Why is the interval solution this?

  • Context: MHB 
  • Thread starter Thread starter find_the_fun
  • Start date Start date
  • Tags Tags
    Interval
Click For Summary

Discussion Overview

The discussion revolves around the solution to the ordinary differential equation (ODE) given by \( x \frac{dy}{dx} - y = x^2 \sin{x} \). Participants are examining the transient term and the interval of the solution, specifically questioning why the interval is given as \( (0, \infty) \) rather than \( (-\infty, \infty) \).

Discussion Character

  • Debate/contested

Main Points Raised

  • Some participants note that \( x=0 \) is eliminated to rewrite the ODE in the form \( \frac{d}{dx}\left(\frac{y}{x}\right)=\sin(x) \).
  • Others express uncertainty about the restriction of \( x \) to only positive values, questioning the reasoning behind this choice.
  • One participant suggests that the interval must be continuous, implying that a break at \( x=0 \) is not permissible.
  • Another participant raises the possibility of choosing the interval \( (-\infty, 0) \) instead of \( (0, \infty) \), indicating a lack of clarity on why both cannot be options.
  • There is a question regarding whether an initial condition exists and if it is defined in \( (-\infty, 0) \) or \( (0, \infty) \).

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the reasons for the interval restriction, and multiple competing views remain regarding the choice of intervals and the implications of continuity.

Contextual Notes

The discussion highlights potential limitations regarding the assumptions made about the continuity of the solution and the role of initial conditions, which remain unresolved.

find_the_fun
Messages
147
Reaction score
0
The question is solve, give transient term and interval of solution for [math]x \frac{dy}{dx}-y = x^2 \sin{x}[/math] and the answer key has [math]y=cx-x\cos{x}[/math] and [math](0, \infty)[/math]. Why wouldn't the interval be [math](-\infty, \infty)[/math]?
 
Physics news on Phys.org
I can see why $x=0$ is eliminated in order to write the ODE in the form:

$$\frac{d}{dx}\left(\frac{y}{x}\right)=\sin(x)$$

But I don't know why they have restricted $x$ only to positive values. :D
 
MarkFL said:
I can see why $x=0$ is eliminated in order to write the ODE in the form:

$$\frac{d}{dx}\left(\frac{y}{x}\right)=\sin(x)$$

But I don't know why they have restricted $x$ only to positive values. :D

Isn't it because interval has to be continuous i.e. can't have a break at 0?
 
find_the_fun said:
Isn't it because interval has to be continuous i.e. can't have a break at 0?

What I mean is I don't see why the interval $(-\infty,0)$ couldn't be chosen either. Not both, but one or the other. :D
 
Was there an initial condition? If so, was it in $(-\infty,0)$ or $(0,\infty)$?
 

Similar threads

MHB -2.2.12 IVP y'=2x/(y+x^2y), y(0)=-2
  • · Replies 2 ·
Replies
2
Views
2K
MHB B.2.2.16 IVP of \dfrac{x(x^2+1)}{4y^3}, y(0)=-\dfrac{1}{\sqrt{2}}
  • · Replies 6 ·
Replies
6
Views
2K
Undergrad General solution of heat equation?
  • · Replies 3 ·
Replies
3
Views
3K
MHB Can You Solve This Initial Value Problem and Determine Its Solution Interval?
  • · Replies 7 ·
Replies
7
Views
4K
Undergrad On vibrating string and differentiating infinite sum
  • · Replies 7 ·
Replies
7
Views
3K
Undergrad Energy of moving Sine-Gordon breather
  • · Replies 2 ·
Replies
2
Views
3K
MHB -2.2.20 IVP interval....trig subst y^2(1-x^2)^{1/2} \,dy=\arcsin{x}\,dx
  • · Replies 4 ·
Replies
4
Views
2K
MHB -02.2.11 initial value, graph, intervals xdx+ye^{-x}dy=0, \quad y(0)=1
  • · Replies 5 ·
Replies
5
Views
2K
MHB -2.1.1 DE Find the general solution
  • · Replies 5 ·
Replies
5
Views
2K
MHB 2.2.16 (a) intial value (b) interval
  • · Replies 9 ·
Replies
9
Views
2K