Lecture Notes in Differential Equations - Bruce E. Shapiro

206 LESSON 24. VARIATION OF PARAMETERS
If we now make the totally arbitrary assumption that
then
and therefore
From equation (24.4)
u ′ y 1 + v ′ y 2 = 0 (24.6)
y ′ P = uy ′ 1 + vy ′ 2 (24.7)
y ′′
P = u ′ y ′ 1 + uy ′′
1 + v ′ y ′ 2 + vy ′′
2 (24.8)
f(t) = a(t)(u ′ y ′ 1 + uy ′′
1 + v ′ y ′ 2 + vy ′′
2 )
+ b(t) (uy ′ 1 + vy ′ 2) + c(t) (uy 1 + vy 2 ) (24.9)
= a(t) (u ′ y ′ 1 + v ′ y ′ 2) + u [a(t)y ′′
1 + b(t)y ′ 1 + c(t)y 1 ]
+ v[a(t)y ′′
2 + b(t)y ′ 2 + c(t)y 2 ] (24.10)
= a(t)(u ′ y ′ 1 + v ′ y ′ 2) (24.11)
Combining equations (24.6) and (24.11) in matrix form
(
y1 y 2
) ( ) (
u
′
v ′ =
0
f(t)/a(t)
)
(24.12)
y 1 ′ y 2
′
The matrix on the left hand side of equation (24.12) is the Wronskian,
which we know is nonsingular, and hence invertible, because y 1 and y 2
form a fundamental set of solutions to a differential equation. Hence
( u
′
v ′ )
=
=
=
where W (t) = y 1 y ′ 2 − y 2 y ′ 1. Hence
( ) −1 ( )
y1 y 2
0
y 1 ′ (
y 2
′ f(t)/a(t)
) ( )
1 y
′
2 −y 2 0
W (t) ( −y 1 ′ y 1 ) f(t)/a(t)
1 −y2 f(t)/a(t)
W (t) y 1 f(t)/a(t)
du
dt = −y 2f(t)
a(t)W (t)
dv
dt = y 1f(t)
a(t)W (t)
(24.13)
(24.14)
(24.15)
Integrating each of these equations,
∫
y2 (t)f(t)
u(t) = −
dt (24.16)
a(t)W (t)
∫
y1 (t)f(t)
v(t) =
dt (24.17)
a(t)W (t)