP =pand therefore two of the terms cancel, giving C(t)eP(t)=q(t),
or
C(t) =e−P(t)q(t).
Integration yields
C(t) =
e−P(t)q(t)dt+K,
1 Another method usingintegrating factors is presented in the Exercises.
whereK is a constant of integration. So we have u(t) =
e−P(t)q(t)dt+K
eP(t)
= KeP(t)+eP(t)
e−P(t)q(t)dt, (2.6) which is the general solution to the general linear, nonhomogeneous equation (2.3). If the antiderivative in the last equation cannot be calculated explicitly, then we write the solution as
u(t) =KeP(t)+eP(t) t
a
e−P(τ)q(τ)dτ.
We urge the reader not to memorize these formulas; rather, remember the method and apply it to each problem as you solve it.
Example 2.3
Find the general solution to
u =1 tu+t3.
The homogeneous equation isu= 1tuand has solution uh(t) =Ce(1/t)dt=Celnt=Ct.
Therefore we vary the parameterC and assume a solution of the original non- homogeneous equation of the form
u(t) =C(t)t.
Substituting into the equation, we get u=C(t) +C(t)t= 1
tC(t)t+t3, or
C(t) =t2. Therefore C(t) =
t2dt = 13t3+K and the general solution to the original equation is
u(t) = 1
3t3+K
t= 1
3t4+Kt.
The arbitrary constantKcan be determined by an initial condition.
Example 2.4
Consider the DE
u = 2u+t. (2.7)
The associated homogeneous equation is u= 2u,
which has solutionuh =Ce2t. Therefore we assume the solution of (2.7) is of the form
u(t) =C(t)e2t. Substituting into the original equation gives
C(t)2e2t+C(t)e2t= 2C(t)e2t+t, or
C(t) =te−2t. Integrating,
C(t) =
te−2tdt+K=−1
4e−2t(2t+ 1) +K.
The integral was calculated analytically using integration by parts. Therefore the general solution of (2.7) is
u(t) =
−1
4e−2t(2t+ 1) +K
e2t
= Ke2t−1
4(2t+ 1).
Notice that the general solution is composed of two terms, uh(t) and up(t), defined by
uh=Ke2t, up=−1
4(2t+ 1).
We knowuhis the general solution to the homogeneous equation, and it is easy to show thatupis a particular solution to the nonhomogeneous equation (2.7).
So, the general solution to the nonhomogeneous equation (2.7) is the sum of the general solution to the associated homogeneous equation and any particular solution to the nonhomogeneous equation (2.7).
Example 2.4 illustrates a general principle that reveals the structure of the solution to a first-order linear DE. The general solution can be written as the sum of the solution to the homogeneous equation and any particular solution of the nonhomogeneous equation. Precisely, the basic structure theorem for first-order linear equations states:
Theorem 2.5
(Structure Theorem) The general solution of the nonhomogeneous equation u =p(t)u+q(t)
is the sum of the general solutionuh of the homogeneous equationu =p(t)u and a particular solutionup to the nonhomogeneous equation. In symbols,
u(t) =uh(t) +up(t), whereuh(t) =KeP(t)andup=eP(t)
e−P(t)q(t)dt, and whereP(t) = p(t)dt.
Example 2.6
Consider an RC electrical circuit where the resistance isR= 1 and the capac- itance isC= 0.5.Initially the charge on the capacitor isq(0) = 5. The current is driven by an emf that generates a variable voltage of sint. How does the circuit respond? The governing DE for the chargeq(t) on the capacitor is
Rq+ 1
Cq= sint, or, substituting the given parameters,
q=−2q+ sint. (2.8)
The homogeneous equationq =−2qhas solutionqh=Ce−2t.We assume the solution to the nonhomogeneous equation has the formq =C(t)e−2t.Substi- tuting into (2.8) gives
C(t)(−2qe−2t) +C(t)qe−2t=−2C(t)e−2t+ sint, or
C(t) =e2tsint.
Integrating,
C(t) =
e2tsintdt+K=e2t(2
5sint−1
5cost) +K,
whereKis a constant of integration. The integral was calculated using software (or, one can use integration by parts). Therefore the general solution of (2.8) is
q(t) =C(t)e−2t=2
5sint−1
5cost+Ke−2t.
Next we apply the initial condition q(0) = 5 to obtain K = 26/5. Therefore the solution to the initial value problem is
q(t) = 2
5sint−1
5cost+26 5 e−2t.
The solution is consistent with Theorem 2.5. Also, there is an important phys- ical interpretation of the solution. The homogeneous solution is thetransient response qh(t) = 265e−2t that depends upon the initial charge and decays over a time; what remains over a long time is the particular solution, which is regarded as the steady-state responseqp(t) = 265 sint−15cost.The ho- mogeneous solution ignores the forcing term (the emf), whereas the particular solution arises from the forcing term. After a long time the applied emf drives the response of the system. This behavior is characteristic of forced linear equa- tions coming from circuit theory and mechanics. The solution is a sum of two terms, a contribution due to the internal system and initial data (the decay- ing transient), and a contribution due to the external forcing term (the steady response). Figure 2.2 shows a plot of the solution.
0 1 2 3 4 5 6 7 8 9 10
−1 0 1 2 3 4 5
t
q
transient response
steady−state response
Figure 2.2 Response of the circuit in Example 2.6 showing the initial tran- sient and the long-time steady-state.
Example 2.7
(Sales Response to Advertising) The field of economics has always been a source of interesting phenomena modeled by differential equations. In this example we set up a simple model that allows management to assess the effectiveness of an advertising campaign. Let S = S(t) be the monthly sales of an item.
In the absence of advertising it is observed from sales history data that the logarithm of the monthly sales decreases linearly in time, or lnS =−at+b.
Thus S = −aS, and sales are modeled by exponential decay. To keep sales up, advertising is required. If there is a lot of advertising, then sales tend to saturate at some maximum valueS=M; this is because there are only finitely many consumers. The rate of increase in sales due to advertising is jointly proportional to the advertising rateA(t) and to the degree the market is not saturated; that is,
rA(t)
M −S M
.
The constant r measures the effectiveness of the advertising campaign. The term M−S
M is a measure of the market share that has still not purchased the product. Then, combining both natural sales decay and advertising, we obtain the model
S=−aS+rA(t)
M −S M
.
The first term on the right is the natural decay rate, and the second term is the rate of sales increase due to advertising, which drives the sales. As it stands, because the advertising rateAis not constant, there are no equilibria (constant solutions). We can rearrange the terms and write the equation in the form
S =−
a+rA(t) M
S+rA(t). (2.9)
Now we recognize that the sales are governed by a first-order linear DE. The Exercises request some solutions for different advertising strategies.
EXERCISES
1. Find the general solution of u =−1tu+t.
2. Find the general solution of u =−u+et.
3. Show that the general solution to the DE u+au=√
1 +t is given by u(t) =Ce−at+
t
0
e−a(t−s)√
1 +sds.
4. A decaying battery generating 200e−5tvolts is connected in series with a 20 ohm resistor, and a 0.01 farad capacitor. Assuming q = 0 at t = 0, find the charge and current for all t > 0. Show that the charge reaches a maximum and find the time it is reached.
5. Solveu+u = 3tby introducingy=u. 6. Solveu= (t+u)2 by lettingy=t+u.
7. Express the general solution of the equation u = 2tu+ 1 in terms of the erf function.
8. Find the solution to the initial value problem u =pu+q, u(0) = u0, where pandqare constants.
9. Find a formula for the general solution to the DEu=pu+q(t), where p is constant. Find the solution satisfyingu(t0) =u0.
10. A differential equation of the form
u=a(t)u+g(t)un
is called aBernoulli equation, and it arises in many applications. Show that the Bernoulli equation can be reduced to the linear equation
y= (1−n)a(t)y+ (1−n)g(t)
by changing the dependent variable fromutoy viay=u1−n. 11. Solve the Bernoulli equations (see Exercise 10).
a) u=3t2u+2tu. b) u=u(1 +uet).
c) u=−1tu+tu12.
12. Initially, a tank contains 60 gal of pure water. Then brine containing 1 lb of salt per gallon enters the tank at 2 gal/min. The perfectly mixed solution is drained off at 3 gal/min. Determine the amount (in lbs) of salt in the tank up until the time it empties.
13. A large industrial retention pond of volumeV, initially free of pollutants, was subject to the inflow of a contaminant produced in the factory’s pro- cessing plant. Over a period ofb days the EPA found that the inflow con- centration of the contaminant decreased linearly (in time) to zero from its initial value of a(grams per volume), its flow rateq(volume per day) be- ing constant. During the b days the spillage to the local stream was also
q. What is the concentration in the pond afterbdays? Do a numerical ex- periment using a computer algebra system where V = 6000 cubic meters, b = 20 days, a = 0.03 grams per cubic meter, and q = 50 cubic meters per day. With this data, how long would it take for the concentration in the pond to get below the required EPA level of 0.00001 grams per cubic meter if fresh water is pumped into the pond at the same flow rate, with the same spillover?
14. Determine the dimensions of the various quantities in the sales–advertising model (2.9). If Ais constant, what is the equilibrium?
15. (Technology Transfer) Suppose a new innovation is introduced at timet= 0 in a community of N possible users (e.g., a new pesticide introduced to a community of farmers). Letx(t) be the number of users who have adopted the innovation at timet. If the rate of adoption of the innovation is jointly proportional to the number of adoptions and the number of those who have not adopted, write down a DE model forx(t). Describe, qualitatively, how x(t) changes in time. Find a formula for x(t).
16. A house is initially at 12 degrees Celsius when its heating–cooling system fails. The outside temperature varies according to Te = 9 + 10 cos 2πt, where time is given in days. The heat loss coefficient is h= 3 degrees per day. Find a formula for the temperature variation in the house and plot it along with Te on the same set of axes. What is the time lag between the maximum inside and outside temperature?
17. In the sales response to advertising model (2.9), assumeS(0) =S0and that advertising is constant over a fixed time period T, and is then removed.
That is,
A(t) =
a, 0≤t≤T 0, t > T
Find a formula for the salesS(t). (Hint: solve the problem on two intervals and piece together the solutions in a continuous way).
18. In a community having a fixed population N, the rate that people hear a rumor is proportional to the number of people who have not yet heard the rumor. Write down a DE for the number of peopleP who have heard the rumor. Over a long time, how many will hear the rumor? Is this a believable model?
19. An object of massm= 1 is dropped from rest at a large height, and as it falls it experiences the force of gravity mg and a time-dependent resistive force of magnitudeFr=t+12 v,wherevis its velocity. Write down an initial value problem that governs its velocity and find a formula for the solution.
20. The MacArthur–Wilson model of the dynamics of species (e.g., bird species) that inhabit an island located near a mainland was developed in the 1960s. LetP be the number of species in the source pool on the mainland, and letS =S(t) be the number of species on the island. Assume that the rate of change of the number of species is
S=χ−à,
where χ is the colonization rate and à is the extinction rate. In the MacArthur–Wilson model,
χ=I(1− S
P) and à= E PS,
whereIandEare the maximum colonization and extinction rates, respec- tively.
a) Over a long time, what is the expected equilibrium for the number of species inhabiting the island? Is this equilibrium stable?
b) GivenS(0) =S0, find an analytic formula for S(t).
c) Suppose there are two islands, one large and one small, with the larger island having the smaller maximum extinction rate. Both have the same colonization rate. Show that the smaller island will eventually have fewer species.
21. (Integrating Factor Method) There is another popular method, called the integrating factor method, for solving first-order linear equations writ- ten in the form
u−p(t)u=q(t).
If this equation is multiplied by e−p(t)dt,called anintegrating factor,
show that
ue−p(t)dt =q(t)e−p(t)dt.
(Note that the left side is a total derivative). Next, integrate both sides and show that you obtain (2.6). Use this method to solve Exercises 1 and 2 above.