Question

I Do We Have the Complete Solution Set? A differential operator in R[D] has order n can be written out in the form o(n-1) with the last coefficient cn (at least) not equal to zero. The key to determining the dimension of these solution spaces is the following existence and uniqueness theorem for initial value problems. 'So it can be efficiently described by giving a basis. ethciently described by giving a basis Theorem 1 (Existence and Unique ness Theorem for IVPs) İfL is a linear ditferential operator in R[D] of order n, then for any reference point a and any initial values bo, bi..bn- in IR, the IVP y(a) bi = L(y) = 0 with initial conditions has one and only one solution (). You should take this for granted and use it as needed. Now let K be the vector space of all solutions to L()0, choose your favorite reference point a, and define a function y(a) r K-R" by the rule r()- yn-(a) Problem (a) Show that your function r K-IR" is linear (b) Using Theorem 1,show that r has an inverse function s : R K Problem I shows if L is of degree n, then the solution space K for the equation l(y) = 0 is n-dimensional. As discussed earlier, it follows that any collection of n solutions บา.y2, . . . ,Un that are linearly independent is a basis for the solution space to L()0

Answer 1

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