What is the constant a for the solution y(x) = e^{ax} in the equation x = x?

  • Thread starter Thread starter chocok
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around finding the constant \( a \) in the function \( y(x) = e^{ax} \) as a solution to the differential equation \( x'' = x \). Participants are exploring the relationship between the function and the equation, questioning the relevance of the proposed solution.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants have attempted different reasoning methods, including examining the implications of the second derivative and the nature of the variable involved. Some question the assumptions made regarding the variable of differentiation and the relevance of the solution form.

Discussion Status

The discussion is ongoing, with participants providing various perspectives on the problem. Some have offered insights into the nature of the equation and the solution, while others express confusion and seek clarification on the assumptions and reasoning presented.

Contextual Notes

There is uncertainty regarding the variable with respect to which the second derivative is taken, as well as the implications of the solution being part of a larger question. Participants are also considering the condition that \( a \) cannot be zero and whether this is necessary for their reasoning.

chocok
Messages
21
Reaction score
0
Question:
I have to find the constant a such that y(x) = e^{ax} is a solution for x" = x

My answer:
I tried 2 ways of reasoning and they both led to my answer that a can be any number (but zero).
Can anyone see if they are correct?? If not, pls give me some hint!

1. x= ln(y)/a but x" is a 2nd derivative with respect to x. so x" = 0 => x=0 =>a\in R \0

2. x has to be a constant for the solution to work, and it has to be specifically 0, so a is any real number but 0.
 
Physics news on Phys.org
I don't see what does y = e^(ax) got to do with x'' = x. The latter is actually a 2nd order derivative with respect to some variable like t (or y) is it not? Because it is evident that d^2/dx^2 x = 0.
 
Thanks for replying!

Actually this is one part of a big question where similar DE are given with the same solution(where others have y) so the solution y=exp(ax) seems irrelevant in this case.

The question didn't state x" is being differentiated w.r.t. what variable, so I just assumed it's dx...

This may sound dumb... but.. :confused::
So you mean as the whole equation is based on x, I could simply jump ahead to conclude that a is unbounded? even without the "not equal to zero" condition? I added the "not equal 0" condition as x = ln(y)/a (is this unnecessary?)

Also does it mean that my #1 or 2 reasoning are totally unnecessary?
 
I'm confused. Can't you just make a characteristic equation and make the equation
r^2 - 1 = 0, and then solve for r? Yielding simply 1 and -1? I haven't done this in a while, sorry.
 
No I still don't follow you at all. You mentioned that it was part of a larger question, so perhaps you could post that part of the question as it is? Because I don't know what you're talking about.
 
sr6622: sorry, I still haven't learned characteristic equation. so I can't use the technique yet.. but thanks!

Defennder:
Ohhhh. I mean it's a big question but each part has nothing to do with each other. They are just unrelated, like part i) is y' + y =0 and we are asked to do the same thing (finding the constant a with y(x)=exp(ax) is the solution).

so I guess as the "solution" y(x)=exp(ax) is irrelevant I can have a to be any number I want then.,,?
 
Last edited:

Similar threads

Replies
2
Views
1K
  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K