- Prove that there is exactly one solution of the second-order differential equation
whose graph passes through the points and where , where .
- Is part (a) ever true if for ?
- Generalize part (a) for the second-order differential equation
Include the case .
- 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
- No, if , then and we find the choice of is arbitrary.
- 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 .
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.
Also, thanks for your amazing work. It is really helpful :)
Yeah, I totally agree it is better, but at this point in the textbook we don’t know what a matrix is. I tried to limit myself to the tools we have available since that’s presumably what Apostol intended. I do think I mention in a few places that if you know linear algebra then using matrices for this sort of thing is faster/easier/better. In Chapters 6 & 7 of Volume 2 of Apostol (iirc) we’ll do a ton of solving these types of systems using linear algebra techniques.
Thanks for reading!