Why Does the Wronskian of Two Functions Yield Zero?

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SUMMARY

The Wronskian of the functions y1(t) = t² + 5t and y2(t) = t² - 5t is calculated to determine their linear independence. Despite the functions being linearly independent, the Wronskian evaluates to zero, which is a common point of confusion. The correct calculation involves differentiating both functions and applying the determinant formula accurately. The discrepancy arises from incorrect differentiation or misapplication of the determinant formula, leading to the incorrect result of 20t instead of the expected 10t².

PREREQUISITES
  • Understanding of Wronskian determinants
  • Knowledge of differentiation techniques
  • Familiarity with linear independence of functions
  • Basic algebraic manipulation skills
NEXT STEPS
  • Review the properties of Wronskian determinants in linear algebra
  • Practice differentiation of polynomial functions
  • Explore examples of linear independence in function spaces
  • Learn about applications of the Wronskian in differential equations
USEFUL FOR

Students studying calculus, particularly those focusing on differential equations, as well as educators teaching linear algebra concepts related to function independence.

blizzard750
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y1(t) = t^(2)+5t, y2(t) = t^(2)-5t

I know that these functions are linearly independent because they are not scalar multiples, but every time that i do the Wronskian i get 0.

[t^(2) + 5t t^(2)-5t]
[2t+5 2t -5]

(2t^(3) -25t) - (2t^3-25t) = 0 my instructor some how gets 20t please tell me how
 
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Did you write that down wrong?
It should be 10t^2
(t^(2) + 5t)(t^(2) - 5t)'=(t^(2) + 5t)(2t -5)=2t^3+5t^2-25!=(2t^(3) -25t)
(t^(2) - 5t)(t^(2) + 5t)=(t^(2) - 5t)(2t +5)=2t^3-5t^2-25!=(2t^(3) -25t)
W=(2t^3+5t^2-25)-(2t^3-5t^2-25)
and one way a person would get 20t is if he/she took
(t^(2) + 5t)'=2t
 

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