Lecture Notes in Differential Equations - Bruce E. Shapiro

172 LESSON 20. THE WRONSKIAN
is also a solution of the homogeneous ODE for all values of A and B. To
see if it is possible to find a set of solutions that satisfy the initial condition,
(20.4) requires that
y(t 0 ) = Ay 1 (t 0 ) + By 2 (t 0 ) = y 0 (20.6)
y ′ (t 0 ) = Ay ′ 1(t 0 ) + By ′ 2(t 0 ) = y ′ 0 (20.7)
If we multiply the first equation by y ′ 2(t 0 ) and the second equation by y 2 (t 0 )
we get
Ay 1 (t 0 )y ′ 2(t 0 ) + By 2 (t 0 )y ′ 2(t 0 ) = y 0 y ′ 2(t 0 ) (20.8)
Ay ′ 1(t 0 )y 2 (t 0 ) + By ′ 2(t 0 )y 2 (t 0 ) = y ′ 0y 2 (t 0 ) (20.9)
Since the second term in each equation is identical, it disappears when we
subtract the second equation from the first:
Ay 1 (t 0 )y ′ 2(t 0 ) − Ay ′ 1(t 0 )y 2 (t 0 ) = y 0 y ′ 2(t 0 ) − y ′ 0y 2 (t 0 ) (20.10)
Solving for A gives
A =
y 0 y ′ 2(t 0 ) − y ′ 0y 2 (t 0 )
y 1 (t 0 )y ′ 2 (t 0) − y ′ 1 (t 0)y 2 (t 0 )
(20.11)
To get an equation for B we instead multiply (20.6) by y ′ 1(t 0 ) and (20.7)
by y 1 (t 0 ) to give
Ay 1 (t 0 )y ′ 1(t 0 ) + By 2 (t 0 )y ′ 1(t 0 ) = y 0 y ′ 1(t 0 ) (20.12)
Ay ′ 1(t 0 )y 1 (t 0 ) + By ′ 2(t 0 )y 1 (t 0 ) = y ′ 0y 1 (t 0 ) (20.13)
Now the coefficients of A are identical, so when we subtract we get
Solving for B,
By 2 (t 0 )y ′ 1(t 0 ) − By ′ 2(t 0 )y 1 (t 0 ) = y 0 y ′ 1(t 0 ) − y ′ 0y 1 (t 0 ) (20.14)
B =
y 0 y ′ 1(t 0 ) − y ′ 0y 1 (t 0 )
y 2 (t 0 )y ′ 1 (t 0) − y ′ 2 (t 0)y 1 (t 0 )
(20.15)
If we define the Wronskian Determinant of any two functions y 1 and y 2
as
W (t) =
∣ y 1(t) y 2 (t)
y 1(t)
′ y 2(t)
′ ∣ = y 1(t)y 2(t) ′ − y 2 (t)y 1(t) ′ (20.16)
then
A = 1
W (t 0 )
∣ y 0 y 2 (t 0 )
y 0 ′ y 2(t ′ 0 ) ∣ , B = 1
W (t 0 ) ∣ y 1(t 0 ) y 0
y 1(t ′ 0 ) y 0
′ ∣ (20.17)
Thus so long as the Wronskian is non-zero we can solve for A and
B.