Matching Equations, Fields, and Curves
A slope field represents a differential equation visually: at each point \((x,y)\), a short segment has slope \(y'\) prescribed by the equation. A solution curve must follow those local directions. Matching an equation to a field, or a field to a curve, does not require inspecting every point. A few well-chosen tests can often settle the match.
This workshop uses three kinds of evidence: the slope at test points, where the slope is zero, and whether a candidate curve passes through the required initial point. These checks work together. A single slope can rule out a candidate, while several compatible checks give strong evidence for a match.
A Reliable Matching Strategy
First identify features that are easy to compare. For each candidate equation, evaluate its right-hand side at selected points. Also find where \(F(x,y)=0\), because the field has horizontal segments there. Then compare these predictions with the candidate field.
Next examine the proposed curves. Their initial values must match the specified point, and their derivatives must agree with the equation’s slopes along the curve. A curve that has the right value at one point but the wrong tangent there cannot be a solution.
Prefer points where candidate equations give clearly different slopes, including points on possible zero-slope lines.
Calculate the slope at those points and identify where the equation predicts horizontal segments.
Keep only curves that pass through the required initial point.
Differentiate it and compare its derivative with the equation along the curve.
Worked Example: Distinguish Three Autonomous Fields
Worked Example: Distinguish Three Autonomous Fields
Match the equations \(A:y'=y\), \(B:y'=-y\), and \(C:y'=1\) to these field descriptions:
- Field I has positive slopes above the \(x\)-axis, negative slopes below it, and horizontal segments along the \(x\)-axis.
- Field II has negative slopes above the \(x\)-axis, positive slopes below it, and horizontal segments along the \(x\)-axis.
- Field III has the same positive slope at every point.
At \((0,2)\), the predicted slopes are:
At \((0,-1)\), they are:
For \(A\), the slope has the same sign as \(y\), and it is zero exactly when \(y=0\). This matches Field I. For \(B\), the slope has the opposite sign to \(y\), with zero slope when \(y=0\), so it matches Field II. Equation \(C\) gives slope \(1\) everywhere, matching Field III.
Now match a curve through the initial point \((0,2)\). The candidate curves are \(g_1(x)=2e^x\), \(g_2(x)=2e^{-x}\), and \(g_3(x)=2+x\). Each has the required initial value: \(g_1(0)=2\), \(g_2(0)=2\), and \(g_3(0)=2\). So the initial value alone does not distinguish them.
Differentiate and compare with the equations. For \(g_1\), \(g_1'(x)=2e^x=g_1(x)\), so it satisfies \(A\). For \(g_2\), \(g_2'(x)=-2e^{-x}=-g_2(x)\), so it satisfies \(B\). For \(g_3\), \(g_3'(x)=1\), so it satisfies \(C\).
As a final check, at \(x=0\) the derivatives are \(2\), \(-2\), and \(1\), respectively. These match the slopes at \((0,2)\) calculated from the three equations. The field and curve matches are consistent.
Why Test Points and Zero Slopes Work
The equation specifies the slope at every point, so matching at a test point is a direct check of the field’s defining information. If two equations give different slopes at even one point, a single field cannot represent both equations there. A zero-slope location is especially useful because horizontal segments are easy to spot and can reveal the equation’s structure.
For a proposed curve \(y=g(x)\), the requirement is stronger than matching one slope. At each \(x\), the point on the curve is \((x,g(x))\). Its tangent slope is \(g'(x)\), while the field prescribes \(F(x,g(x))\). Thus the curve follows the field precisely when those quantities agree throughout its interval.
This condition explains why curve verification involves substituting the curve into the equation, rather than comparing its derivative with \(F(x,y)\) at an unrelated point.
Worked Example: Use a Zero-Slope Line and Curves
Worked Example: Use a Zero-Slope Line and Curves
Match \(A:y'=x\), \(B:y'=-x\), and \(C:y'=x+y\) to candidate fields. Field I has horizontal segments all along the vertical line \(x=0\). Field II also has horizontal segments along \(x=0\), but its slopes reverse sign on opposite sides of that line compared with Field I. Field III does not have horizontal segments all along \(x=0\).
At \((0,1)\), the equations give:
Therefore Field III must match \(C\): at \((0,1)\) its segment is not horizontal. The test point \((0,1)\) does not distinguish \(A\) from \(B\), so use \((1,0)\):
Field I has positive slopes at \((1,0)\), so it matches \(A\). Field II has negative slopes there, so it matches \(B\). The equations \(A\) and \(B\) have zero slope exactly when \(x=0\); \(C\) has zero slope when \(x+y=0\), not all along \(x=0\).
Consider candidate curves \(g_1(x)=1+\frac{x^2}{2}\) and \(g_2(x)=1-\frac{x^2}{2}\), both through \((0,1)\). Their derivatives are \(g_1'(x)=x\) and \(g_2'(x)=-x\). Thus \(g_1\) satisfies \(A\), and \(g_2\) satisfies \(B\).
Check by substitution along each curve. For \(g_1\), the right-hand side of \(A\) is \(x\), equal to \(g_1'(x)\). For \(g_2\), the right-hand side of \(B\) is \(-x\), equal to \(g_2'(x)\). At \(x=0\), both derivatives are \(0\), agreeing with the horizontal field segments at their common initial point.
Matching a Curve Is a Point-by-Point Check
A candidate curve may pass through an initial point and still belong to the wrong equation. It may even have the correct tangent at that initial point, yet fail elsewhere. Verification along the curve avoids both errors: substitute \(y=g(x)\) into the right-hand side and compare the result with \(g'(x)\) as an expression in \(x\).
Worked Example: Separate Similar Equations
Worked Example: Separate Similar Equations
Match \(A:y'=x-y\), \(B:y'=y-x\), and \(C:y'=xy\) with the curves below, each starting at \((0,1)\):
- \(g_1(x)=x-1+2e^{-x}\)
- \(g_2(x)=x+1\)
- \(g_3(x)=e^{x^2/2}\)
First check the initial values. \(g_1(0)=0-1+2=1\), \(g_2(0)=1\), and \(g_3(0)=e^0=1\). All three pass through \((0,1)\), so this check alone cannot select the matches.
For \(g_1\), differentiate:
Substitute into the right-hand side of \(A\):
Thus \(g_1\) matches \(A\). At the initial point, the equation predicts slope \(0-1=-1\), and \(g_1'(0)=1-2=-1\), confirming the initial tangent as well.
For \(g_2\), \(g_2'(x)=1\), and the right-hand side of \(B\) is \(g_2(x)-x=(x+1)-x=1\). So \(g_2\) matches \(B\). At \((0,1)\), both the predicted slope and the curve’s derivative are \(1\).
For \(g_3\), the chain rule gives \(g_3'(x)=xe^{x^2/2}\). The right-hand side of \(C\) along the curve is \(xg_3(x)=xe^{x^2/2}\). The expressions agree, so \(g_3\) matches \(C\). At \(x=0\), both give slope \(0\).
Each match passes two checks: the curve has the required initial value, and its derivative equals the equation’s right-hand side along the entire curve.
A Common Mistake: Checking Only the Initial Point
Suppose someone claims that \(y=1+x\) solves \(y'=y-x\) with initial value \(y(0)=1\). The initial value is correct: \(1+0=1\). The tangent at the initial point also agrees, since the equation gives \(y'=1-0=1\), and the candidate curve has derivative \(1\).
But that is not enough. Along the candidate curve, the equation’s right-hand side is:
Here the derivatives happen to agree for every \(x\), so this candidate does solve the equation. To see why a point-only check is unreliable, consider instead the curve \(h(x)=1+x+x^2\). It also has \(h(0)=1\) and \(h'(0)=1\), so it passes both initial-point checks. But \(h'(x)=1+2x\), while the equation requires \(h(x)-x=1+x^2\). These are not equal for all \(x\); for instance, at \(x=1\), they are \(3\) and \(2\). Thus \(h\) is not a solution.
Why These Matches Matter
Matching fields, equations, and curves is a way to connect symbolic and visual information. The equation predicts local slopes; the field displays those predictions; and a solution curve must remain tangent to the displayed directions. Initial values then select which curve is being considered. Keeping those roles distinct prevents a plausible-looking sketch from substituting for a mathematical check.
When choices are provided, a useful order is to eliminate impossible equations using zero-slope locations, distinguish remaining choices with a test point, and only then verify curves and initial values. If two candidates agree at one test point, choose another point where their formulas differ.
Check Your Understanding
Use test points and direct checks to answer these questions.
- For \(y'=2-y\), what is the slope at \((0,2)\), and where does this equation predict horizontal field segments?
- At \((1,-1)\), compare the slopes predicted by \(y'=x+y\) and \(y'=x-y\). What does the result tell you about choosing a test point?
- Does \(y=3e^{-2x}\) satisfy \(y'=-2y\) and the initial condition \(y(0)=3\)? Show both checks.
- A proposed curve passes through the correct initial point and has the correct initial slope. What further check is needed to establish that it is a solution?
- For \(y'=xy\), is the slope zero at every point on \(x=0\), every point on \(y=0\), or both? Use the equation to justify your answer.
