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# Prove some facts about solutions of y′′ + y = 0

1. Prove that there is exactly one solution of the second-order differential equation

whose graph passes through the points and where , where .

2. Is part (a) ever true if for ?
3. Generalize part (a) for the second-order differential equation

Include the case .

1. Proof. The equation is of the form

Therefore, using Theorem 8.7 (pages 326-327 of Apostol) we have and . Since we have solutions given by

The conditions in the problem tell us and . From the first condition we have

Using this expression for and the second condition we have

where we use that so . Thus, and are uniquely determined, so the solution is unique

2. No, if , then and we find the choice of is arbitrary.
3. If , then and so the first condition implies

Then, the second condition implies

Thus, is uniquely determined.

If , then implies . Therefore,

and

Thus, and are uniquely determined as long as .

1. Anonymous says:

In a) it seems you forgot to multiply by at the end.

• Anonymous says:

Yeah forgot the closing bracket. b\frac{\cos(a_2)}{\cos(a_1)}. Sorry for spam.

2. Artem says:

I think there is a mistake in Apostol in (b). Actually, there are 2 distinct solutions. In case n = 1 mod 2, then we can write , and we get a system where you simply need to set , in the case n = 1 mod 2. If you do the same with n = 2 mod 2, then setting you get another solution.

• S says:

I suppose the question asks in case the bs of the points are arbitrary.

• S says:

Plus, as RoRi wrote, the solution, if exists in case bs satisfy the needed condition, is not unique.

3. Kaitei says:

Rori, I think this type of exercise is better solved with a matrix and linear equation systems solutions analysis. The solution is much more clean, and much more easy to visualize the necessary conditions to the existence of an unique solution.
Just let A be the matrix with the coefficients of c1 and c2. If det A != 0, then there is an unique solution of the linear equation system.