Lecture Notes in Differential Equations - Bruce E. Shapiro

250 LESSON 27. HIGHER ORDER EQUATIONS
∣ ∣∣∣∣∣∣∣∣∣∣∣∣∣ y 1 · · · y n
y 1
′ y ′ n
dW
dx = .
.
y (n−2)
1 y n
(n−2)
− 1
n−1
∑
a i (t)y (i)
1 · · · − 1
n−1
∑
a i (t)y m
(i)
a n (t)
a n (t)
∣
i=0
i=0
(27.182)
The value of a determinant is unchanged if we add a multiple of one to another.
So multiply the first row bya 0 (t)/a n (t), the second row bya 2 (t)/a n (t),
etc., and add them all to the last row to obtain
dW
dt
=
∣
y 1 · · · y n
y 1
′ y n
′
.
.
y (n−2)
1 y n
(n−2)
−p(t)y (n−1)
1 · · · −p(t)y n
(n−1)
∣
(27.183)
where p(t) = a n−1 (t)/a n (t). We can factor a constant out of every element
of a single row of the determinant if we multiply the resulting (factored)
determinant by the same constant.
dW
dt
= −p(t)
∣
y 1 · · · y n
y 1
′ y n
′
.
.
y (n−2)
1 y n
(n−2)
y (n−1)
1 · · · y n
(n−1)
= −p(t)W (t) (27.184)
∣
Integrating this differential equation achieves the desired formula for W .
Example 27.11. Calculate the Wronskian of y ′′ − 9y = 0.
Since p(t) = 0, Abel’s formula gives W = Ce − ∫ 0·dt = C.
Example 27.12. Use Abel’s formula to compute the Wronskian of y ′′′ −
2y ′′ − y ′ − 3y = 0
This equation has p(t) = −2, and therefore W = Ce − ∫ (−2)dt = Ce 2t .
Example 27.13. Compute the Wronskian of x 2 y (??) +xy (??) +y ′′ −4x = 0.
We have p(t) = t/t 2 = 1/t. Therefore W = Ce − ∫ (1/t)dt = Ce − ln t = C/t.