Page 73 - Advanced engineering mathematics

2.2 The Constant Coefficient Case 53

Then
y(x) = e αx
c 1 e iβx + c 2 e −iβx
αx
αx
= c 1 e (cos(βx) + i sin(βx)) + c 2 e (cos(βx) − i sin(βx))
αx
αx
= (c 1 + c 2 )e cos(βx) + i(c 1 − c 2 )e sin(βx).
Here c 1 and c 2 are arbitrary real or complex numbers. If we choose c 1 = c 2 = 1/2, we obtain
αx
the particular solution e cos(βx). And if we choose c 1 = 1/2i =−c 2 , we obtain the particular
αx
solution e sin(βx). Since these solutions are linearly independent, we can write the general
solution in this complex root case as
αx
αx
y(x) = c 1 e cos(βx) + c 2 e sin(βx) (2.6)
in which c 1 and c 2 are arbitrary constants. We may also write this general solution as
αx
y(x) = e (c 1 cos(βx) + c 2 sin(βx)). (2.7)
Either of equations (2.6) or (2.7) is the preferred way of writing the general solution in Case 3,
although equation (2.5) also is correct.
We do not repeat this derivation each time we encounter Case 3. Simply write the general
solution (2.6) or (2.7), with α ± iβ the roots of the characteristic equation.

EXAMPLE 2.8
Solve y + 2y + 3y = 0. The characteristic equation is

2
λ + 2λ + 3 = 0
√ √
with complex conjugate roots −1 ± i 2. With α =−1 and β = 2, the general solution is
√ √
y = c 1 e −x cos( 2x) + c 2 e −x sin( 2x).

EXAMPLE 2.9

Solve y + 36y = 0. The characteristic equation is
2
λ + 36 = 0
with complex roots λ =±6i.Now α = 0 and β = 6, so the general solution is
y(x) = c 1 cos(6x) + c 2 sin(6x).

We are now able to solve the constant coefficient homogeneous equation

y + ay + by = 0
in all cases. Here is a summary.
Let λ 1 and λ 2 be the roots of the characteristic equation
2
λ + aλ + b = 0.
Then:
1. If λ 1 and λ 2 are real and distinct,

y(x) = c 1 e λ 1 x + c 2 e λ 2 x .

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October 14, 2010 14:12 THM/NEIL Page-53 27410_02_ch02_p43-76