Constant Solutions and Direction Fields
A direction field represents a first-order differential equation by drawing short segments whose slopes match the equation at selected points. At a point where the equation's right-hand side is zero, the corresponding segment is horizontal. A row of horizontal segments across the field can therefore reveal a constant solution—but a single horizontal segment, or even a diagonal curve of such segments, is not enough.
The key test is whether the same constant value of the dependent variable makes the right-hand side zero for every input value in the interval under consideration. This condition connects an algebraic calculation with a visible feature of the direction field.
Here, a solution means a differentiable function that satisfies the differential equation at every input value in its interval. The word “equilibrium” emphasizes that the solution stays at one fixed \(y\)-value. Its graph is the horizontal line \(y=c\).
Why Equilibria Appear as Horizontal Rows
The connection between the algebraic test and the picture follows directly from the derivative. The derivative of a constant function is zero everywhere. Thus, if \(y=c\) is a solution, the right-hand side must also be zero at each point \((x,c)\) on its graph. In a direction field, every segment along that horizontal line therefore has slope zero.
The converse holds as well. If the right-hand side is zero at every point \((x,c)\) as \(x\) varies across the interval, then the constant function \(y(x)=c\) has derivative zero and satisfies the equation at every input value. This gives a complete test:
Set \(y=c\), where \(c\) does not depend on \(x\).
Evaluate \(F(x,c)\) while allowing \(x\) to vary throughout the interval.
If \(F(x,c)=0\) for every \(x\) in the interval, then \(y=c\) is an equilibrium solution. Otherwise, it is not.
Proof. Suppose first that \(y(x)=c\) is a solution. Since \(c\) is constant, \(y'(x)=0\) for every \(x\in I\). The differential equation then gives \(F(x,c)=y'(x)=0\) for every \(x\in I\). Conversely, suppose \(F(x,c)=0\) for every \(x\in I\). The constant function \(y(x)=c\) has derivative \(y'(x)=0\), so at each \(x\in I\), \(y'(x)=0=F(x,c)=F(x,y(x))\). It therefore satisfies the equation throughout \(I\). This proves both directions.
Worked Examples
Worked Example: Finding Two Equilibria
Find the equilibrium solutions of \(y'=y(3-y)\), and describe their appearance in the direction field.
For a constant solution \(y=c\), substitute \(c\) into the right-hand side. The required condition is
A product is zero when at least one factor is zero. Therefore, \(c=0\) or \(3-c=0\), giving \(c=0\) or \(c=3\).
Check both candidates in the equation. For \(y(x)=0\), the derivative is \(0\), and the right-hand side is \(0(3-0)=0\). For \(y(x)=3\), the derivative is \(0\), and the right-hand side is \(3(3-3)=0\). Both functions satisfy the equation.
Since the right-hand side does not depend on \(x\), the slope is zero at every point with \(y=0\) and at every point with \(y=3\). The direction field shows two horizontal rows of horizontal segments, and each row traces one equilibrium solution.
Worked Example: An Equilibrium with an \(x\)-Dependent Factor
Find the equilibrium solution of \(y'=(1+x^2)(y+2)\), for real \(x\), and identify its row in the direction field.
For a constant solution \(y=c\), the condition is
For every real \(x\), \(x^2\geq 0\), so \(1+x^2\geq 1\); in particular, this factor is never zero. Thus the product can be zero for every \(x\) only if \(c+2=0\), which gives \(c=-2\).
Check the candidate: \(y(x)=-2\) has derivative \(0\), and substituting into the right-hand side gives
So \(y(x)=-2\) is an equilibrium. In the direction field, all the segments along the horizontal line \(y=-2\) are horizontal, even though the slopes away from that row vary with \(x\).
Worked Example: A Zero-Slope Curve That Is Not an Equilibrium
Does \(y'=y-x\) have an equilibrium solution? What does its zero-slope set look like?
If \(y=c\) were an equilibrium, substitution would require
This cannot hold throughout any interval containing more than one value of \(x\). For example, at \(x=0\) it would require \(c=0\), while at \(x=1\) it would require \(c=1\). A single constant \(c\) cannot meet both requirements. Therefore, there is no equilibrium solution on an interval containing both \(0\) and \(1\).
At an individual point, the slope is zero when \(y-x=0\), or \(y=x\). This is a diagonal set of zero-slope points, not a horizontal row at one fixed height. Check the distinction: along \(y=x\), the equation's right-hand side is \(x-x=0\), but the function \(y(x)=x\) has derivative \(1\), not \(0\). Hence this zero-slope set does not itself describe a solution curve.
Reading the Field Carefully
A direction field can suggest candidate equilibria visually. Look for horizontal marks that continue across the field at a fixed height. Then verify the candidate using the equation: visual evidence is limited to the points shown, while the equilibrium condition must hold throughout the interval being considered.
| What the field shows | What it means |
|---|---|
| A horizontal row of zero-slope marks across the interval | A candidate equilibrium; confirm that \(F(x,c)=0\) throughout the interval. |
| One or a few horizontal marks | The slope is zero at those points; this alone does not establish a constant solution. |
| A diagonal or curved set of zero-slope marks | The right-hand side vanishes along a changing \(y\)-value; this need not be a solution curve. |
The equation and the field serve different purposes. Factoring or otherwise solving \(F(x,c)=0\) can identify the possible constant values algebraically. The field then displays those values as horizontal rows. Neither the fact that a row is horizontal nor the fact that a point has zero slope should be confused with a conclusion about nonconstant solution curves.
A Common Mistake: Checking Only One Input Value
A common error is to find an \(x\)-value where the right-hand side vanishes and infer that a constant solution exists. The equilibrium condition is stronger: the same height \(c\) must make the right-hand side zero for every \(x\) in the interval.
Worked Example: Why a Single Vertical Slice Is Not Enough
Consider \(y'=x(y-1)\). At \(x=0\), the right-hand side is \(0\) for every value of \(y\), since \(0(y-1)=0\). It would be incorrect to conclude that every constant function is an equilibrium.
Test the constant candidate \(y=0\). Its derivative is \(0\), but the right-hand side after substitution is
This is not zero for every \(x\); at \(x=1\), it equals \(-1\). Thus \(y=0\) is not an equilibrium. For \(y=1\), the derivative is \(0\), and the right-hand side is \(x(1-1)=0\) for every \(x\). Therefore \(y(x)=1\) is an equilibrium, represented by the horizontal row \(y=1\).
The vertical slice \(x=0\) contains horizontal marks at every height, but that fact applies only at that one input value. The equilibrium test follows a fixed height across the interval, rather than scanning a single vertical slice.
Why Equilibrium Solutions Matter
Equilibria provide fixed reference levels in a direction field. They help organize what the field shows and provide exact solutions that can be checked directly. Finding them also forces a useful distinction: a zero of the right-hand side at one point is a local fact about the slope, while an equilibrium is a whole solution defined by a condition that holds throughout an interval.
When examining a proposed equilibrium, keep the test precise:
- Choose a constant value \(c\), not a value that changes with \(x\).
- Substitute \(y=c\) into the full right-hand side.
- Check whether the result is zero for every \(x\) in the interval.
- Confirm that the constant function has derivative zero and satisfies the equation.
This procedure links symbolic work to the horizontal rows in a direction field and prevents isolated zero-slope marks from being mistaken for equilibrium solutions.
Check Your Understanding
For each question, apply the condition that the right-hand side must vanish at one fixed height throughout the interval.
- Find the equilibrium solutions of \(y'=y(y-4)\). Verify each by substitution.
- Does \(y'= (x^2+1)(y-5)\) have an equilibrium solution? If so, state it and check it.
- For \(y'=2x+y\), determine whether any constant function is an equilibrium on an interval containing more than one \(x\)-value.
- Explain why a horizontal direction-field segment at one point does not by itself establish an equilibrium.
- For \(y'=x(y+3)\), identify the equilibrium and explain why the zero slopes at \(x=0\) do not make every constant function an equilibrium.
