# 4.6: Second-Order Differential and Difference Equations

With our understanding of the functions $$e^x$$, $$e^{jΘ}$$, and the quadratic equation $$z^2 + \frac b a z + /frac c a =0$$, we can undertake a rudimentary study of differential and difference equations.

## Differential Equations

In your study of circuits and systems you will encounter the homogeneous differential equation

$\frac {d^2} {dt^2} x(t)+a_1\frac d {dt} x(t)+a_2=0$

Because the function $$e^{st}$$ reproduces itself under differentiation, it is plausible to assume that x(t)=est is a solution to the differential equation. Let's try it:

$\frac {d^2} {dt^2}(e^{st})+a_1\frac d {dt}(e^{st})+a_2(e^{st})=0$

$(s^2+a_1s+a_2)e^{st}=0$

If this equation is to be satisfied for all $$t$$, then the polynomial in $$s$$ must be zero. Therefore we require

$s^2+a_1s+a_2=0$

As we know from our study of this quadratic equation, the solutions are

$s_{1,2}=−\frac {a_1} 2 ± \frac 1 2 \sqrt{a^2_1−4a_2}$

This means that our assumed solution works, provided $$s=s_1$$ or $$s_2$$. It is a fundamental result from the theory of differential equations that the most general solution for x(t) is a linear combination of these assumed solutions:

$x(t)=A_1e^{s_1t}+A_2e^{s_2t}$

If $$a^2_1−4a_2$$ is less than zero, then the roots $$s_1$$ and $$s_2$$ are complex:

$s_{1,2}=−\frac {a_1} 2 ± j\frac 1 2 \sqrt{4a_2−a^2_1}$

Let's rewrite this solution as

$s_{1,2}=σ±jω$

where σ and ω are the constants

$σ=−\frac {a_1} 2$

$ω=\frac 1 2 \sqrt{4a_2−a^2_1}$

With this notation, the solution for x(t) is

$x(t)=A_1e^{σt}e^{jωt}+A_2e^{σt}e^{−jωt}$

If this solution is to be real, then the two terms on the right-hand side must be complex conjugates. This means that $$A_2=A^∗_1$$ and the solution for x(t) is

$x(t)=A_1e^{σt}e^{jωt}+A^∗_1e^{σt}e^{−Jωt} = 2\mathrm{Re} \{A_1e^{σt} e^{jωt}\}$

The constant $$A_1$$ may be written as $$A_1=|A|e^{jφ}$$. Then the solution for x(t) is

$x(t)=2|A|e^{σt}\cos(ωt+φ)$

This “damped cosinusoidal solution” is illustrated in the Figure.

Exercise $$\PageIndex{1}$$

Find the general solutions to the following differential equations:

a. $$\frac {d^2} {dt^2} X(t)+2\frac d {dt} x(t)+2=0$$
b. $$\frac {d^2} {dt^2} x(t)+2\frac d {dt} x(t)−2=0$$
c. $$\frac {d^2} {dt^2} x(t)+2=0$$

## Difference Equations

In your study of digital filters you will encounter homogeneous difference equations of the form

$x_n+a_1x_n−1+a_2x_{n−2}=0$

What this means is that the sequence $$\{x_n\}$$ obeys a homogeneous recursion:

$x_n=−a_1x_{n−1}−a_2x_{n−2}$

A plausible guess at a solution is the geometric sequence $$x_n=z^n$$. With this guess, the difference equation produces the result

$z^n+a_1z^n−1+a_2z^{n−2}=0$

$(1+a_1z^{−1}+a_2z^{−2})z^n=0$

If this guess is to work, then the second-order polynomial on the left-hand side must equal zero:

$1+a_1z^{−1}+a_2z^{−2}=0$

$z^2+a_1z+a_2=0$

The solutions are

$z_{1,2}=−\frac {a_1} 2 ± j\frac 1 2 \sqrt{4a_2−a^2_1} = re^{jθ}$

The general solution to the difference equation is a linear combination of the assumed solutions:

$x_n=A_1z^n_1+A_2(z^∗_1)^n$

$=A_1z^n_1+A^∗_1(z^∗_1)$

$=2\mathrm{Re}{A_1z^n_1}$

$=2|A|r^n\cos(θn+φ)$

This general solution is illustrated in the Figure.

Exercise $$\PageIndex{2}$$

Find the general solutions to the following difference equations:

a. x_n+2x_{n−1}+2=0
b. x_n−2x_{n−1}+2=0
c. x_n+2x_{n−2}=0