# Slope Fields and horizontal segments

• Apr 23rd 2009, 12:45 PM
Moderatelyconfused
Slope Fields and horizontal segments
I am not really sure how to start this problem:

The slope field of the differential equation is dy/dx=(x^2y+y^2x)/3x+y will have horizontal segments when...

How do I find the horizontal segments without a calculator. I think I need to integrate, but I'm not sure.
• Apr 23rd 2009, 01:07 PM
redsoxfan325
Quote:

Originally Posted by Moderatelyconfused
I am not really sure how to start this problem:

The slope field of the differential equation is dy/dx=(x^2y+y^2x)/3x+y will have horizontal segments when...

How do I find the horizontal segments without a calculator. I think I need to integrate, but I'm not sure.

Do you mean this:
$\displaystyle \frac{dy}{dx} = \frac{x^2 y+y^2 x}{3x}+y$

or this:
$\displaystyle \frac{dy}{dx} = \frac{x^{2y}+y^{2x}}{3x}+y$

?
• Apr 23rd 2009, 01:50 PM
Moderatelyconfused
I meant: dy/dx=(x^2y+y^2x)/(3x+y)

I apologize, the y was actually part of the denominator.
• Apr 23rd 2009, 02:04 PM
mr fantastic
Quote:

Originally Posted by Moderatelyconfused
I am not really sure how to start this problem:

The slope field of the differential equation is dy/dx=(x^2y+y^2x)/3x+y will have horizontal segments when...

How do I find the horizontal segments without a calculator. I think I need to integrate, but I'm not sure.

dy/dx = 0 when the numerator is equal to zero (but check that the denominator doesn't also equal zero when this happens).