# Using an ODE to show a local minimum

• ver_mathstats
In summary, the conversation revolved around solving an ODE with the hint to use the first derivative test to show a local minimum at x = 0. The suggested method involved taking partial derivatives, equating them to zero, and calculating second partial derivatives. However, it was later mentioned that the value of y(0) was given, which is necessary to prove that y''(0) > 0. The final solution involved differentiating the ODE once and using this value to prove the minimum.
ver_mathstats
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.

Delta2
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).

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.

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))##

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.

## 1. What is an ODE?

An ODE (ordinary differential equation) is a mathematical equation that describes the relationship between a function and its derivatives. It is often used to model physical systems and predict their behavior over time.

## 2. How can an ODE be used to show a local minimum?

An ODE can be used to show a local minimum by representing the function as a differential equation and then solving for the critical points, which are points where the derivative of the function is equal to zero. If the second derivative at a critical point is positive, then it is a local minimum.

## 3. What is a local minimum?

A local minimum is a point on a graph where the function has the lowest value within a small interval around that point. It is lower than all other nearby points, but it may not be the absolute lowest point on the entire graph.

## 4. Why is it important to use an ODE to show a local minimum?

Using an ODE to show a local minimum allows us to mathematically prove that the point is indeed a minimum and not just an estimate. This is especially useful in scientific research and engineering, where precise calculations and predictions are necessary.

## 5. Can an ODE be used to show a global minimum?

No, an ODE can only be used to show a local minimum. To determine a global minimum, the entire function must be analyzed and compared to other functions. It is possible for a local minimum to also be a global minimum, but this is not always the case.

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