Lecture Notes in Differential Equations - Bruce E. Shapiro

246 LESSON 27. HIGHER ORDER EQUATIONS
0. Since y 1 , ..., y n form a fundamental set of solutions, any solution to the
initial value problem (27.151) must have the form
φ(t) = C 1 y 1 + C 2 y 2 + · · · + C n y n (27.156)
for some set of constants {C 1 , ..., C n }. Hence there must exist constants
C 1 , ..., C n such that
C 1 y 1 (t 0 ) + · · · + C n y n (t 0 ) = y 0
C 1 y 1(t ′ 0 ) + · · · + C n y n(t ′ 0 ) = y 1
(27.157)
.
C 1 y (n−1)
1 (t 0 ) + · · · + C n y n (n−1) (t 0 ) = y n−1
i.e., there is a solution {C 1 , ..., C n } to
⎛
⎜
⎝
y 1 (t 0 ) · · · y n (t 0 )
⎞ ⎛
.
.
⎟ ⎜
⎠ ⎝
y (n−1)
1 (t 0 ) · · · y n (n−1) (t 0 )
This is only true if the matrix
W[y 1 , ..., y n ](t 0 ) =
⎛
⎜
⎝
C 1
.
C n
⎞
⎟
⎠ =
⎛
⎜
⎝
y 1 (t 0 ) · · · y n (t 0 )
y 0
.
y n
.
.
y (n−1)
1 (t 0 ) · · · y n (n−1) (t 0 )
⎞
⎟
⎠ (27.158)
⎞
⎟
⎠ (27.159)
is invertible, which in turn is true if and only if its determinant is nonzero.
But the determinant is the Wronskian, hence there exists a number t 0 such
that the Wronskian W [y 1 , ..., y n ](t 0 ) ≠ 0.
Theorem 27.9. Suppose y 1 , ..., y n are solutions L n y = 0 on an interval
(a, b), and let their Wronskian be denoted by W [y 1 , ..., y n ](t). Then the
following are equivalent:
1. W [y 1 , ..., y n ](t) ≠ 0 ∀t ∈ (a, b)
2. ∃t 0 ∈ (a, b) such that W [y 1 , ..., y n ](t 0 ) ≠ 0
3. y 1 , ..., y n are linearly independent functions on (a, b).
4. y 1 , ..., y n are a fundamental set of solutions to L n y = 0 on (a, b).
Example 27.10. Two solutions of the differential equation 2t 2 y ′′ + 3ty ′ −
y = 0 are y 1 = t 1/2 and y 2 = 1/t; this can be verified by substitution into