Using an ODE to show a local minimum

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The discussion revolves around using the ordinary differential equation (ODE) y' = xcos(y) to demonstrate that the solution has a local minimum at x = 0. Participants express confusion about applying the first derivative test to a function of two variables and the necessity of using the initial condition y(0) = 0. It is clarified that to establish a local minimum, one must show that the second derivative y''(0) is greater than zero, which involves differentiating the ODE. The final consensus emphasizes the importance of the initial condition in proving the local minimum. Overall, the conversation highlights the steps needed to analyze the ODE effectively.
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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.
 
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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).
 
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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.
 
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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.
 
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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))##
 
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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