- #1

- 16

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During the derivation we assume that: u

_{1}'y

_{1}

^{(k)}+ u

_{2}'y

_{2}

^{(k)}+ . . . + u

_{n}'y

_{n}

^{(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 u

_{i}, 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: u

_{1}'y

_{1}

^{(k)}+ u

_{2}'y

_{2}

^{(k)}+ . . . + u

_{n}'y

_{n}

^{(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?