Lecture Notes in Differential Equations - Bruce E. Shapiro

137
Hence both r = 1 and r = 3. This tells us each of the following functions
are solutions of (16.14):
y = e t (16.20)
y = e 3t (16.21)
We will see shortly how to combine these to get a more general solution.
We can generalize the last example as follows.
Theorem 16.4. If r is a root of the characteristic equation of Ly = 0,
then e rt is a solution of Ly = 0.
Proof.
L[e rt ] = a(e rt ) ′′ + b(e rt ) ′ + c(e rt ) (16.22)
Since r is a root of the characteristic equation,
Hence
Thus y = e rt is a solution of Ly = 0.
= ar 2 e rt + bre rt + ce rt (16.23)
= (ar 2 + br + c)e rt (16.24)
ar 2 + br + c = 0 (16.25)
L[e rt ] = 0 (16.26)
Theorem 16.5. If the characteristic polynomial has a repeated root r,
then y = te rt is a solution of Ly = 0.
Proof.
L(te rt ) = a(te rt ) ′′ + b(te rt ) ′ + c(te rt ) (16.27)
= a(e rt + rte rt ) ′ + b(e rt + rte rt ) + cte rt (16.28)
= a(2re rt + r 2 te rt ) + b(e rt + rte rt ) + cte rt (16.29)
= e rt (2ar + b + (ar 2 + br + c)t) (16.30)
Since r is a root, r 2 + br + c = 0. Hence
Since r is root, from the quadratic equation,
L(te rt ) = e rt (2ar + b) (16.31)
r = −b ± √ b 2 − 4ac
2a
(16.32)