Lecture Notes in Differential Equations - Bruce E. Shapiro

136 LESSON 16. LINEAR EQS. W/ CONST. COEFFICENTS
Example 16.1. Show that L is a linear operator.
Recall from definition (15.3) we need to show the following to demonstrate
that L is linear
L(αy + βz) = αLy + βLz (16.6)
where α and β are constants and y(t) and z(t) are functions. But
L(αy + βz) = (aD 2 + bD + c)(αy + βz) (16.7)
= aD 2 (αy + βz) + bD(αy + βz) + c(αy + βz) (16.8)
= a(αy + βz) ′′ + b(αy + βz) ′ + c(αy + βz) (16.9)
= a(αy ′′ + βz ′′ ) + b(αy ′ + βz ′ ) + c(αy + βz) (16.10)
= α(ay ′′ + by ′ + cy) + β(az ′′ + bz ′ + cz) (16.11)
= αLy + βLz (16.12)
Definition 16.3. The characteristic polynomial corresponding to characteristic
polynomial (16.1) is
ar 2 + br + c = 0 (16.13)
Equation (16.13) is also called the characteristic equation of the differential
equation.
The following example illustrates why the characteristic equation is useful.
Example 16.2. For what values of r does y = e rt satisfy the differential
equation
y ′′ − 4y ′ + 3y = 0 (16.14)
Differentiating y = e rt ,
Plugging both expressions into (16.14),
y ′ = re rt (16.15)
y ′′ = r 2 e rt (16.16)
r 2 e rt − 4re rt + 3e r t = 0 (16.17)
Since e rt can never equal zero we can cancel it out of every term,
r 2 − 4r + 3 = 0 (16.18)
Equation (16.18) is the characteristic equation of (16.14). Factoring it,
(r − 3)(r − 1) = 0 (16.19)