Lecture Notes in Differential Equations - Bruce E. Shapiro

245
Certainly every φ(t) given by (27.150) satisfies the differential equation; we
need to show that for some set of constants C 1 , ..., C n it also satisfies the
initial conditions.
Differentiating (27.150) n − 1 times and combining the result into a matrix
equation,
⎛
⎜
⎝
φ(t 0 )
φ ′ (t 0 )
.
φ (n−1) (t 0 )
⎞ ⎛
⎟
⎠ = ⎜
⎝
y 1 (t 0 ) y 2 (t 0 ) · · · y n (t 0 )
y 1(t ′ 0 ) y 2(t ′ 0 ) y n(t ′ 0 )
.
.
y (n−1)
1 (t 0 ) y (n−1)
2 (t 0 ) · · · y n (n−1) (t 0 )
⎞ ⎛
⎟ ⎜
⎠ ⎝
(27.152)
The matrix on the right hand side of equation (27.152) is W[y 1 , ..., y n ](t 0 ).
By assumption, the determinant W [y 1 , ..., y n ](t 0 ) ≠ 0, hence the corresponding
matrix W[y 1 , ..., y n ](t 0 ) is invertible. Since W[y 1 , ..., y n ](t 0 ) is
invertible, there is a solution {C 1 , ..., C n } to the equation
given by
⎛
⎜
⎝
C 1
.
C n
⎛
⎜
⎝
y 0
.
y n
⎞
⎛
⎟
⎜
⎠ = W[y 1 , ..., y n ](t 0 ) ⎝
⎞
⎛
⎟
⎠ = {W[y 1 , .., y n ](t 0 )} −1 ⎜
⎝
C 1
.
C n
y 0
.
y n
⎞
⎞
C 1
C 2
.
C n
⎞
⎟
⎠
⎟
⎠ (27.153)
⎟
⎠ (27.154)
Hence there exists a non-trivial set of numbers {C 1 , ..., C n } such that
φ(t) = C 1 y 1 + C 2 y 2 + · · · + C n y n (27.155)
satisfies the initial value problem (27.151).
By uniqueness, every solution of this initial value problem must be identical
to (27.155), and this means that it must be a linear combination of the
{y 1 , ..., y n }.
Thus every solution of the differential equation is also a linear combination
of the {y 1 , ..., y n }, and hence {y 1 , ..., y n } must form a fundamental set of
solutions.
To prove the converse, suppose that y 1 , ..., y n are a fundamental set of solutions.
We need to show that for some number t 0 ∈ (a, b), W [y 1 , ..., y n ](t 0 ) ≠