Why Compare Slopes Across Vertical Lines?
A direction field displays the slope prescribed by a differential equation at points in the \(xy\)-plane. To identify the structure of the equation, compare points that have the same \(y\)-coordinate but different \(x\)-coordinates. If their slopes always match, the slope rule does not depend on \(x\). If two such points have different slopes, the rule cannot be a function of \(y\) alone.
This distinction is the difference between an autonomous and a nonautonomous differential equation. It describes the rule that generates the field, not whether a particular solution is constant or changing.
What the Classification Means
Write a general first-order slope rule as \(y'=F(x,y)\). At any point \((x,y)\), the direction field assigns the slope \(F(x,y)\). In an autonomous equation, fixing \(y\) fixes the slope: moving horizontally to another \(x\)-coordinate leaves it unchanged. Consequently, the pattern of slopes repeats across vertical lines.
For a nonautonomous equation, the slope may change as \(x\) changes while \(y\) is held fixed. The field may therefore have different slope marks at points on the same horizontal level. It is not necessary for the rule to vary with both coordinates: for example, \(y'=x\) is nonautonomous even though \(y\) does not appear in its right-hand side.
To classify a rule correctly, consider all relevant \(y\)-values in the domain. Finding a pair of different slopes at the same \(y\)-value proves that the equation is nonautonomous. Finding one pair of matching slopes does not prove that it is autonomous; other \(y\)-values may behave differently.
A Characterization Using Vertical Lines
The repeated-vertical-pattern test follows directly from the definition. Suppose the slope rule is \(F(x,y)\) on a domain where points with the same \(y\)-coordinate can be compared. Then it is autonomous exactly when, for each fixed \(y\), the value of \(F(x,y)\) is the same for every allowed \(x\).
Justification. If the equation is autonomous, then \(F(x,y)=f(y)\) for some function \(f\). Thus, for any two allowed \(x\)-values at the same height \(y\),
So every vertical line has the same slope at that height. Conversely, suppose the slope is independent of \(x\) at every fixed \(y\). Define \(f(y)\) to be this common slope. The common value is well-defined because all allowed \(x\)-values give the same result. Then \(F(x,y)=f(y)\) throughout the domain, so the equation can be written as \(y'=f(y)\), which is autonomous. This proves both directions.
The condition must hold at every relevant height. A coincidence at one height is only one piece of evidence; it cannot establish the required independence for all \(y\).
Worked Example: A Rule Depending Only on \(y\)
Worked Example: Classify \(y'=y(2-y)\)
The right-hand side contains \(y\), but not \(x\). It is already written as a function of \(y\) alone:
To see how this appears in the field, evaluate the slope at several heights. At \(y=-1\),
At \(y=0\), \(f(0)=0(2-0)=0\). At \(y=1\), \(f(1)=1(2-1)=1\). At \(y=3\), \(f(3)=3(2-3)=3(-1)=-3\). Each value is the same for every \(x\). For example, at the two points \((-2,1)\) and \((4,1)\), the slopes are both \(f(1)=1\).
Conclusion and check: The equation is autonomous because \(y(2-y)\) depends only on \(y\). Substituting any \(x\) into \(F(x,y)=y(2-y)\) makes no change to the computed slope; the vertical repetition follows from the formula.
Worked Example: A Rule Depending on Both Variables
Worked Example: Classify \(y'=x+y\)
Compare points at the same height \(y=2\). At \(x=-1\), the slope is
At \(x=0\), it is
At \(x=3\), it is
All three points have \(y=2\), but their slopes are \(1\), \(2\), and \(5\). In particular, the first two provide a direct test: \(F(-1,2)\ne F(0,2)\).
Conclusion and check: The equation is nonautonomous. If its right-hand side were a function of \(y\) alone, it would have to give the same value at every point with \(y=2\). The substitutions above contradict that requirement.
Worked Example: Nonautonomous Without \(y\)
Worked Example: Classify \(y'=x^2-1\)
This slope rule contains \(x\), but no \(y\). Check two points with the same height \(y=4\). At \(x=0\),
At \(x=2\),
The slopes differ even though the two points have the same \(y\)-coordinate. Thus the slope cannot be represented by one value \(f(4)\) for both points.
Conclusion and check: The equation is nonautonomous. The absence of \(y\) does not make it autonomous: the definition requires the slope rule to be expressible as a function of \(y\) alone, and these calculations show that it is not.
Why One Matching Height Is Not Enough
A fixed-height comparison is useful in one direction. If you find two \(x\)-values giving different slopes at the same \(y\), you have proved nonautonomy. But matching slopes at a particular height does not establish autonomy, because the rule might depend on \(x\) at other heights.
Worked Example: A Matching Pair That Does Not Prove Autonomy
Consider \(y'=xy\). At height \(y=0\), the slope is \(x(0)=0\) for every \(x\). In particular, at \(x=0\) and \(x=1\),
Those slopes match, but now compare two points at height \(y=2\):
These slopes differ. The agreement at \(y=0\) was a special case caused by the factor \(y\), not evidence that the rule is independent of \(x\).
Conclusion and check: The equation is nonautonomous, as confirmed by \(F(0,2)=0\ne2=F(1,2)\). To prove autonomy, independence from \(x\) must hold at every relevant height, not merely at one selected height.
A Reliable Classification Process
Use the following process when an equation or direction field is presented. The aim is to test dependence on \(x\) systematically rather than to judge from a single visual feature.
Identify \(F(x,y)\) in \(y'=F(x,y)\), or read the slopes from the field if the equation is not given.
Compare slopes at different \(x\)-coordinates on the same horizontal level.
One pair with unequal slopes proves the rule is nonautonomous.
Verify that the slope is independent of \(x\) throughout the domain, so it can be written as \(f(y)\).
When inspecting a drawn field, compare marks on the same horizontal level, not on a diagonal or a vertical line. Equal-looking marks can be difficult to judge precisely, so use the equation or numerical slope values when available. The mathematical condition concerns exact equality, not approximate visual resemblance.
Common Mistake: Looking Only for \(y\)
A tempting shortcut is to call an equation autonomous whenever its right-hand side includes \(y\), or whenever one horizontal level appears to repeat. Neither shortcut is valid. The question is whether the entire slope rule can be expressed as a function of \(y\) alone.
For example, the incorrect reasoning “\(y'=xy\) is autonomous because the slopes at \(y=0\) are all zero” checks only one height. It overlooks that at \(y=2\), the slopes at \(x=0\) and \(x=1\) are \(0\) and \(2\), respectively. The corrected conclusion is nonautonomous.
Likewise, the equation \(y'=x\) is nonautonomous even though it contains no \(y\). At a fixed height, its slope changes with \(x\); for instance, at \(y=1\), the slopes at \(x=0\) and \(x=2\) are \(0\) and \(2\). The presence or absence of a particular variable in the written expression is not the test by itself. The test is whether the slope can be represented using \(y\) alone across the domain.
Why This Distinction Matters
Autonomy lets you recognize a global pattern in a direction field: each horizontal level has one slope value repeated across \(x\). Nonautonomy signals that horizontal position matters to the rate rule, so two points with the same dependent-variable value may have different slopes. This is a structural description of a differential equation and its field; it does not by itself determine the shape of any particular solution.
Check Your Understanding
For each question, justify the classification using the slope rule or a same-height comparison.
- Is \(y'=3-y\) autonomous or nonautonomous? State why.
- For \(y'=x-y\), compare the slopes at \((1,2)\) and \((4,2)\). What does the comparison establish?
- Why does the fact that \(y'=xy\) has slope zero everywhere on the line \(y=0\) fail to prove it is autonomous?
- Is \(y'=5x\) autonomous, even though \(y\) does not appear? Explain with two points at the same height.
- What must be checked across the domain to establish that an equation is autonomous?
