Can Linearly Dependent Solutions Form a Fundamental Set?

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Discussion Overview

The discussion centers on the conditions under which linearly dependent solutions of a second order linear differential homogeneous equation can form a fundamental set of solutions. Participants explore the implications of linear dependence, the Wronskian determinant, and specific cases of solution relationships.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant asserts that if two solutions, y1 and y2, are multiples of one another, they are linearly dependent and cannot form a fundamental set of solutions.
  • Another participant agrees that linearly dependent solutions imply that the Wronskian is zero, providing a mathematical justification for this claim.
  • A different viewpoint is introduced, questioning whether a solution of the form y2 = U(t)y1 could still be considered a fundamental solution, suggesting that the Wronskian might not be zero in this case.
  • One participant responds by clarifying that if U(t) is not a constant, then y1 and y2 are not linearly dependent.

Areas of Agreement / Disagreement

Participants generally agree that linearly dependent solutions cannot form a fundamental set, but there is disagreement regarding specific cases, particularly concerning the relationship between y1 and y2 when one is expressed as a function of the other.

Contextual Notes

The discussion includes assumptions about the nature of U(t) and its implications for linear dependence, which remain unresolved. The relationship between the Wronskian and the linear dependence of solutions is also a point of contention.

d_b
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For a second order linear differential homogeneous equation, if the two solution y1 and y2 is a multiple of one another. It means that it is linearly dependent which mean they can not form a fundamental set of solutions to second order differential homogeneous equation.

Am I correct?? or could it be any cases where y1 and y2 is a mulitple of one another and still can form a fundamental set of solutions.

Also if y1 and y2 are L.D is that mean wronskian equals zero?
 
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You are correct, they cannot be multiples, they cannot be linearly dependent, as that means they are the same solution. (u=au1+bu1=cu1.)
 
And yes, if y1 and y2 are multiples of each other, then their Wronskian is equal to 0:
Specifically, if y1(t)= ay2(t) for some number a, then it is also true that y1'= ay2' so the Wronskian is
\left|\begin{array}{cc}y1 & y2 \\ y1' & y2'\end{array}\right|= \left|\begin{array}{cc} y1 & ay1 \\ y1' & ay1'\end{array}\right|= a(y1)(y1')- a(y1)(y1')= 0
 
what about y2=U(t)y1 ? Isn't y1 and y2 is the set of fundamental solution?? why is wronskian is not equal to zero then?
 
those are not linearly dependent.If U was a constant instead of U(t), they would be.
 

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