Using an ODE to show a local minimum

  • Thread starter Thread starter ver_mathstats
  • Start date Start date
  • Tags Tags
    Local Minimum Ode
Click For Summary

Homework Help Overview

The problem involves analyzing the ordinary differential equation (ODE) given by y' = xcos(y). The goal is to demonstrate that the solution has a local minimum at x = 0, utilizing concepts related to the first derivative test.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to understand how to apply the first derivative test in the context of the ODE and questions whether they need to compute partial derivatives to find critical points. Other participants suggest evaluating the ODE at x = 0 to analyze y'(0) and discuss the implications of the value of y(0) on the second derivative.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the problem. Some guidance has been provided regarding the need to differentiate the ODE to find the second derivative and its relationship to determining a local minimum. However, there is no explicit consensus on the approach yet.

Contextual Notes

Participants note that the value of y(0) is crucial for determining the sign of the second derivative, which is necessary for concluding the presence of a local minimum. There seems to be some confusion regarding the use of y(0) and its implications for the analysis.

ver_mathstats
Messages
258
Reaction score
21
Homework Statement
Using the ODE, show that the solution has a local minimum at x = 0.
Relevant Equations
First derivative test
The ODE given to us is y' = xcosy. I am having a bit of trouble when it comes to solving this problem. We are supposed to show that the solution has a local minimum at x = 0 with the hint to think of the first derivative test. However, I am only really familiar with the first derivative test when it comes to a function like f(x) not y' with two variables.

In order to solve this would be have to first take f(x,y) = xcosy and then solve the partial derivatives fx(x,y) and fy(x,y), then equate them both to zero, solve for x and y to obtain a critical point, and finally calculate the second partial derivatives to prove there is a local minimum at x = 0? Or is there a much more efficient way to do this?

Any help would be appreciated as I'm a bit confused about how to approach this, thank you.
 
  • Like
Likes   Reactions: Delta2
Physics news on Phys.org
Let me rewrite the ode to make some things more clear to you:
The ODE is $$y'(x)=x\cos(y(x))$$. So what can you say about ##y'(0)##(Hint: Apply the ODE for x=0).
 
  • Like
Likes   Reactions: ver_mathstats
The sign of the second derivative depends on the value of y(0), so I don't think you can conclude that it's a minimum if you aren''t given that information.
 
  • Like
Likes   Reactions: ver_mathstats and Delta2
pasmith said:
The sign of the second derivative depends on the value of y(0), so I don't think you can conclude that it's a minimum if you aren''t given that information.
We were given y(0) = 0, sorry told not to use initially but now apparently we can use it.
 
Last edited:
ver_mathstats said:
We were given y(0) = 0, sorry told not to use initially but now apparently we can use it.
You need it to prove that ##y''(0)>0## hence it is a minimum. By differentiating the given ODE one time you end up with ##y''(x)=\cos(y(x))-xy'(x)\sin(y(x))##
 
  • Like
Likes   Reactions: ver_mathstats
Delta2 said:
You need it to prove that ##y''(0)>0## hence it is a minimum. By differentiating the given ODE one time you end up with ##y''(x)=\cos(y(x))-xy'(x)\sin(y(x))##
Thank you, this makes so much sense now! I appreciate it.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
30
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
1K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
1K