Using an ODE to show a local 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 f_x(x,y) and f_y(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).

 

[pasmith]

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]

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.

 

[Delta2]

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]

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.