Why Slope Signs Matter
A differential equation of the form \(y'=f(x,y)\) assigns a slope to each point \((x,y)\) in its domain. The sign of that slope tells whether the tangent direction at that point points upward, downward, or horizontally as \(x\) increases.
Reading these signs helps you interpret a slope field without calculating a complete solution. It can show where a solution curve is locally rising or falling and where the field assigns horizontal tangent segments. The sign must be checked in each region: a zero-slope location does not by itself guarantee that the signs on its two sides differ.
Reading the Sign of the Right-Hand Side
At a point \((x_0,y_0)\), the differential equation assigns the slope \(f(x_0,y_0)\). If this number is positive, the small line segment drawn at that point rises from left to right. If it is negative, the segment falls from left to right. If it is zero, the segment is horizontal.
These descriptions concern the tangent direction as \(x\) increases. They do not mean that every solution curve rises or falls everywhere: the assigned sign may depend on both \(x\) and \(y\), and a curve may pass through regions with different signs.
The collection of all zero-slope locations is found by solving \(f(x,y)=0\). It may be a line, several curves, isolated points, or an empty set, depending on the equation. To classify nearby regions, test the sign of \(f\) there; do not infer the sign from the zero set alone.
Worked Example: A Line Separating Positive and Negative Slopes
Worked Example: A Line Separating Positive and Negative Slopes
Consider the differential equation \(y'=x-y\). Find the zero-slope locations and determine the slope sign on each side.
Step 1: Find where the slope is zero. Set the right-hand side equal to zero:
Solving gives \(y=x\). Thus every point on the line \(y=x\) has an assigned slope of zero.
Step 2: Determine the signs. If \(y<x\), then \(x-y>0\), so the assigned slope is positive. If \(y>x\), then \(x-y<0\), so the assigned slope is negative.
Step 3: Check representative points. At \((2,1)\), \(f(2,1)=2-1=1>0\). At \((2,3)\), \(f(2,3)=2-3=-1<0\). At \((2,2)\), \(f(2,2)=2-2=0\).
Check: The three substitutions agree with the classifications: below \(y=x\) the slope is positive, above it the slope is negative, and on it the slope is zero.
Worked Example: Testing the Sign of a Product
Worked Example: Testing the Sign of a Product
For \(y'=(x-1)(y+2)\), find the zero-slope locations and classify the signs in the regions they determine.
Step 1: Solve for zero slope. A product is zero when at least one factor is zero:
The zero-slope locations are the vertical line \(x=1\) and the horizontal line \(y=-2\).
Step 2: Check the signs of the factors. For \(x<1\), \(x-1<0\); for \(x>1\), \(x-1>0\). For \(y<-2\), \(y+2<0\); for \(y>-2\), \(y+2>0\). Multiplying the factor signs gives the sign of the slope.
| Region | Sign of \(x-1\) | Sign of \(y+2\) | Sign of slope |
|---|---|---|---|
| \(x<1,\ y<-2\) | Negative | Negative | Positive |
| \(x<1,\ y>-2\) | Negative | Positive | Negative |
| \(x>1,\ y<-2\) | Positive | Negative | Negative |
| \(x>1,\ y>-2\) | Positive | Positive | Positive |
Step 3: Verify with test points. At \((0,-3)\), the slope is \((-1)(-1)=1>0\). At \((0,0)\), it is \((-1)(2)=-2<0\). At \((2,-3)\), it is \((1)(-1)=-1<0\). At \((2,0)\), it is \((1)(2)=2>0\).
Check: Each test value matches the sign table, and substitution into the original expression gives zero whenever \(x=1\) or \(y=-2\).
Why a Sign Test Describes Rising or Falling
The sign of \(f(x,y)\) gives the slope at one point. To make a claim about how a solution changes over an interval, the sign must hold along the solution throughout that interval. The Mean Value Theorem explains this connection.
For the positive case, the Mean Value Theorem gives a number \(c\) with \(a<c<b\) such that
Because \(b-a>0\) and \(y'(c)>0\), it follows that \(y(b)-y(a)>0\), so \(y(b)>y(a)\). For the negative case, \(y'(c)<0\); since \(b-a>0\), the equation gives \(y(b)-y(a)<0\), so \(y(b)<y(a)\). This justifies calling a solution rising or falling on an interval when its derivative has the corresponding strict sign throughout that interval.
For a differential equation, the derivative along a solution is \(y'(x)=f(x,y(x))\). Therefore, a region where \(f\) is positive indicates that any solution segment lying entirely in that region has positive derivative there; a region where \(f\) is negative indicates a negative derivative. A pointwise sign alone does not establish what happens over an entire interval unless the sign remains consistent along the segment.
Worked Example: A Zero-Slope Set Without a Sign Change
Worked Example: A Zero-Slope Set Without a Sign Change
Consider \(y'=y^2\). Find the zero-slope locations and compare the signs above and below them.
Step 1: Solve for zero slope.
Thus the line \(y=0\) consists of zero-slope locations.
Step 2: Check both sides. If \(y>0\), then \(y^2>0\). If \(y<0\), then \(y^2>0\) as well. At \(y=0\), \(y^2=0\).
Step 3: Verify with values. At \(y=2\), the slope is \(2^2=4>0\). At \(y=-2\), it is \((-2)^2=4>0\). At \(y=0\), it is \(0^2=0\).
Check: Substitution confirms that the slope is positive on both sides of the zero-slope line and zero on the line itself. The zero-slope set does not mark a change from positive to negative here.
A Common Mistake: Assuming Every Zero Separates Opposite Signs
A tempting but incorrect shortcut is to assume that wherever \(f(x,y)=0\), the slope must switch from positive to negative or from negative to positive. The example \(y'=y^2\) disproves that assumption: at \(y=0\), the slope is zero, but at both \(y=1\) and \(y=-1\), the slope is positive.
Another error is to classify a whole region from one test point when the expression may change sign inside it. A sign test is useful only after the zero set has been found and the tested region contains no additional zeros that could divide it. For products, factoring often makes the zero set and the sign possibilities clear. For other expressions, solve or analyze the equation \(f(x,y)=0\) carefully before drawing a sign conclusion.
A Reliable Sign-Analysis Process
Identify the right-hand side \(f(x,y)\) in \(y'=f(x,y)\).
Solve \(f(x,y)=0\) for all points in the equation's domain.
Use algebra or test points, and check each region separately.
Positive means an upward tangent as \(x\) increases; negative means a downward tangent; zero means a horizontal tangent.
This process separates two questions that are easy to confuse: where the field assigns a horizontal tangent, and what sign the field assigns nearby. The first comes from solving an equation. The second comes from checking signs.
Why These Locations Are Useful
Zero-slope locations and sign regions provide a compact description of important local features of a slope field. They let you anticipate the direction of tangent segments in different parts of the plane and locate where a solution passing through a point has a horizontal tangent. They also help prevent a common visual mistake: interpreting a line of horizontal segments as necessarily separating upward-pointing segments from downward-pointing ones.
Keep the interpretation precise. The equation determines the slope at each point. A solution curve, if it passes through that point, has the corresponding tangent slope there. Whether the solution rises or falls over a whole interval depends on the sign of its derivative along that interval, not merely on the sign at one isolated point.
Check Your Understanding
For each question, identify the zero-slope locations and use sign checks rather than assuming a sign change.
- For \(y'=x+2y\), find the zero-slope line and determine the sign at \((0,1)\) and \((0,-1)\).
- For \(y'=(y-3)^2\), where is the slope zero, and what is its sign on either side of that location?
- For \(y'=(x+1)(y-2)\), list the zero-slope lines and determine the slope sign in the region \(x> -1,\ y<2\).
- What additional condition is needed before a positive derivative at points of a solution implies that the solution rises throughout an interval?
- Why is it incorrect to claim that every zero-slope location separates positive slopes from negative slopes?
