Question #282304

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


{x2, x+1, x-3} ans, W=8, linearly independent

{3e2x, e2x} ans, W=0, linearly dependent

{x2, x3, x4} ans, W=2x^6, linearly independent


1
Expert's answer
2021-12-30T11:54:26-0500

1.

W(x2, x+1, x3)=det(x2x+1x32x11200)=2(x+1(x3))=24=80,W(x^2,~x+1,~x-3)=det \begin{pmatrix} x^2 & x+1 & x-3\\ 2x & 1&1\\ 2&0&0 \end{pmatrix}=2\cdot(x+1-(x-3))=2\cdot4=8\not =0, linearly independent.

2.

W(3e2x, e2x)=det(3e2xe2x6e2x2e2x)=6e4x6e4x=0,W(3e^{2x},~e^{2x})=det \begin{pmatrix} 3e^{2x} & e^{2x} \\ 6e^{2x}& 2e^{2x} \end{pmatrix}=6e^{4x}- 6e^{4x}=0, linearly dependent.

3.

W(x2, x3, x4)=det(x2x3x42x3x24x326x12x2)=x2(36x424x4)x3(24x38x3)+x4(12x26x2)=12x616x6+6x6=2x60,W(x^2,~x^3,~x^4)=det \begin{pmatrix} x^2& x^3&x^4\\ 2x& 3x^2&4x^3\\ 2&6x&12x^2 \end{pmatrix}=x^2(36x^4-24x^4)-x^3(24x^3-8x^3)+x^4(12x^2-6x^2)=12x^6-16x^6+6x^6=2x^6\not=0, linearly independent.


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