Complex solutions to a differential equation a vector space?

Click For Summary

Homework Help Overview

The discussion revolves around determining whether the set of all complex solutions to a specific second-order differential equation constitutes a vector space. The equation in question is \(\frac{d^2 y}{d x^2} + 2\frac{d y}{d x} - 3 y = 0\), and participants are exploring the implications of complex coefficients in the general solution.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are examining the definition of a vector space and questioning whether the set of solutions, characterized by complex coefficients, meets the necessary axioms. There is a focus on closure under addition and scalar multiplication, as well as the nature of the "set" being discussed.

Discussion Status

The conversation is ongoing, with participants providing guidance on how to approach the problem by referencing the axioms of vector spaces. Some participants are clarifying the nature of the solutions and the implications of complex coefficients.

Contextual Notes

There is a noted ambiguity regarding the term "complex solution," with participants seeking to clarify whether it refers to the coefficients of the solutions or the solutions themselves. The original poster expresses uncertainty about identifying the set in question.

csnsc14320
Messages
57
Reaction score
1

Homework Statement



Is the set of all complex solutions to the differential equation \frac{d^2 y}{d x^2} + 2\frac{d y}{d x} - 3 y = 0

If so, find a basis, the dimension, and give the zero vector

Homework Equations





The Attempt at a Solution



I solved the equation and got the answer:

y(x) = C_1 e^{-3x} + C_2e^x

I know how to test if a set is a vector space but I'm not really seeing the "set" here. Is it because C_1 and C_2 can be complex numbers? In which case, wouldn't any complex number work so would I get the set of all complex numbers?

any help is appreciated
 
Physics news on Phys.org
i think you missed out somthing in your question -

I'm assuming it is, "is the set of all complex solutions to the differential equation - a vector space"

I would start with the axioms for a vector space - what are they?
Then what is the general form of your solution? This will generally have some undetermined constants to give a family of solutions

"the space" is then the set of all solutions. Is it a vector space?

In short, a vector space is closed under scalar multiplication and addition, with some other axioms, so check:
closure under scalar multiplication - so given any solution is a scalar times the solution also a solution
closure under addition - so given 2 solutions is their sum also a solution

then fill out the other axioms
 
Last edited:
updated post above
 
csnsc14320 said:

Homework Statement



Is the set of all complex solutions to the differential equation \frac{d^2 y}{d x^2} + 2\frac{d y}{d x} - 3 y = 0

If so, find a basis, the dimension, and give the zero vector

Homework Equations





The Attempt at a Solution



I solved the equation and got the answer:

y(x) = C_1 e^{-3x} + C_2e^x

I know how to test if a set is a vector space but I'm not really seeing the "set" here. Is it because C_1 and C_2 can be complex numbers? In which case, wouldn't any complex number work so would I get the set of all complex numbers?

any help is appreciated
First, exactly what do you mean by "complex solution"? If you mean simply that the coefficients C1 and C2 are complex, as you say, they the "set" asked about is NOT "the set of all complex numbers". It is the set of all such functions:
\{ f(x)= C_1 e^{-3x}+ C_2 e^x : C_1, C_2 \in \math{C}\}.

If you add two such functions is the sum also a function of that kind? If you multiply such a function by a complex number is the product also a function of that kind?
 

Similar threads

  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
1K