Can Linearly Dependent Solutions Form a Fundamental Set?

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SUMMARY

In the context of second order linear differential homogeneous equations, if two solutions y1 and y2 are multiples of one another, they are linearly dependent and cannot form a fundamental set of solutions. The Wronskian of linearly dependent solutions is zero, confirming that if y1(t) = ay2(t) for some constant a, then the Wronskian determinant is also zero. The discussion clarifies that if U(t) is not a constant, y1 and y2 can still be considered a fundamental set of solutions, as they are not linearly dependent in that case.

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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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