Question #274737

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

  1. {ln x, ln x2}                   ans, W=0, linearly dependent
  2. {2+x, 1-x, 3+x2}           ans, W=-6, linearly independent 

Expert's answer

1.

{ln⁡x,ln⁡x2}\{\ln x, \ln x^2\}

{ln⁡x,2ln⁡x},x>0\{\ln x, 2\ln x\}, x>0

(ln⁡x)′=1/x,(ln⁡(x2))′=2/x(\ln x)'=1/x, (\ln(x^2))'=2/x

W(ln⁡x,ln⁡x2)=∣ln⁡xln⁡x21/x2/x∣W(\ln x, \ln x^2)=\begin{vmatrix} \ln x & \ln x^2 \\ 1/x & 2/x \end{vmatrix}

=(2/x)ln⁡x−(1/x)ln⁡x2=(2/x)\ln x-(1/x)\ln x^2

=(2/x)ln⁡x−(2/x)ln⁡x=0=(2/x)\ln x-(2/x)\ln x=0

ln⁡x,ln⁡x2\ln x, \ln x^2 are linearly dependent on (0,∞).(0, \infin).


2.


{2+x,1−x,3+x2}\{2+x, 1-x, 3+x^2\}

(2+x)′=1,(1−x)′=−1,(3+x2)′=2x(2+x)'=1,(1-x)'=-1, (3+x^2)'=2x

(2+x)′′=0,(1−x)′′=0,(3+x2)′′=2(2+x)''=0,(1-x)''=0, (3+x^2)''=2

W(2+x,1−x,3+x2)=∣2+x1−x3+x21−12x002∣W(2+x, 1-x,3+x^2)=\begin{vmatrix} 2+x & 1-x & 3+x^2 \\ 1 & -1 & 2x \\ 0 & 0 & 2 \end{vmatrix}


=2∣2+x1−x1−1∣=2(−2−x−1+x)=2\begin{vmatrix} 2+x & 1-x \\ 1 & -1 \end{vmatrix}=2(-2-x-1+x)

=−6≠0=-6\not=0

2+x,1−x,3+x22+x, 1-x, 3+x^2 are linearly independent on (−∞,∞).(-\infin, \infin).



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