Lecture Notes in Differential Equations - Bruce E. Shapiro

189
where r 1 and r 2 are the roots of the characteristic polynomial
ar 2 + br + c = 0 (22.15)
Proof. Using the quadratic equation, the roots of the characteristic polynomial
satisfy
r 1 + r 2 = −b + √ b 2 − 4ac
+ −b − √ b 2 − 4ac
= − b (22.16)
(
2a
2a
a
−b + √ ) (
b
r 1 r 2 =
2 − 4ac −b − √ )
b 2 − 4ac
= b2 − (b 2 − 4ac)
2a
2a
4a 2 = c a
(22.17)
hence
b = −a(r 1 + r 2 ) (22.18)
c = ar 1 r 2 (22.19)
Thus
Ly = ay ′′ + by ′ + cy
= ay ′′ − a(r 1 + r 2 )y ′ + ar 1 r 2 y
= a[y ′′ − (r 1 + r 2 )y ′ + r 1 r 2 y]
= a(D − r 1 )(y ′ − r 2 y)
= a(D − r 1 )(D − r 2 )y
(22.20)
Hence L = a(D − r 1 )(D − r 2 ) which is the desired factorization.
Theorem (22.4) provides us with the following simple algorithm for solving
any second order linear differential equation or initial value problem with
constant coefficients.