How to find critical points for a differential equation?

  • Thread starter CW83
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  • #1
CW83
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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
 

Answers and Replies

  • #2
joshmccraney
Gold Member
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start by setting both equal to 0! then try quadratic factoring for the dx. the dy should be obvious.
 
  • #3
joshmccraney
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also, you ask for critical points and where the equation equals zero...are you sure both occur simultaneously? i havent checked but dont assume this to be the case
 
  • #4
CW83
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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
epenguin
Homework Helper
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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.
 
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