MATH ECON II

HIGHER ORDER ODES

1. 1-Dimensional, 2nd order ODEs

1.1. A Motivating Example. Let us return to the market equilib-rium model. We took one step away from the classical static model,by introducing the price updating ODE p(t) = (D(p(t)) S(p(t)))with D(p) and S(p) being the demand and supply at price p respec-tively, and > 0 some constant. This equation obviously assumesthat the supply and demand depend only on the price at the currenttime t. This may be reasonable in a static model, where we only havethe present time. However, agents in a dynamic market may developexpectation regarding the price, according to its past trend. The ideais that agents may assume price convey information on other agentsbehaviour (and they may be right doing so). So, for example, we mayhave agents who look at the price, and see that the price is going up(p > 0), and going up quicker every minute (p > 0) and so may changetheir demand accordingly. So it may be that in this case change inprice is again due to change in excess demand, only now, demand andsupply are functions of p, p, and p (the second derivative w.r.t time ofp), hence we will get the equation

p(t) = (D(p(t), p(t), p(t)) S(p(t), p(t), p(t)))(1.1.1)This is a second order differential equation, as it functionally relatestogether p(t) and its derivatives up to the second order. In many casesof interest, such an equation can be re-written so that it presents thesecond order derivative as a function of lower order derivatives, andtime. Namely, an expression of the form

x(t) = F (x(t), x(t), t).(1.1.2)

A general solution is a function x(t) = g(t, A,B) that satisfies theequation x = F (x, x, t) given that x(0) = A and x(0) = B. A solutionwill be determined by the initial conditions x(0) = x0 and x(0) = x0(or, more generally, x(t0) and x(t0) for some initial time t0).

1.2. Reducible equations. These are equations which are indepen-dent of x, namely x(t) = F (x(t), t). By introducing z = x(t) we obtain

1

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the equation

z(t) = F (z(t), t)(1.2.1)

which is a 1st order equation. If we are able to solve this equation (bythe methods we introduced above), then we can write

x(t) =

z(t)dt.(1.2.2)

Example 1.2.1. x = x. Denote z = x. Then we obtain the equation

z = z z(t) = Aet,(1.2.3)and therefore x(t) =

z(t)dt = Aet + B. Notice that indeed, x(t) is

determined by a specification of two initial conditions. For example,suppose x(0) = 1 and x(0) = 1. Then 1 = x(0) = Ae0 + B = A + Band 1 = x(0) = Ae0 = A, and hence A = 1 and B = 0 and the solutionis given by x(t) = et.

1.3. Homogeneous Linear Equations. These are equations of theform

x+ a(t)x+ b(t)x = 0.(1.3.1)

The term linear follows from the fact that if x1(t) and x2(t) satisfyequation (1.3.1) then so does rx1(t) + sx2(t) for every r, s R. Indeed,if r, s R and x1(t) and x2(t) satisfy equation (1.3.1) then

d2

dt2(rx1 + sx2) + a(t)

d

dt(rx1 + sx2) + b(t)(rx1 + sx2) =

r(x1 + a(t)x1 + b(t)x1) + s(x2 + a(t)x2 + b(t)x2) = r 0 + s 0 = 0.Theorem 1.3.1. If x1 and x2 are linearly independent solutions ofequation (1.3.1) then every solution x of equation (1.3.1) has the formx = rx1 + sx2 for some r, s R, with r, s R determined by the initialconditions.

Theorem 1.3.1 states, in fact, that the general solution of equation(1.3.1) is of the form x(t) = Ax1(t) + Bx2(t), where x1, x2 are somelinearly independent solutions of equation (1.3.1).

Example 1.3.2. Consider the equation x + x = 0. What is the generalsolution of this equation? Theorem 1.3.1 states that it is sufficient tofind two linearly independent solutions. Can we think of functions thatsatisfy x + x = 0? We shall use a trick - multiply the equation by 2x.We thus obtain 2xx+ 2xx = 0. Integration w.r.t. t yields:

C =

(2xx+ 2xx)dt = x2 + x2.(1.3.2)

MATH ECON II HIGHER ORDER ODES 3

Do we a function x s.t., for example, x2+x2 = 1? This might remind usof the trigonometric identity sin2 t+cos2 t = 1. Indeed, set x1(t) = cos tand x2(t) = sin t. Then x

2i + x

2i = 1 for i = 1, 2. We can also verify

now that xi + xi = 0 for every i = 1, 2. But are x1(t) and x2(t)linearly independent? Linear independence means that there is no R s.t. x1(t) = x2(t) for every t. Suppose by contradictionthat there was such . Then for every t we have cos t = sin t. Inparticular this must hold for t = 0, in which case we obtain the equality1 = cos 0 = sin 0 = 0, a contradiction. Thus, the assumption thatx1 and x2 are linearly dependent leads to a contradictions, meaningthat they are independent. Hence it follows by Theorem 1.3.1 that thegeneral solution for the equation x+ x = 0 is given by

x(t) = A cos t+B sin t.(1.3.3)

1.4. Homogeneous equations with constant coefficients. Sup-pose that a(t) = a and b(t) = b are constants. We wish to find thegeneral solution of the equation

x+ ax+ bx = 0.(1.4.1)

The characteristic equation of equation (1.4.1) is the quadratic equa-tion

r2 + ar + b = 0.(1.4.2)

The roots of this equation are given by

r1,2 = a2a2

4 b.(1.4.3)

Denote = a2

4 b. This is the discriminant of the characteristic

equation.

Theorem 1.4.1. 1. If > 0 then r1 6= r2 are real numbers, andthe general solution x to equation (1.4.1) is given by:

x(t) = Aer1t +Ber2t.(1.4.4)

2. If = 0 then r1 = r2 = r is a real number, and the generalsolution x to equation (1.4.1) is given by:

x(t) = (A+Bt)ert.(1.4.5)

1. If < 0 then r1 6= r2 are complex numbers of the form r1,2 =a

2+ i with =

, and the general solution x to equation(1.4.1) is given by:

x(t) = eat2 (A cos(t) +B sin(t)) .(1.4.6)

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Example 1.4.2. We wish to find the general solution of x+5x+4x = 0.The first step is to compute . In our case a = 5 and b = 4 and so = 5

2

4 4 = 25

4 4 = 9

4> 0. In this case we know that the general

solution for the equation is given by (see Theorem 1.4.1, section (1))

x(t) = Aer1t +Ber2t,(1.4.7)

where r1, r2 are the roots of the characteristic equation r2 + 5r+ 4 = 0.

Computing the roots yields r1 = 1 and r2 = 4, and thus the generalsolution of the equation is

x(t) = Aet +Be4t.(1.4.8)

Example 1.4.3. We wish to find the general solution of x+ 2x+ x = 0.The first step is to compute . In our case a = 2 and b = 1 and so = 2

2

4 1 = 4

4 4 = 0. In this case we know that the general solution

for the equation is given by (see Theorem 1.4.1, section (2))

x(t) = (A+Bt)ert,(1.4.9)

where r is the unique solution for the characteristic equation r2 + 2r+1 = 0. Computing the root yields r = 1, and thus the general solutionof the equation is

x(t) = (A+Bt)et.(1.4.10)

Example 1.4.4. We wish to find the general solution of x+4x+5x = 0.The first step is to compute . In our case a = 4 and b = 5 and so = 4

2

4 5 = 4 5 = 1 < 0. In this case we know that the general

solution for the equation is given by (see Theorem 1.4.1, section (3))

x(t) = eat2 (A cos(t) +B sin(t)),(1.4.11)

where =. So in our case = = 1 = 1, and as a = 4

we find that general solution of the equation is

x(t) = e2t(A cos t+B sin t).(1.4.12)

1.4.1. Nonhomogeneous equations with constant coefficients.

Example 1.4.5. Price adjustment model with speculative sellers andbuyers and temporal fluctuations. In this model the demand isD(p, p, p) =D0(t)p+p+p, and the supply is S(p, p, p) = S0(t)ap+bp+cp.The price adjustment equation (the law of supply and demand) is

p = D(p, p, p) S(p, p, p) =(1.4.13)D0(t) S0(t) ( + a)p+ ( + b)p+ ( + c)p.

MATH ECON II HIGHER ORDER ODES 5

The equation (1.4.13) has the form

x+ ax+ bx = f(t),(1.4.14)

where a, b R are constants, and f is some continuous function of t.We refer to these equations as nonhomogeneous linear equations withconstant coefficients. In general (namely, for a general f), it mightbe quite difficult to solve equation (1.4.14). However, we will be ableto accomplish that in some specific, yet interesting, examples. Thefundamental theorem here is:

Theorem 1.4.6. Let xh be the general solution for the homogeneousequation x + ax + bx = 0, and let xp be a particular solution for thenonhomogeneous equation (1.4.14). Then the general solution for theequation (1.4.14) is x(t) = xh(t) + xp(t).

The theorem suggests that finding the general solution for the non-homogeneous equation (1.4.14) boils down into 1. finding the generalsolution for the homogeneous equation x+ax+bx = 0, and 2. finding aparticular solution for the equation (1.4.14). As we already know howto find the general solution for the homogeneous equation, we shall nowexplore methods for finding a particular solution in various cases.

Example 1.4.7. f(t) = C, with C being a constant. We will be lookingfor a particular solution xp(t) = B with B being a constant. It is easyto verify that the particular solution is in fact xp(t) =

Cb.

Example 1.4.8. f is a polynomial. Namely, f(t) =ni=0

citi, with c0, c1, ..., cn

R. Notice that ddt

(tn) = ntn1 and d2dt2

(tn) = n(n1)tn2. This leads usto look for a polynomial particular solution. Namely, we assume that

xp(t) =ni=0

diti, where the degree n of the polynomial xp is the same as

the degree n of the polynomial f . For example, suppose that we wishto solve x + 2x + x = t2. We will be looking for a particular solutionof the form xp(t) = at

2 + bt + c. Thus xp(t) = 2at + b and xp(t) = 2a.Substituting this back into the equation yields:

2a+ 2(2at+ b) + (at2 + bt+ c) = t2 at2 + (4a+ b)t+ 2a+ 2b+ c = t2 = t2,

for every t. The only possible way to for the equation above to holdfor every t is that the coefficients of the same power of t on both sidesmust be equal. Hence

a = 1, 4a+ b = 0, 2a+ 2b+ c = 0 a = 1, b = 4, c = 10.

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Therefore the particular solution is given by xp(t) = t2 4t 10. The

general solution xh of the homogeneous equation x+2x+x = 0 is givenby (verify that!)

xh(t) = (A+Bt)et,

and thus, the general solution for the nonhomogeneous equation x +2x+ x = t2 is (by Theorem 1.4.6)

x(t) = (A+Bt)et + t2 4t 10.Example 1.4.9. f(t) = A exp(t). We shall look for particular solutionof the form

xp(t) = B exp(t).(1.4.15)

For example consider the equation x + 2x + 4x = exp(2t). So we arinterested in a particular solution of the form xp(t) = B exp(2t), and soxp(t) = 2B exp(2t), and xp(t) = 4B exp(2t). Thus (substituting backin the equation):

4B exp(2t) + 4B exp(2t) + 4B exp(2t) = exp(2t) B = 112.

(1.4.16)

Therefore, xp(t) =112

exp(2t) is a particular solution of the equationx+ 2x+ 4x = exp(2t).

Example 1.4.10. f(t) = A sin(t) +B cos(t). We shall look for a solu-tion xp(t) = C cos(t) +D sin(t). For example, consider the equationx+ x+ x = cos t. Here = 1, and we are looking for a particular solu-tion of the form xp(t) = C cos t+D sin t. Thus xp(t) = D cos tC sin t,and xp(t) = C cos t D cos t = xp(t). Therefore, by substitutingback into the equation we obtain:

cos t = xp + xp + xp = xp(t) + xp + xp = xp(t) = D cos t C sin tcos t = D cos t C sin t.(1.4.17)

Now, the only way to obtain an equality above, the coefficient of sin onboth sides of the equation should be equal ,and the same should holdfor cos. Therefore, D = 1 and C = 0 and we find that xp(t) = cos t isa particular solution of the equation.

1.5. Market Model with Price Expectations. Lets go back to theexample of price adjustment with speculation and fluctuation. SupposeD0(t) = F0 cos(t), S0(t) = 0, and that = 1 +a = k b+ b = d+ 1, + c = m Then the price adjustment equation transforms into

F0 cos(t) = kp+ dp+mp(1.5.1)

MATH ECON II HIGHER ORDER ODES 7

For the particular solution we look for a function of the form pp(t) =A cos(t) + B sin(t). Then pp(t) = A sin(t) + B cos(t) andpp(t) = A2 cos(t)B2 sin(t) = 2pp(t). SoF0 cos(t) = kp+ dp+mp = kp+ dpm2p = dp+ (k m2)p =d(B cos(t) A sin(t)) + (k m2)(A cos(t) +B sin(t))

(k m2)A+ dB = F0, dwA+ (k m2)B = 0

A =(k m2)F0

(k m2)2 + (d)2 , B =dF0

(k m2)2 + (d)2Let 1 =

km2[(km2)2+(d)2]1/2 and 2 =

d[(km2)2+(d)2]1/2 . Notice that

21 +22 = 1, so (1, 2, 1) is a Pythagorean triplet (i.e., these three non-

negative numbers obey the Pythagorean theorem). We thus deducethat there is a [0, pi

2

]s.t. 1 = cos and 2 = sin. We can now

write the particular solution of the equation as

pp(t) =F0

[(k m2)2 + (d)2]1/2 (cos cos(t) + sin sin(t)) =F0

[(k m2)2 + (d)2]1/2 cos(t ),(1.5.2)

where the last equality above follows from the trigonometric identitycos(x+ y) = cos x cos y sinx sin y. So prices oscillate with amplitude

F0(km2)2+(d)2 (after the introduction of a phase). What happens as

k m2 approaches 0? In this case the amplitude F0[(km2)2+(d)2]1/2

will increase and in the limit 2 km

it will approach F0d

.

1.6. Inflation and Unemployment. The Philips relations depicts,empirically, the relation between the rate of growth w in wages Wand unemployment U . In other words

w = f(U)(1.6.1)

with f being some monotonically decreasing function (so more unem-ployment means slower growth rate w). A similar relation, baring thesame name, can be written for the empirical relationship between un-employment and the rate of inflation p, which is the rate of growthof price index P . The relating economic factor between the two (wand p) is the labor productivity `, which is given exogenously, and forsimplicity assume p = w ` (so the higher productivity is the lowerthe inflation is, as you may produce the demand for lower costs, and ifwages increase sharply it will create inflation). So we can now write

p = f(U) `.(1.6.2)

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A variant on Equation (1.6.1) is to add an expectation factor to it.Namely, employers expect a certain rate pi of inflation to kick into thefuture wage, and account for that. If the employer future discount rate > 0 then Equation (1.6.1) transforms into

w(t) = f(U(t)) + pi(t).(1.6.3)

For simplicity of calculation, suppose f(U) = a + bU . So p + ` = w =a + bU + pi. We should also say something about the expectationvariable pi. We will assume that employers adapt their expectationaccording to the evolution of the real inflation, namely, of p. So if theyovershoot, they will lower expectations and decrease pi, and vice versa.In other words, for some > 0

pi(t) = (p(t) pi(t)).(1.6.4)Finally, we shall need some feedback between U and p. This is given inthe most simple way (not the only way of course, just the most simpleand even simplistic) by the monetary policy, namely, if the moneygrowth rate is m = m(t), which is exogenously given to us, then weshall assume

U(t) = (p(t)m(t))(1.6.5)with > 0, which means that if money supply growth exceeds inflationrate then it happened as a result in direct saving in labor cost1, hencemore unemployment. So now, revisiting Equation (1.6.4 we have

pi(t) = (p(t) pi(t)) = (a+ bU + pi(t) ` pi(t)).(1.6.6)Taking the time derivative in Equation (1.6.6) we obtain

pi(t) = (bU + pi(t) ` pi(t)) =(1.6.7)(b(p(t)m(t)) + pi(t) ` pi(t)),

with the last equality following from Equation (1.6.5). Now, as byEquation (1.6.4) we have p(t) = 1

pi(t) + pi(t) then we can substitute

that back into Equation (1.6.7) to obtain

pi(t) = (b(1

pi(t) + pi(t)m(t)) + pi(t) ` pi(t)),(1.6.8)

which is a differential equation of the form

pi + Api +Bpi = f(t)(1.6.9)

with A = b , B = b, and f(t) = bm(t) `. Given `, acentral bank can control m(t) and thus influence inflation expectations,

1Notice, that a first possible tweak of the model would be to account for indirect changes, likechanging wages for hour, shorter hours, seasonal workers etc.

MATH ECON II HIGHER ORDER ODES 9

and through that (given that we solve the second order equation (1.6.9)affect inflation rates (Equation 1.6.4) and unemployment (1.6.5).