Lecture Notes in Differential Equations - Bruce E. Shapiro

149
where I is the identity matrix.
Since the inverse of the derivative is the integral, then for any function f(t),
∫
D −1 f(t) = f(t)dt + C (17.54)
Hence
(∫
DD −1 f(t) = D
)
f(t) + C = f(t) (17.55)
and
∫
D −1 Df(t) =
Df(t)dt + C = f(t) + C (17.56)
We write the general linear first order initial value problem in terms of the
linear differential operator D = d/dt as
From the basic properties of derivatives,
[D + p(t)] y = q(t), y(t 0 ) = y 0 (17.57)
De f(t) = d dt ef(t) = e f(t) d dt f(t) = ef(t) Df(t) (17.58)
From the product rule,
Df(t)g(g) = f(t)Dg(t) + g(t)Df(t) (17.59)
Hence
D(e f(t) y) = e f(t) Dy + yDe f(t) (17.60)
= e f(t) Dy + ye f(t) Df(t) (17.61)
= e f(t) (Dy + yDf(t)) (17.62)
If we let
∫
f(t) =
p(t)dt = D −1 p(t)dt (17.63)
then (using our previous definition of µ),
(∫
e f(t) = exp
)
p(t)dt = µ(t) (17.64)