Question on assumptions made during variation of parameters

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I was recently trying to prove the variation of parameters formula for an nth degree equation, and I have come up with a question about the assumptions made during the derivation.

During the derivation we assume that: u1'y1(k) + u2'y2(k) + . . . + un'yn(k) = 0
for k < n-1.

It leads to the matrix form: WU' = X, where W is the Wronskian, U' is a column vector consisting of the derivatives of each ui, and X is the solution vector that has f(t) in the nth row, and 0 in all the others.


My question is on the assumption: u1'y1(k) + u2'y2(k) + . . . + un'yn(k) = 0, k < n-1.

How would this the results of our solutions change if we instead assumed the left hand sum equaled some constant? It would have the same effect when the derivative is taken at each step, as the constant would go to 0 consistently.

I see it would lead to different values for u, as the solution to the system would be different. Would the answers simplify to the same values as assuming the sums are 0? Would it lead to different answers, and then be forced correct when applying initial conditions? Any thoughts?
 

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HallsofIvy
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I was recently trying to prove the variation of parameters formula for an nth degree equation, and I have come up with a question about the assumptions made during the derivation.

During the derivation we assume that: u1'y1(k) + u2'y2(k) + . . . + un'yn(k) = 0
for k < n-1.

It leads to the matrix form: WU' = X, where W is the Wronskian, U' is a column vector consisting of the derivatives of each ui, and X is the solution vector that has f(t) in the nth row, and 0 in all the others.


My question is on the assumption: u1'y1(k) + u2'y2(k) + . . . + un'yn(k) = 0, k < n-1.

How would this the results of our solutions change if we instead assumed the left hand sum equaled some constant? It would have the same effect when the derivative is taken at each step, as the constant would go to 0 consistently.

I see it would lead to different values for u, as the solution to the system would be different. Would the answers simplify to the same values as assuming the sums are 0? Would it lead to different answers, and then be forced correct when applying initial conditions? Any thoughts?
It would make the equations more difficult to solve! The point is that there are an infinite number of solutions to the original differential equation and we are looking for one. We can make pretty nearly any simplifying assumptions we want. That just reduces the set of "possible" solutions we are looking for.
 

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