## A V4 at

We can now express the solution as

As shown in the appendix, we can use Euler's formula to express the imaginary exponential functions. ev" and e'J", as

Using these relationships, we write the solution to the homogeneous form of the differential equation as

\Vi = i'/"[C|(cos vt + / sin vt) + C;(cos vt — i sin U)|

or as yh = ^'(Ci -F C2)cos vt + e'"(C, - C2)i sin vt

Since (C| + C2) and (C| — Co)/ are arbitrary constants, we can rename them as Aiand A2

An important point is that A i and A2 are real-valued. The reason is that C\ and C2, like the roots, are conjugate complex numbers. As we show in the appendix, the sum of two conjugate complex numbers is always a real number, and the product of / and the difference between two conjugate complex numbers is also a real number. Consequently we obtain a real-valued solution to the differential equation even though the roots are complex numbers.

Theorem 23.3 If the roots of the characteristic equation are complex numbers, the solution to the homogeneous form of die linear, secondordcr differential equation wilh constant coefficients can be expressed as yh = A\ehl cos vt -F A2e'" sin vt (23.15)

where

When the roots of the characteristic equation are complex-valued, the solution involves circular, or trigonometric, functions (sine and cosine) off. This has interesting applications to economics because the circular functions are oscillating functions off, which leads to cyclical behavior for y(f), much like many real-world economic variables, yhW

Example 23.4 Solve the following homogeneous differential equation:

Solution

The characteristic equation is rl + 2r + 5 = 0

for which the roots are

Figure 23.4 Representative trajectory for example 23.4

In this case h = — I and v - 2i. By theorem 23.3, the solution is

>'/,(/ ) = A\e~' cos2/ 4- A2e~' sin 2/ Figure 23.4 shows a representative trajectory for this solution. ■

### The Particular Solution

We have found the solution to the homogeneous form of equation (23.1). If we can now find a particular solution to equation (23.1), we will be able to obtain the complete solution by adding these two solutions together.

As in earlier chapters, the particular solution we look for in the case of an autonomous equation is the steady-state solution for v. If y is a steady-state value, it must be true that v = y = 0 at this value. Set y - v -= 0 in the complete differential equation in equation (23.1 ) to solve for v. This gives v = —, Û2

The steady-state solution exists as long as a2 ^ 0. For the remainder of this section, we assume a2 # 0. The steady-state solution then serves as the particular solution we require:

yr = bj ¡ii where lire /j subscript reminds us that this is ihe particular solution. If ih = 0. the steady-state value is undefined and we must instead use a different technique, explained in section 23.2, to find a particular solution.

Example 23.5 Find the particular solution for die following differential equation y - 5 y + 2y = 10

Solution

Find the steady-state value by setting y = y = 0. and use this as the particular solution. This gives

Example 23.6 Find the particular solution for the following differential equation:

Solution

Find the steady-state value by setting y = 0, and use this as the particular solution This gives

The Complete Solution

The complete solution to a second-order differential equation is the sum of the homogeneous solution and a particular solution

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