## 8.6 Second-order differential equations

### General form

A second-order 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 first-order equation by making the substitution z(t) = x'(t).

### Linear second-order equations with constant coefficients

A linear second-order 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 x1 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 = xx1. 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)] − [x1"(t) + ax1'(t) + bx1(t)] =  f (t) −  f (t) = 0). Further, for every solution z of the homogeneous equation, x1 + 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 second-order 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 first-order equations, that the homogeneous equation has a solution of the form x(t) = Aert. Let's see if it does. If x(t) = Aert then x'(t) = rAert and x"(t) = r2Aert, so that
 x"(t) + ax'(t) + bx(t) = r2Aert + arAert + bAert = Aert(r2 + ar + b).
Thus for x(t) to be a solution of the equation we need
r2 + 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 a2 > 4b then it has two distinct real roots, say r and s, and we have shown that both x(t) = Aert and x(t) = Best, for any values of A and B, are solutions of the equation. Thus also x(t) = Aert + Best 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 (a2 = 4b) or complex roots (a2 < 4b) require a slightly different analysis, with the following conclusions.

#### 2. Finding a solution of a nonhomogeneous equation

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 − 6t2, then try to find values of A, B, and C such that A + Bt + Ct2 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) = 2eBt for some value of B, then try to find a value of A such that AeBt is a solution.

#### 3. Add together the solutions found in steps 1 and 2

This step is quite trivial!

#### Stability of solutions of homogeneous equation

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 Aert + Best, 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, tkert 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 Aeαt 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.