8.6 Secondorder differential equations
General form
A secondorder ordinary differential equation is a differential equation of the form
G(t, x(t), x'(t), x"(t)) = 0 for all t,
involving only t, x(t), and the first and second derivatives of x. We can write such an equation in the form
x"(t) = F (t, x(t), x'(t)).
Equations of the form x"(t) = F (t, x'(t))
An equation of the form
x"(t) = F (t, x'(t)),
in which x(t) does not appear, can be reduced to a firstorder equation by making the substitution z(t) = x'(t).
 Example

Let u(w) be a utility function for wealth w. The function
ρ(w) = −wu"(w)/u'(w)
is known as the ArrowPratt measure of relative risk aversion. (If ρ_{u}(w) > ρ_{v}(w) for two utility functions u and v then u reflects a greater degree of riskaversion than does v.)
What utility functions have a degree of riskaversion that is independent of the level of wealth? That is, for what utility functions u do we have
a = −wu"(w)/u'(w) for all w?
This is a secondorder differential equation in which the term u(w) does not appear. (The variable is w, rather than t.) Define z(w) = u'(w). Then we have
a = −wz'(w)/z(w)
or
az(w) = −wz'(w),
a separable equation that we can write as
a·dw/w = −dz/z.
The solution is given by
a·ln w = −ln z(w) + C,
or
z(w) = Cw^{−a}.
Now, z(w) = u'(w), so to get u we need to integrate:
u(w) = 

Cln w + B 
if a = 1 
Cw^{1−a}/(1 − a) + B 
if a ≠ 1 

We conclude that a utility function with a constant degree of riskaversion equal to a takes this form.

A linear secondorder differential equation with constant coefficients takes the form
x"(t) + ax'(t) + bx(t) = f (t)
for constants a and b and a function f . Such an equation is homogeneous if f (t) = 0 for all t.
Let x_{1} be a solution of the equation
x"(t) + ax'(t) + bx(t) = f (t),
to which I refer subsequently as the "original equation". For any other solution of this equation x, define z = x − x_{1}. Then z is a solution of the homogeneous equation
x"(t) + ax'(t) + bx(t) = 0
(because z"(t) + az'(t) + bz(t) = [x"(t) + ax'(t) + bx(t)] − [x_{1}"(t) + ax_{1}'(t) + bx_{1}(t)] = f (t) −
f (t) = 0).
Further, for every solution z of the homogeneous equation, x_{1} + z is clearly a solution of original equation.
We conclude that the set of all solutions of the original equation may be found by finding one solution of this equation and adding to it the general solution of the homogeneous equation. That is, we may use the following procedure to solve the original equation.
 Procedure
for finding general solution of linear secondorder differential equation with constant coefficients

The general solution of the differential equation
x"(t) + ax'(t) + bx(t) = f (t)
may be found as follows.
 Find the general solution of the associated homogeneous equation x"(t) + ax'(t) + bx(t) = 0.
 Find a single solution of the original equation x"(t) + ax'(t) + bx(t) = f (t).
 Add together the solutions found in steps 1 and 2.

I now explain how to perform to the first two steps and illustrate step 3.
1. Finding the general solution of a homogeneous equation
You might guess, based on the solutions we found for firstorder equations, that the homogeneous equation has a solution of the form x(t) = Ae^{rt}. Let's see if it does. If x(t) = Ae^{rt} then x'(t) =
rAe^{rt} and x"(t) = r^{2}Ae^{rt}, so that
x"(t) + ax'(t) + bx(t) 
= r^{2}Ae^{rt} + arAe^{rt} +
bAe^{rt} 

= Ae^{rt}(r^{2} + ar + b). 
Thus for x(t) to be a solution of the equation we need
r^{2} + ar + b = 0.
This equation is knows as the characteristic equation of the differential equation. It has either two distinct real roots, a single real root, or two complex roots. If a^{2} > 4b then it has two distinct real roots, say r and s, and we have shown that both x(t) =
Ae^{rt} and x(t) = Be^{st}, for any values of A and B, are solutions of the equation. Thus also x(t) = Ae^{rt} + Be^{st} is a
solution. It can be shown, in fact, that every solution of the equation takes this form.
The cases in which the characteristic equation has a single real root (a^{2} = 4b) or complex roots (a^{2} < 4b) require a slightly different analysis, with the following conclusions.
 Proposition

Consider the homogeneous linear secondorder differential equation with constant coefficients
x"(t) + ax'(t) + bx(t) = 0 for all t,
where a and b are numbers. The general solution of this equation depends on the character of the roots of the characteristic equation r^{2} + ar + b = 0 as follows.
 Distinct real roots
 If a^{2} > 4b, in which case the characteristic equation has distinct real roots, the general solution of the equation is
Ae^{rt} + Be^{st}.
 Repeated real root
 If a^{2} = 4b, in which case the characteristic equation has a single root, the general solution of the equation is
(A + Bt)e^{rt},
where r = −(1/2)a is the root.
 Complex roots
 If a^{2} < 4b, in which case the characteristic equation has complex roots, the general solution of the equation is
(Acos(βt) + Bsin(βt))e^{αt},
where α = −a/2 and β = √(b − a^{2}/4). This solution may alternatively be expressed as
Ce^{αt} cos(βt + ω),
where the relationships between the constants C, ω, A, and B are A = C cos ω and B = −C sin ω.

 Example

Consider the differential equation
x"(t) + x'(t) − 2x(t) = 0.
The characteristic equation is
r^{2} + r − 2 = 0
so the roots are 1 and −2. That is, the roots are real and distinct. Thus the general solution of the differential equation is
x(t) = Ae^{t} + Be^{−2t}.

 Example

Consider the differential equation
x"(t) + 6x'(t) + 9x(t) = 0.
The characteristic equation has a repeated real root, equal to −3. Thus the general solution of the differential equation is
x(t) = (A + Bt)e^{−3t}.

 Example

Consider the equation
x"(t) + 2x'(t) + 17x(t) = 0.
The characteristic roots are complex. We have a = 2 and b = 17, so α = −1 and β = 4, so the general solution of the differential equation is
[A cos(4t) + B sin(4t)]e^{−t}.

A good technique to use to find a solution of a nonhomogeneous equation is to try a linear combination of f (t) and its first and second derivatives. If, for example, f (t) = 3t − 6t^{2}, then try to find values of A, B, and C such that A + Bt +
Ct^{2} is a solution. Or if f (t) = 2sin t + cos t, then try to find values of A and B such that f (t) = Asin t + Bcos t is a solution. Or if f (t) = 2e^{Bt} for some value
of B, then try to find a value of A such that Ae^{Bt} is a solution.
 Example

Consider the differential equation
x"(t) + x'(t) − 2x(t) = t^{2}.
The function on the righthand side is a seconddegree polynomial, so to find a solution of the equation, try a general seconddegree polynomialthat is, a function of the form x(t) = C + Dt + Et^{2}. (The important point to note is that you should not restrict to a function of the form x(t) =
Et^{2}.) For this function to be a solution,
2E + D + 2Et − 2C − 2Dt − 2Et^{2} = t^{2} for all t,
so we need (equating the coefficients of t^{2}, t, and the constant on both sides),
2E + D − 2C 
= 0 
2E − 2D 
= 0 
−2E 
= 1. 
We deduce that E = −1/2, D = −1/2, and C = −3/4. That is,
x(t) = −3/4 − t/2 − t^{2}/2
is a solution of the differential equation.

This step is quite trivial!
 Example

Consider the equation from the previous example, namely
x"(t) + x'(t) − 2x(t) = t^{2}.
We saw above that the general solution of the associated homogeneous equation is
x(t) = Ae^{t} + Be^{−2t}.
and that
x(t) = −3/4 − t/2 − t^{2}/2
is a solution of the original equation.
Thus the general solution of the original equation is
x(t) = Ae^{t} + Be^{−2t} − 3/4 − t/2 − t^{2}/2.

Consider the homogeneous equation
x"(t) + ax'(t) + bx(t) = 0.
If b ≠ 0, this equation has a single equilibrium, namely 0. (That is, the only constant function that is a solution is equal to 0 for all t.) To consider the stability of this equilibrium, consider separately the three possible forms of the general solution of the equation.
 Characteristic equation has two real roots
 If the characteristic equation has two real roots, r and s, so that the general solution of the equation is Ae^{rt} + Be^{st}, then the equilibrium is stable if and only if r < 0 and s < 0.
 Characteristic equation has a single real root
 If the characteristic equation has a single real root, then the equilibrium is stable if and only if this root is negative. (Note that if r < 0 then tor any value of k, t^{k}e^{rt} converges to 0 as t increases without bound.)
 Characteristic equation has complex roots
 If the characteristic equation has complex roots, the form of the solution of the equation is Ae^{αt} cos(βt + ω), where α = −a/2, the real part of each root. Thus the equilibrium is stable if and only if the real part of each root is negative.
The real part of a real root is simply the root, so we can combine the three cases: the equilibrium is stable if and only if the real parts of both roots of the characteristic equation are negative. A bit of algebra shows that this condition is equivalent to a > 0 and b > 0.
If b = 0, then every number is an equilibrium, and none of these equilibria is stable.
In summary, we have the following result.
 Proposition

An equilibrium of the homogeneous linear secondorder differential equation x"(t) + ax'(t) + bx(t) = 0 is stable if and only if the real parts of both roots of the characteristic equation r^{2} + ar + b = 0 are negative, or, equivalently, if and only if a > 0 and b >
0.

 Example

Consider the following macroeconomic model. Denote by Q aggregate supply, p the price level, and π the expected rate of inflation. Assume that aggregate demand is a linear function of p and π, equal to a − bp + cπ where a > 0, b > 0, and c > 0. An equilibrium condition is
Q(t) = a − bp(t) + cπ(t).
Denote by Q* the longrun sustainable level of output, and assume that prices adjust according to the equation
p'(t) = h(Q(t) − Q*) + π(t),
where h > 0. Finally, suppose that expectations are adaptive:
π'(t) = k(p'(t) − π(t))
for some k > 0. Is this system stable?
One way to answer this question is to reduce the system to a single secondorder differential equation by differentiating the equation for p'(t) to obtain p"(t) and then substituting in for π'(t) and π(t). We obtain
p"(t) − h(kc − b)p'(t) + khbp(t) = kh(a − Q*).
We conclude the system is stable if and only if kc < b. (Since k > 0, h > 0, and b > 0, we have khb > 0.)
In particular, if c = 0 (i.e. expectations are ignored) then the system is stable. If expectations are taken into account, however, and respond rapidly to changes in the rate of inflation (k is large), then the system may be unstable.

Exercises
Copyright © 19972003 by Martin J. Osborne