Determining whether two functions are linear independent via wronskianby ysebastien Tags: determining, functions, independent, linear, wronskian 

#1
Mar412, 12:59 PM

P: 6

1. The problem statement, all variables and given/known data
Determine which of the following pairs of functions are linearly independent. (a) [itex]f(t)=3t,g(t)=t[/itex] (b) [itex]f(x)=x^{2},g(x)=4x^{2}[/itex] 2. Relevant equations the Wronskian is defined as, W=Det{{f(u),g(u)},{f'(u),g'(u)}} if {f(u),g(u)} are linearly dependent, W=0 if W=/=0, {f(u),g(u)} are linearly independent 3. The attempt at a solution The assumed interval for the independent variables t,x are [itex] x,t \in (\infty,\infty) [/itex] for (a), I determined [itex] W(t)=3t3t [/itex], which for x>0 is [itex] W(t)=3t3t=0 [/itex]. for x<0 we have [itex] W(t)=3t3t=6t [/itex]. Since for some value of [itex] t \in (\infty,\infty)[/itex] I found a [itex]W(t) \neq 0[/itex] I can conclude that the functions f(t) and g(t) are linearly independent. Now for (b), Similarly to (a), I find a [itex] W(x)=8x^{2}x8x^{2}x[/itex] for [itex] x>0 : W(x)=8x^{3}8x^{3}=0[/itex] for [itex] x<0 : W(x)=8(x)^{2}x8x^{2}(x)=8x^{3}+8x^{3}=16x^{3}\neq 0[/itex] Similarly I conclude that f(x) and g(x) are linearly independent since I found values of x which make the wronskian not equal to 0. However, while my conclusion is correct for (a), (b) is supposedly linearly dependent. Is my method correct? if so what mistake did I make in concluding that the functions of (b) were linearly independent? Thanks 



#2
Mar412, 01:36 PM

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x^2 is equal to x^2.




#3
Mar412, 01:43 PM

P: 6

Yes,
I use that fact, but I still find that for [itex] x<0 : W(x)=8(x)^{2}x8(x)^{3}=8x^{3}+8x^{3}=16x^{3}[/itex] Am I just completely missing something here? 



#4
Mar412, 03:53 PM

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Determining whether two functions are linear independent via wronskian
Your calculation of [itex]W(x)=8x^{2}x8x^{2}x[/itex] is wrong. Just use that x^2=x^2 from the beginning, so g(x)=4x^2.




#5
Mar412, 03:57 PM

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P: 7,226

The specific place where your calculations are incorrect is$$
\frac d {dx}x^2 = 2x$$is false. 



#6
Mar412, 03:57 PM

P: 6

Ah, I see. Thank you



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