What Makes a Solution an Equilibrium?
A direction field shows the slope prescribed by a differential equation at many points in the plane. Some rows of these marks may be horizontal throughout the field. A horizontal row can indicate a solution that stays at one fixed value of the dependent variable, rather than increasing or decreasing as the independent variable changes.
To decide whether such a row represents a solution, it is not enough to find a point where the slope is zero. A constant solution must have zero slope at every point along its graph. This all-inputs condition is the key to recognizing an equilibrium.
Here \(x\) is the independent variable and \(y\) is the dependent variable. The graph of \(y(x)=c\) is the horizontal line \(y=c\). If the equation gives slope zero all along that line, the line itself is a solution curve.
Check the Whole Horizontal Line
For a proposed constant value \(c\), substitute \(y=c\) into the right-hand side of the equation. The result must be zero for every relevant \(x\), not just for one particular input. This gives a direct test:
Solving \(f(x,y)=0\) may locate points or curves where the field has horizontal marks. But a zero-slope location is not automatically an equilibrium solution. For an equilibrium, one fixed value \(y=c\) must make the slope zero across the full interval of \(x\)-values under consideration.
Worked Example: Finding Equilibria from a Factored Equation
Identify the equilibrium solutions of \(y'=y(2-y)\), then describe the slope signs immediately above and below each one.
For a constant solution \(y=c\), the required condition is \(c(2-c)=0\). Factoring gives
Neither expression depends on \(x\). Substituting each value into the differential equation confirms the result:
Thus \(y=0\) and \(y=2\) are equilibrium solutions. To read the neighboring slopes, check the sign of \(y(2-y)\). For \(y=-1\), the slope is \((-1)(3)=-3\), so slopes below \(0\) are negative. For \(y=1\), the slope is \(1(1)=1\), so slopes between \(0\) and \(2\) are positive. For \(y=3\), the slope is \(3(-1)=-3\), so slopes above \(2\) are negative.
The field therefore has horizontal traces at \(y=0\) and \(y=2\), with negative marks below \(0\), positive marks between the two traces, and negative marks above \(2\). Substitution checks both equilibrium values, and the sign checks describe the neighboring pattern.
Why the Test Works
The test follows directly from what it means to be a solution. If \(y(x)=c\) is constant, its derivative is zero at every input. The differential equation requires that derivative to equal \(f(x,c)\), so \(f(x,c)\) must be zero for every \(x\) in the interval.
In the other direction, suppose \(f(x,c)=0\) for every \(x\in I\). The constant function \(y(x)=c\) has derivative \(y'(x)=0\). At each input, the right-hand side is also \(f(x,c)=0\). Thus the function satisfies the differential equation throughout the interval.
This reasoning also explains the visual test. Each mark in a direction field represents a slope prescribed by the equation. Along the graph \(y=c\), every mark must be horizontal for the constant graph to follow the field.
A Zero Slope at One Input Is Not Enough
Some equations have horizontal marks at points that do not lie on any equilibrium trace. If \(f(x,y)=0\) only for a particular \(x\), that tells you the slope is zero at that input; it does not show that a constant function satisfies the equation over an interval. The distinction is especially important when the right-hand side depends on both \(x\) and \(y\).
Worked Example: Separating a Zero-Slope Location from an Equilibrium
For \(y'=x(y-1)^2\), find the equilibrium solution and explain why the zero slopes at \(x=0\) do not make every constant value an equilibrium.
Set \(y=c\). The right-hand side becomes
For this expression to be zero for every \(x\) in an interval, the factor \((c-1)^2\) must be zero. Hence \(c=1\). Substitution verifies it:
At \(x=0\), however, the equation gives \(f(0,y)=0\) for every \(y\). This is a vertical set of zero-slope marks at one input, not a horizontal equilibrium trace. For example, the constant candidate \(y=2\) has right-hand side \(x(2-1)^2=x\), which is not zero for every \(x\). It is not a solution.
For neighboring values of \(y=1\), the squared factor is positive whenever \(y\ne1\). Thus the slope has the sign of \(x\): it is positive when \(x>0\) and negative when \(x<0\), on either side of the equilibrium trace. At \(x=0\), all these slopes are zero. Substitution confirms that only \(y=1\) is an equilibrium, while the sign check describes the nearby marks.
Reading the Pattern Around an Equilibrium
Once an equilibrium has been identified, inspect the field just above and below its horizontal trace. The signs of the slopes tell you whether nearby solution curves are locally increasing or decreasing as \(x\) increases. The lengths or steepness of the marks indicate the relative size of those slopes.
This description concerns the direction indicated by the field. A positive slope means a solution curve rises as \(x\) increases at that point; a negative slope means it falls. A zero slope means the tangent is horizontal there. Do not infer more than the equation supports from a few marks, but do use a consistent nearby sign pattern to describe what the field shows.
Look for a horizontal row of zero-slope marks, or solve the equation \(f(x,c)=0\) for a constant \(c\).
Substitute \(y=c\) into \(f(x,y)\) and verify that the result is zero throughout the interval, not merely at one \(x\)-value.
Check that the constant function has derivative zero and that the right-hand side is also zero along the entire horizontal line.
Read the slope signs above and below the trace, and note whether the marks are relatively shallow or steep.
Worked Example: Reading Two Equilibrium Traces and Their Neighbors
Consider \(y'=(x^2+1)(y+2)(y-1)\). Find its equilibrium solutions and describe representative slopes in the regions around them.
For \(y=c\), the condition is
Since \(x^2+1>0\) for every real \(x\), the constant factors must satisfy \((c+2)(c-1)=0\). Therefore \(c=-2\) or \(c=1\). Substituting gives
Both are equilibrium solutions. To check the neighboring signs, note that the factor \(x^2+1\) is positive. When \(y<-2\), both \(y+2\) and \(y-1\) are negative, so their product is positive. When \(-2<y<1\), \(y+2\) is positive and \(y-1\) is negative, so the product is negative. When \(y>1\), both are positive, so the product is positive.
For numerical checks, at \(x=0\) the slopes at \(y=-3,0,2\) are respectively \(4,-2,4\). At \(x=2\), the multiplier \(x^2+1\) is \(5\), so the corresponding slopes are \(20,-10,20\). These values confirm the positive, negative, positive pattern and show that the slopes are steeper at the sampled points when \(x=2\). The equilibrium traces themselves remain horizontal for all \(x\), as the substitutions verify.
A Common Mistake
A frequent error is to identify every zero of the right-hand side as an equilibrium without checking whether it persists for every input. For \(y'=x(y-1)^2\), one might set \(x=0\), observe that the expression is zero for every \(y\), and conclude that every horizontal line is an equilibrium. That conclusion confuses zero slope at one input with zero slope along an entire solution.
Why Equilibria Matter in a Direction Field
Equilibrium traces give fixed reference levels in a slope field. They help organize the rest of the picture: you can identify where slopes vanish persistently, then compare the signs and steepness of marks in the regions between and around those levels. In a model, such a level may represent a quantity that remains unchanged when it is already at that value.
The central discipline is to distinguish three observations: a horizontal mark at a point, a set of points with zero slope, and a horizontal line of zero slopes that is itself a solution. Only the last meets the definition of an equilibrium solution. Substitution into the differential equation is the reliable check.
Check Your Understanding
For each question, use the condition that an equilibrium value must make the right-hand side zero for every relevant \(x\).
- For \(y'=y(y-4)\), what are the equilibrium solutions?
- For \(y'=x(y+2)\), which constant value is an equilibrium? Why is the zero slope at \(x=0\) insufficient to make every constant an equilibrium?
- For \(y'=(y-3)^2\), what is the equilibrium trace, and what is the sign of the slope on either side of it?
- In a direction field, what is the difference between a horizontal mark at one point and an equilibrium solution?
- If the slopes just above and below an equilibrium trace have opposite signs, what information does that give about the local directions of nearby solution curves?
