The Wronskian of a Set of Three Functions
The Wronskian of three functions is given by:
y1
y2
y3
W(y1, y2 , y3 ; t) = y1 ′
y2 ′
y3 ′ .
y1 ′′
y 2 ′′
y3 ′′
For the Cauchy-Euler differential equation: t 3 y ′′′− 3t 2 y ′′+ 6 t y ′− 6 y = 0 ,
three linearly independent solutions are: y = t, t 2 and t 3 .
 y = t + t 2 + 2 t3
 1

Therefore,  y 2 = 3t + t 2 − t 3

2
3
 y3 = 11t + 7 t + 8 t



 are also solutions,


but are they linearly independent or linearly dependent? Their Wronskian is:
t + t 2 + 2 t3
W (t) = 1 + 2 t + 6 t 2
2 + 12 t
Substitute t = 1: W (1) =
3t + t 2 − t 3
11t + 7 t 2 + 8 t 3
3 + 2 t − 3t 2 11 + 14 t + 24 t 2 .
2 − 6t
14 + 48 t
4
3 26
9
2 49 = 0 .
14 − 4 6 2
Therefore, the set
{ y1, y2 , y3 }
is linearly dependent.
We reduce the Wronskian matrix (at t = 1) to reduced-row-echelon-form:
3 26   1 0 5 
 4
 9
2 49  ~  0 1 2  .

 14 − 4 6 2   0 0 0 
This states that: column3 = 5 column1 + 2 column2.
The set of our three solutions will satisfy this same linear dependence.
Therefore, y3 = 5 y1 + 2 y 2 :
(11t + 7 t 2 + 8 t3 ) = 5( t + t 2 + 2 t3 ) + 2 ( 3t + t 2 − t3 ) .