Let us find the Wronskian of the following functions and determine whether it is linearly dependent or linearly independent on ( − ∞ , ∞ ) . (-∞,∞). ( − ∞ , ∞ ) .
1. { x 2 , x + 1 , x − 3 } \{x^2, x+1, x-3\} { x 2 , x + 1 , x − 3 }
Since
W ( x 2 , x + 1 , x − 3 ) = ∣ x 2 x + 1 x − 3 2 x 1 1 2 0 0 ∣ = 2 ( x + 1 ) − 2 ( x − 3 ) = 2 x + 2 − 2 x + 6 = 8 ≠ 0 , W(x^2, x+1, x-3)=
\begin{vmatrix}
x^2 & x+1 & x-3\\
2x & 1 & 1\\
2 & 0 & 0
\end{vmatrix}
\\=2(x+1)-2(x-3)=2x+2-2x+6=8\ne 0, W ( x 2 , x + 1 , x − 3 ) = ∣ ∣ x 2 2 x 2 x + 1 1 0 x − 3 1 0 ∣ ∣ = 2 ( x + 1 ) − 2 ( x − 3 ) = 2 x + 2 − 2 x + 6 = 8 = 0 ,
we conclude that the functions are linearly independent.
2. { 3 e 2 x , e 2 x } \{3e^{2x}, e^{2x}\} { 3 e 2 x , e 2 x }
Since
W ( 3 e 2 x , e 2 x ) = ∣ 3 e 2 x e 2 x 6 e 2 x 2 e 2 x ∣ = 6 e 4 x − 6 e 4 x = 0 , W(3e^{2x}, e^{2x})=
\begin{vmatrix}
3e^{2x} & e^{2x}\\
6e^{2x} & 2e^{2x}
\end{vmatrix}
=6e^{4x}-6e^{4x}= 0, W ( 3 e 2 x , e 2 x ) = ∣ ∣ 3 e 2 x 6 e 2 x e 2 x 2 e 2 x ∣ ∣ = 6 e 4 x − 6 e 4 x = 0 ,
we conclude that the functions are linearly dependent.
3. { x 2 , x 3 , x 4 } \{x^2, x^3, x^4\} { x 2 , x 3 , x 4 }
Since
W ( x 2 , x 3 , x 4 ) = ∣ x 2 x 3 x 4 2 x 3 x 2 4 x 3 2 6 x 12 x 2 ∣ = 36 x 6 + 8 x 6 + 12 x 6 − 6 x 6 − 24 x 6 − 24 x 6 = 2 x 6 ≠ 0 , W(x^2, x^3, x^4)=
\begin{vmatrix}
x^2 & x^3 & x^4\\
2x & 3x^2 & 4x^3\\
2 & 6x & 12x^2
\end{vmatrix}
\\=36x^6+8x^6+12x^6-6x^6-24x^6-24x^6=2x^6\ne 0, W ( x 2 , x 3 , x 4 ) = ∣ ∣ x 2 2 x 2 x 3 3 x 2 6 x x 4 4 x 3 12 x 2 ∣ ∣ = 36 x 6 + 8 x 6 + 12 x 6 − 6 x 6 − 24 x 6 − 24 x 6 = 2 x 6 = 0 ,
we conclude that the functions are linearly independent.
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