Lecture Notes in Differential Equations - Bruce E. Shapiro

244 LESSON 27. HIGHER ORDER EQUATIONS
the determinant will be zero. Thus the Wronskian of a linearly dependent
set of functions will always be zero. In fact, as we show in the following
theorems, the Wronskian will be nonzero if and only if the functions form
a complete set of solutions to the same differential equation.
Example 27.8. Find the Wronskian of y ′′′ − 4y ′ = 0.
The characteristic equation is 0 = r 3 − 4r = r(r 2 − 4) = r(r − 2)(r + 2) and
a fundamental set of solutions are y 1 = 1, y 2 = e 2t , and y 3 = e −2t . Their
Wronskian is
y 1 y 2 y 3
W (t) =
y 1 ′ y 2 ′ y 3
′ (27.144)
∣y 1 ′′ y 2 ′′ y 3
′′ ∣
∣ 1 e 2t e −2t ∣∣∣∣∣
=
0 2e 2t −2e −2t
(27.145)
∣0 4e 2t 4e −2t ∣ =
∣ 2e2t −2e −2t ∣∣∣
4e 2t 4e −2t = 16. (27.146)
Example 27.9. Find the Wronskian of y ′′′ − 8y ′′ + 16y ′ = 0.
The characteristic equation is 0 = r 3 −8r 2 +16r = r(r 2 −8r+16) = r(r−4) 2 ,
so a fundamental set of solutions is y 1 = 1, y 2 = e 4t and y 3 = te 4t . Therefore
1 e 4t te 4t
W (t) =
0 4e 4t e 4t (1 + 4t)
(27.147)
∣0 16e 4t 8e 4t (1 + 2t) ∣
= (4e 4t )[8e 4t (1 + 2t)] − [e 4t (1 + 4t)](16e 4t ) (27.148)
= 16e 8t (27.149)
Theorem 27.8. Suppose that y 1 , y 2 , ..., y n all satisfy the same higher
order linear homogeneous differential equation L n y = 0 on (a, b) Then
y 1 , y 2 , ..., y n form a fundamental set of solutions if and only if for some
t 0 ∈ (a, b), W [y 1 , ..., y n ](t 0 ) ≠ 0.
Proof. Let y 1 , ..., y n be solutions to L n y = 0, and suppose that W [y 1 , ..., y n ](t 0 ) ≠
0 for some number t 0 ∈ (a, b).
We need to show that y 1 , ..., y n form a fundamental set of solutions. This
means proving that any solution to L n y = 0 has the form
Consider the initial value problem
φ(t) = C 1 y 1 + C 2 y 2 + · · · + C n y n (27.150)
L n y = 0, y(t 0 ) = y 0 , ..., y (n−1) (t 0 ) = y n (27.151)