Question #274739

Find the Wronskian of the following functions and determine whether it is linearly dependent or linearly independent on (-∞,∞).

  1. {x2, x+1, x-3}                ans, W=8, linearly independent
  2. {3e2x, e2x}                     ans, W=0, linearly dependent
  3. {x2, x3, x4}                    ans, W=2x^6, linearly independent
1
Expert's answer
2021-12-23T18:08:15-0500

1.



W(y1,y2,y3,x)=x2x+1x32x11200W(y_1, y_2, y_3, x)=\begin{vmatrix} x^2 & x+1 & x-3 \\ 2x & 1 & 1 \\ 2 & 0 & 0 \end{vmatrix}=2x+1x311=2(x+1x+3)=80=2\begin{vmatrix} x+1 & x-3 \\ 1 & 1 \end{vmatrix}=2(x+1-x+3)=8\not=0

Therefore, the set {x2,x+1,x3}\{x^2, x+1, x-3\} is linearly independent on (,).(-\infin, \infin).


2.



W(y1,y2,x)=3e2xe2x6e2x2e2xW(y_1, y_2, x)=\begin{vmatrix} 3e^{2x} & e^{2x} \\ 6e^{2x} & 2e^{2x} \end{vmatrix}=6e2x6e2x=0=6e^{2x}-6e^{2x}=0

Therefore, the set {3e2x,e2x}\{3e^{2x}, e^{2x}\} is linearly dependent on (,).(-\infin, \infin).


3.



W(y1,y2,y3,x)=x2x3x42x3x24x326x12x2W(y_1, y_2, y_3, x)=\begin{vmatrix} x^2 & x^3 & x^4 \\ 2x & 3x^2 & 4x^3 \\ 2 & 6x & 12x^2 \end{vmatrix}=x2(36x424x4)x3(24x38x3)=x^2(36x^4-24x^4)-x^3(24x^3-8x^3)+x4(12x26x2)=12x616x6+6x6+x^4(12x^2-6x^2)=12x^6-16x^6+6x^6=2x60,except at  x=0=2x^6\not=0, except\ at\ \ x=0

Therefore, the set {x2,x3,x4}\{x^2, x^3, x^4\} is linearly independent on (,).(-\infin, \infin).

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