How to find critical points for a differential equation?

In summary, the student is trying to find the critical points where the given equation equals zero. They have found the partial derivatives and set them equal to zero, resulting in four lines through the origin. They also mention using another approach of finding the points where both (dy/dx)f and its reciprocal are equal to zero. The student has also identified an error in their second partial derivative calculation.
  • #1
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Homework Statement



f(x,y)= 3x5y2 - 30x3y2 + 60xy2 +150

How do I find the critical points where this equation equals zero?

Homework Equations



None?

The Attempt at a Solution



I found the partial derivatives, but I'm stuck now and don't know where to go from here.

dx = 15x4y2 - 90x2y2 + 60y2
dy = 6x5 - 60x3y + 10xy + 240
 
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  • #2
start by setting both equal to 0! then try quadratic factoring for the dx. the dy should be obvious.
 
  • #3
also, you ask for critical points and where the equation equals zero...are you sure both occur simultaneously? i haven't checked but don't assume this to be the case
 
  • #4
Okay so I did that. I got for dx, x = 3 +- sqrt(20)/2
The first part of dy should actually be 6yx^5.

After I have the dx value, I can sub them into the dy equation which gives approximately y = -0.1545 and y = -3.6099. Where do I go from here? There should be 4 separate (x,y) points.
 
  • #5
Spell things out a little more carefully. What you quote a solution for I think is really saying

∂f/∂x = 0 at all points along (x/y)2 = those numbers, which is actually four lines through the origin I think.

Maybe there is a nice algebraic breakup of your ∂f/∂y but anyway what you are doing should lead you to the four points along the four lines.

Alternative approach is to find (dy/dx)f and its reciprocal and find the points where both are 0. Well same thing really, -(dy/dx)f is the ratio of your two derived polynomials.

Also I think the second of yours, what you call "dy" is mistaken and it works out fairly simply.

You might as well have taken the factor 3 out of your starting equation.
 
Last edited:

1. What is a critical point in a differential equation?

A critical point in a differential equation is a point where the derivative of the function is equal to zero. In other words, it is a point where the slope of the function is flat or horizontal.

2. Why is finding critical points important in solving differential equations?

Finding critical points allows us to identify important characteristics of the function, such as its maximum and minimum values. This information is crucial in understanding the behavior of the function and solving the differential equation.

3. How do you find critical points for a differential equation?

To find critical points, you need to set the derivative of the function equal to zero and solve for the variable. This will give you the x-values of the critical points. You can then substitute these values into the original function to find the corresponding y-values.

4. Are there any other methods for finding critical points besides setting the derivative equal to zero?

Yes, there are other methods such as using the first and second derivative tests, where you analyze the sign and concavity of the derivative to determine the location and nature of the critical points.

5. Can a differential equation have multiple critical points?

Yes, a differential equation can have multiple critical points. This occurs when the function has multiple horizontal tangents or when the derivative changes sign at different points. It is important to consider all critical points when solving the differential equation.

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