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 
1
Expert's answer
2021-12-03T11:06:13-0500

1.

{lnx,lnx2}\{\ln x, \ln x^2\}

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

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

W(lnx,lnx2)=lnxlnx21/x2/xW(\ln x, \ln x^2)=\begin{vmatrix} \ln x & \ln x^2 \\ 1/x & 2/x \end{vmatrix}

=(2/x)lnx(1/x)lnx2=(2/x)\ln x-(1/x)\ln x^2

=(2/x)lnx(2/x)lnx=0=(2/x)\ln x-(2/x)\ln x=0

lnx,lnx2\ln x, \ln x^2 are linearly dependent on (0,).(0, \infin).


2.


{2+x,1x,3+x2}\{2+x, 1-x, 3+x^2\}

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

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

W(2+x,1x,3+x2)=2+x1x3+x2112x002W(2+x, 1-x,3+x^2)=\begin{vmatrix} 2+x & 1-x & 3+x^2 \\ 1 & -1 & 2x \\ 0 & 0 & 2 \end{vmatrix}


=22+x1x11=2(2x1+x)=2\begin{vmatrix} 2+x & 1-x \\ 1 & -1 \end{vmatrix}=2(-2-x-1+x)

=60=-6\not=0

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



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