An equilibrium solution is a solution to a d.e. whose derivative is zero everywhere. On a graph an equilibrium solution looks like a horizontal line.
Given a slope field, we can find equilibrium solutions by finding everywhere a horizontal line fits into the slope field.
Equilibrium solutions come in two flavors: stable and unstable. These terms are easiest to understand by looking at slope fields.
A stable equilibrium solution is one that other solutions are trying to get to. If we pick a point a little bit off the equilibrium in either direction, the solution that goes through that point tries to snuggle up to the equilibrium solution.
An unstable equilibrium solution is one that the other solutions are trying to get away from. If we pick a point a little bit off the equilibrium, the solution that goes through that point is trying to run away from the equilibrium solution.
If the solutions are trying to get away on one side and snuggle up on the other side, the equilibrium is still unstable.
If we're given a differential equation instead of a slope field, we can determine whether each equilibrium is stable or unstable by using the differential equation to sketch a very rough slope field.
Practice:
Given the following slope field, draw all equilibrium solutions: 
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Determine all equilibrium solutions to the differential equation 
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In order for  to be zero, we must have y = 5. The constant function y = 5 is the only equilibrium solution.
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Find all equilibrium solutions of the differential equation 
Determine if each equilibrium solution is stable or unstable. | |
To find equilibrium solutions we set the differential equation equal to 0 and solve for y. 0 = y2 - y = y(y - 1) so the equilibrium solutions are y = 0 and y = 1. Now to figure out if the other solutions are trying to snuggle up to or run away from each of these equilibrium solutions. When y > 1 the quantity 
is positive, which means the slopes on the slope field will be positive when y > 1. Let's sketch this into our very rough slope field. When y is between 0 and 1, y is positive but (y - 1) is negative, so the product y(y - 1) is negative. This means the slopes on the slope field will be negative for 0 Finally, when y is less than 0 both y and (y - 1) are negative, so the product y(y - 1) is positive. This means the slopes will be positive for y < 0. The solutions above y = 1 are trying to get away from the equilibrium, so that's an unstable equilibrium. The other solutions are trying to get closer to y = 0, so that's a stable equilibrium: 
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For the slope field, draw all equilibrium solutions (if any).
Answer
The only equilibrium solution is y = 0:
For the slope field, draw all equilibrium solutions (if any).
Answer
There are no places on this slope field where a horizontal line fits in, so there are no equilibrium solutions.
For the slope field, draw all equilibrium solutions (if any).
Answer
For the slope field, draw all equilibrium solutions (if any).
Answer
There are no places on this slope field where a horizontal line fits in, so there are no equilibrium solutions.
Given a differential equation, we can find equilibrium solutions by setting the derivative equal to zero and solving.
For the differential equation, find all equilibrium solutions (if any) and determine if each is stable or unstable.
Answer
We factor:
The equilibrium solutions are y = 0, y = 1, and y = -1.
When y > 1 the quantity (1 - y
When 0 < y < 1 all quantities are positive, so the slopes are positive.
When -1 < y < 0 the quantity y is negative, so the slopes are negative.
When y < -1 the quantities y and (1 + y) are negative, so the slopes are positive.
We end up with y = 1 and y = -1 being stable equilibrium solutions, while y = 0 is unstable:
For each of the following differential equations, determine all equilibrium solutions (if any).
Answer
These don't require any calculus except to understand what the question is asking. To answer the question, set each derivative equal to zero and solve for the dependent variable.
0 = y
For each of the following differential equations, determine all equilibrium solutions (if any).
Answer
These don't require any calculus except to understand what the question is asking. To answer the question, set each derivative equal to zero and solve for the dependent variable.
0 = P(1 - P)P = 0, 1
For each of the following differential equations, determine all equilibrium solutions (if any).
Answer
These don't require any calculus except to understand what the question is asking. To answer the question, set each derivative equal to zero and solve for the dependent variable.
0 = Q2 + 5Q + 6 = (Q + 2)(Q + 3)Q = -2, -3
For each of the following differential equations, determine all equilibrium solutions (if any).
Answer
These don't require any calculus except to understand what the question is asking. To answer the question, set each derivative equal to zero and solve for the dependent variable.
0 = y2 - y = y( 1 - y )y = 0,1
For each of the following differential equations, determine all equilibrium solutions (if any).
Answer
These don't require any calculus except to understand what the question is asking. To answer the question, set each derivative equal to zero and solve for the dependent variable.
This d.e. has no equilibrium solutions because the equation
0 = R2 + 4
has no solutions.
For the differential equation, find all equilibrium solutions (if any) and determine if each is stable or unstable.
Answer
First we factor:
The equilibrium solutions are y = -2 and y = 3.
For y > 3, both (y - 3) and (y + 2) are positive. The product (y - 3) (y + 2) is positive, as are the slopes.
For -2 < y < 3 the quantity (y - 3) is negative and (y + 2) is positive, so the slopes are negative.
For y < -2, both (y - 3) and (y + 2) are negative, so the slopes are positive.
We conclude that y = -2 is a stable equilibrium and y = 3 is an unstable equilibrium:
For the differential equation, find all equilibrium solutions (if any) and determine if each is stable or unstable.
Answer
Let's factor:
The equilibrium solutions are y = 0, y = -1, and y = 1.
When y > 1 all quantities are positive, so the slopes are positive also.
When 0 < y < 1 the quantity (y - 1) is negative and the others are positive, so the slopes are negative.
When -1 < y < 0 the quantities y and (y - 1) are negative, so the slopes are positive.
Finally, when y < -1 all quantities are negative, so the slopes are negative.
We conclude that 0 is the only stable equilibrium solution:
For the differential equation, find all equilibrium solutions (if any) and determine if each is stable or unstable.
Answer
First we factor:
The equilibrium solutions are y = 0 and y = 1.
Since the quantity (y - 1)2 is always positive it won't affect the signs of the slopes.
When y > 0, the slopes are positive.
When y < 0, the slopes are negative:
Both equilibrium solutions are unstable.
For the differential equation, find all equilibrium solutions (if any) and determine if each is stable or unstable.
Answer
First we factor:
The equilibrium solutions are y = 0 and y = -1:
Since (y + 1)2 is always positive, this factor won't affect the signs of the slopes.
When y > 0 the slopes are positive, and when y < 0 the slopes are negative:
Both equilibrium solutions are unstable.
