In this final section of the chapter we consider two general types of differential equation that are frequently encountered in economics. The discussion will be brief—for a more extensive treatment we refer to FMEA.
Separable Equations
A differential equation of the type
˙
x =f (t )g(x) (1)
is calledseparable. The unknown function isx =x(t ), and its rate of changex˙is given as the product of a function only oftand a function only ofx. A simple case isx˙ =t x, which is obviously separable, whilex˙=t+xis not. In fact, all the differential equations studied in the previous section were separable equations of the typex˙ =g(x), withf (t )≡1. Equation (9.8.10), for instance, is separable, sincex˙+ax=bx2can be rewritten asx˙ =g(x)where g(x)= −ax+bx2.
The following general method for solving separable equations is justified in FMEA.
Method for Solving Separable Differential Equations:
A. Write equation (1) as
dx
dt =f (t )g(x) (∗)
B. Separate the variables:
dx
g(x)=f (t ) dt C. Integrate each side:
dx g(x)=
f (t ) dt
D. Evaluate the two integrals (if possible) and you obtain a solution of (∗) (possibly in implicit form). Solve forx, if possible.
NOTE 1 In step B we divided byg(x). In fact, ifg(x)has a zero atx =a, so thatg(a)=0, thenx(t )≡awill be a particular solution of the equation, because the right- and left-hand sides of the equation are both 0 for allt. For instance, in the logistic equation (9.8.5), both x(t )=0 andx(t )=Kare particular solutions.
E X A M P L E 1 Solve the differential equation
dx dt =etx2
and find the solution curve (often called theintegral curve) which passes through the point (t, x)=(0,1).
S E C T I O N 9 . 9 / S E P A R A B L E A N D L I N E A R D I F F E R E N T I A L E Q U A T I O N S 337 Solution: We observe first thatx(t )≡0 is one (trivial) solution. To find the other solutions we follow the recipe:
dx x2 =etdt Separate:
dx x2 =
etdt Integrate:
−1
x =et+C Evaluate:
It follows that
x = −1
et+C (∗)
To find the integral curve through(0,1), we must determine the correct value ofC. Because we requirex =1 fort =0, it follows from(∗)that 1= −1/(1+C), soC = −2. Thus, the integral curve passing through(0,1)isx =1/(2−et).
E X A M P L E 2 (Economic Growth)5 LetX=X(t )denote the national product,K=K(t )the capital stock, andL =L(t )the number of workers in a country at timet. Suppose that, for all t≥0,
(a) X=√ K√
L (b) K˙ =0.4X (c) L=e0.04t
Derive from these equations a single differential equation for K = K(t ), and find the solution of that equation whenK(0)=10000. (In (a) we have a Cobb–Douglas production function, (b) says that aggregate investment is proportional to output, whereas (c) implies that the labour force grows exponentially.)
Solution: From equations (a)–(c), we derive the single differential equation
˙ K= dK
dt =0.4√ K√
L=0.4e0.02t√ K This is clearly separable. Using the recipe yields:
√dK
K =0.4e0.02tdt dK
√K =
0.4e0.02tdt 2√
K=20e0.02t+C IfK=10 000 fort =0, then 2√
10 000=20+C, soC=180. Then√
K=10e0.02t+90, and so the required solution is
K(t )=(10e0.02t+90)2 =100(e0.02t+9)2
The capital–labour ratio has a somewhat bizarre limiting value in this model:K(t )/L(t )= 100(e0.02t+9)2/e0.04t =100[(e0.02t+9)/e0.02t]2=100[1+9e−0.02t]2 →100 ast → ∞.
5 This is a special case of the Solow–Swan growth model. See Example 5.7.3 in FMEA.
E X A M P L E 3 Solve the separable differential equation(lnx)x˙ =e1−t. Solution: Following the recipe yields
lnx dx dt =e1−t lnx dx=e1−tdt
lnx dx=
e1−tdt xlnx−x = −e1−t+C
Here we used the result in Example 9.1.3. The desired functionsx =x(t )are those that satisfy the last equation for allt.
NOTE 2 We usually say that we have solved a differential equation even if the unknown function (as shown in Example 3) cannot be expressed explicitly. The important point is that we have expressed the unknown function in an equation that does not include the derivative of that function.
First-Order Linear Equations
Afirst-order linear differential equationis one that can be written in the form
˙
x+a(t )x=b(t ) (2)
where a(t )and b(t ) denote known continuous functions of t in a certain interval, and x =x(t )is the unknown function. Equation (2) is called “linear” because the left-hand side is a linear function ofxandx.˙
Whena(t )andb(t )are constants, the solution was given in (9.8.9):
˙
x+ax=b ⇐⇒ x =Ce−at+b
a (Cis a constant) (3) We found the solution of this equation by introducing a new variable. In fact, the equation is separable, so the recipe for separable equations will also lead us to the solution. If we let C =0 we obtain the constant solutionx(t )=b/a. We say thatx =b/ais anequilibrium state, or astationarystate, for the equation. Observe how this solution can be obtained from
˙
x+ax =bby lettingx˙ =0 and then solving the resulting equation forx. If the constant ais positive, then the solutionx =Ce−at+b/aconverges tob/aast → ∞. In this case, the equation is said to bestable, because every solution of the equation converges to an equilibrium ast approaches infinity. (Stability theory is an important issue for differential equations appearing in economics. See e.g. FMEA for an extensive discussion.)
E X A M P L E 4 Find the solution of
˙
x+3x = −9 and determine whether the equation is stable.
S E C T I O N 9 . 9 / S E P A R A B L E A N D L I N E A R D I F F E R E N T I A L E Q U A T I O N S 339 Solution: By (3), the solution isx =Ce−3t −3. Here the equilibrium state isx = −3, and the equation is stable becausea=3>0, andx → −3 ast→ ∞.
E X A M P L E 5 (Price Adjustment Mechanism) LetD(P )=a−bP denote the demand quantity and S(P )=α+βP the supply of a certain commodity when the price isP. Herea,b,α, and β are positive constants. Assume that the priceP =P (t )varies with time, and thatP˙ is proportional to excess demandD(P )−S(P ). Thus,
˙
P =λ[D(P )−S(P )]
whereλis a positive constant. Inserting the expressions forD(P )andS(P )into this equation givesP˙ =λ(a−bP −α−βP ). Rearranging, we then obtain
P˙+λ(b+β)P =λ(a−α) According to (3), the solution is
P =Ce−λ(b+β)t+a−α b+β
Becauseλ(b+β)is positive, ast tends to infinity, P converges to the equilibrium price Pe=(a−α)/(b+β), for whichD(Pe)=S(Pe). Thus, the equation is stable.
Variable Right-Hand Side
Consider next the case where the right-hand side is not constant:
˙
x+ax=b(t ) (4)
Whenb(t )is not constant, this equation is not separable. A clever trick helps us find the solu- tion. We multiply each side of the equation by the positive factoreat, called anintegrating factor. This gives the equivalent equation
˙
xeat+axeat =b(t )eat (∗)
This idea may not be obvious beforehand, but it works well because the left-hand side of (∗)happens to be the exact derivative of the productxeat. Thus(∗)is equivalent to
d
dt(xeat)=b(t )eat (∗∗)
According to the definition of the indefinite integral, equation(∗∗)holds for all t in an interval iffxeat=
b(t )eatdt+Cfor some constantC. Multiplying this equation bye−at gives the solution forx. Briefly formulated:
˙
x+ax =b(t ) ⇐⇒ x =Ce−at+e−at
eatb(t ) dt (5)
E X A M P L E 6 Find the solution of
˙
x+x =t and determine the solution curve passing through(0,0).
Solution: According to (5), witha =1 andb(t )=t, the solution is given by x=Ce−t+e−t
t etdt =Ce−t+e−t(t et−et)=Ce−t+t−1 where we used integration by parts to evaluate
t etdt. (See Example 9.5.1.) Ifx=0 when t =0, we get 0=C−1, soC=1 and the required solution isx=e−t+t−1.
E X A M P L E 7 (Economic Growth) Consider the following model of economic growth in a developing country (see FMEA, Example 5.4.3 for a more general model):
(a) X(t )=0.2K(t ) (b) K(t )˙ =0.1X(t )+H (t ) (c) N (t )=50e0.03t HereX(t )is total domestic product per year,K(t )is capital stock,H (t )is the net inflow of foreign investment per year, andN (t )is the size of the population, all measured at timet.
In (a) we assume that the volume of production is simply proportional to the capital stock, with the factor of proportionality 0.2 being called theaverage productivityof capital. In (b) we assume that the total growth of capital per year is equal to internal savings plus net foreign investment. We assume that savings are proportional to production, with the factor of proportionality 0.1 being called thesavings rate. Finally, (c) tells us that population increases at a constant proportional rate of growth 0.03.
Assume thatH (t ) = 10e0.04t and derive from these equations a differential equation for K(t ). Find its solution given that K(0) = 200. Find also an expression forx(t ) = X(t )/N (t ), which is domestic product per capita.
Solution: From (a) and (b), it follows thatK(t )must satisfy the linear equation
˙
K(t )−0.02K(t )=10e0.04t Using (5) we obtain
K(t )=Ce0.02t+e0.02t
e−0.02t10e0.04tdt=Ce0.02t+10e0.02t
e0.02tdt
=Ce0.02t+(10/0.02)e0.04t =Ce0.02t+500e0.04t
Fort =0,K(0)=200=C+500, soC = −300. Thus, the solution is
K(t )=500e0.04t−300e0.02t (∗) Per capita production isx(t )=X(t )/N (t )=0.2K(t )/50e0.03t =2e0.01t−1.2e−0.01t.
The solution procedure for the general linear differential equation (2) is somewhat more complicated, and we refer to FMEA.
R E V I E W P R O B L E M S F O R C H A P T E R 9 341
P R O B L E M S F O R S E C T I O N 9 . 9
1. Solve the equationx4x˙=1−t. Find the integral curve through(t, x)=(1,1).
⊂SM⊃2. Solve the following differential equations
(a) x˙=e2t/x2 (b) x˙=e−t+x (c) x˙−3x=18 (d) x˙=(1+t )6/x6 (e) x˙−2x= −t (f) x˙+3x=t et2−3t
3. Suppose thaty=αkeβtdenotes production as a function of capitalk, where the factoreβt is due to technical progress. Suppose that a constant fractions∈(0,1)is saved, and that capital accumulation is equal to savings, so that we have the separable differential equation
˙
k=sαkeβt, k(0)=k0 The constantsα,β, andk0are positive. Find the solution.
4. (a) SupposeY=Y (t )is national product,C(t )is consumption at timet, andI¯is investment, which is constant. SupposeY˙ =α(C+ ¯I−Y )andC =aY+b, wherea,b, andαare positive constants witha <1. Derive a differential equation forY.
(b) Find its solution whenY (0)=Y0is given. What happens toY (t )ast→ ∞?
⊂SM⊃5. (a) In a growth model productionQis a function of capitalK and labourL. Suppose that (i)K˙ =γ Q, (ii)Q=KαL, (iii)L/L˙ =βwithL(0)=L0. Assume thatβ=0 and α∈(0,1). Derive a differential equation forK.
(b) Solve this equation whenK(0)=K0.
6. Findx(t )when Eltx(t )= afor allt. Assume that bothtandxare positive and thatais a constant. (Recall that Eltx(t )is the elasticity ofx(t )w.r.t.t.)
R E V I E W P R O B L E M S F O R C H A P T E R 9
1. Find the following integrals:
(a)
(−16) dx (b)
55dx (c)
(3−y) dy (d)
(r−4r1/4) dr (e)
x8dx (f)
x2√
x dx (g) 1
p5dp (h)
(x3+x) dx 2. Find the following integrals:
(a)
2e2xdx (b)
(x−5e25x) dx (c)
(e−3x+e3x) dx (d) 2
x+5dx 3. Evaluate the following integrals:
(a) 12
0
50dx (b)
2 0
(x−12x2) dx (c) 3
−3
(u+1)2du (d)
5 1
2
zdz (e)
12 2
3dt
t+4dt (f)
4 0
v
v2+9dv
⊂SM⊃4. Find the following integrals:
(a) ∞
1
5
x5dx (b)
1 0
x3(1+x4)4dx (c) ∞
0
−5t
et dt (d) e
1
(lnx)2dx
(e) 2
0
x2
x3+1dx (f) 0
−∞
e3z
e3z+5dz (g) e/2
1/2
x3ln(2x) dx (h) ∞
1
e−√x
√x dx
⊂SM⊃5. Find the following integrals:
(a) 25
0
1 9+√
xdx (b)
7 2
t√
t+2dt (c) 1
0
57x23
19x3+8dx
6. FindF(x)if (a)F (x)= x
4
√ u+ x
√u du (b)F (x)= x
√x
lnu du
7. WithC(Y )as the consumption function, suppose the marginal propensity to consume isC(Y )= 0.69, withC(0)=1000. FindC(Y ).
8. In the manufacture of a product, the marginal cost of producingxunits isC(x)=αeβx+γ, withβ=0, and fixed costs areC0. Find the total cost functionC(x).
9. Supposef andgare continuous functions on [−1,3] and that 3
−1
(f (x)+g(x)) dx =6 and 3
−1
(3f (x)+4g(x)) dx=9. FindI= 3
−1
(f (x)+g(x)) dx.
⊂SM⊃10. For the following two cases, find the equilibrium price and quantity and calculate the consumer and producer surplus when the demand curve isf (Q)and the supply curve isg(Q):
(a) f (Q)=100−0.05Q and g(Q)=10+0.1Q.
(b) f (Q)= 50
Q+5 andg(Q)=4.5+0.1Q.
⊂SM⊃11. (a) Definef fort >0 byf (t )=4(lnt )2
t . Findf(t )andf(t ).
(b) Find possible local extreme points, and sketch the graph off. (c) Calculate the area below the graph off over the interval
1, e2 . 12. Solve the following differential equations:
(a) x˙= −3x (b) x˙+4x=12 (c) x˙−3x=12x2 (d) 5x˙= −x (e) 3x˙+6x=10 (f) x˙−12x=x2
⊂SM⊃13. Solve the following differential equations:
(a) x˙=t x2 (b) 2x˙+3x= −15 (c) x˙−3x=30 (d) x˙+5x=10t (e) x˙+12x=et (f) x˙+3x=t2
R E V I E W P R O B L E M S F O R C H A P T E R 9 343 14. (a) LetV (x)denote the number of litres of fuel left in an aircraft’s fuel tank if it has flownx
km. Suppose thatV (x)satisfies the following differential equation:
V(x)= −aV (x)−b
(The fuel consumption per km is a constantb >0. The term−aV (x), witha >0, is due to the weight of the fuel.) Find the solution of the equation withV (0)=V0.
(b) How many km,x∗, can the plane fly if it takes off withV0litres in its tank?
(c) What is the minimum number of litres,Vm, needed at the outset if the plane is to flyxˆkm?
(d) Putb=8,a=0.001,V0=12 000, andxˆ=1200. Findx∗andVmin this case.
15. A population ofnindividuals has an income density functionf (r) =(1/m)e−r/mforr in [0,∞), wheremis a positive constant. (See Section 9.4.)
(a) Show thatmis the mean income.
(b) Suppose the demand function isD(p, r)=ar−bp. Compute the total demandx(p)when the income distribution is as in (a).
HARDER PROBLEM
⊂SM⊃16. A probability density functionf is defined for allxby
f (x)= λae−λx
(e−λx+a)2 (aandλare positive constants) (a) Show thatF (x)= a
e−λx+ais an indefinite integral off (x), and determine lim
x→∞F (x)and
x→−∞lim F (x).
(b) Show that x
−∞
f (t ) dt=F (x), and thatF (x)is strictly increasing.
(c) ComputeF(x)and show thatFhas an inflection pointx0. ComputeF (x0)and sketch the graph ofF.
(d) Compute ∞
−∞
f (x) dx.
10 T O P I C S I N F I N A N C I A L E C O N O M I C S
I can calculate the motions of heavenly bodies, but not the madness of people.1
—I. Newton (attr.)
This chapter treats some basic topics in the mathematics of finance. The main concern is how the values of investments and loans at different times are affected by interest rates.
Sections 1.2 and 4.9 have already discussed some elementary calculations involving interest rates. This chapter goes a step further and considers different interest periods. It also discusses in turn effective rates of interest, continuously compounded interest, present values of future claims, annuities, mortgages, and the internal rate of return on investment projects. The calcu- lations involve the summation formula for geometric series, which we therefore derive.
In the last section we give a brief introduction to difference equations.