Using an ODE to show a local minimum

  • #1
ver_mathstats
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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.
 

Answers and Replies

  • #2
Delta2
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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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  • #3
pasmith
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The sign of the second derivative depends on the value of [itex]y(0)[/itex], so I don't think you can conclude that it's a minimum if you aren''t given that information.
 
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  • #4
ver_mathstats
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The sign of the second derivative depends on the value of [itex]y(0)[/itex], 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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  • #5
Delta2
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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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  • #6
ver_mathstats
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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.
 

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