From a Differential Equation to a Slope
A first-order differential equation relates an unknown function to its first derivative. In the common form \(y'=f(x,y)\), the right-hand side gives a rule for finding a slope from the coordinates \(x\) and \(y\).
For instance, if the equation is \(y'=2x-y+1\), then at the point with coordinates \((0,0)\), the rule assigns slope \(1\). At \((1,3)\), it assigns slope \(0\). The equation is therefore more than a statement about a single number: it specifies a slope at every point where its right-hand side is defined.
The coordinates matter in a particular way: the first input to \(f\) is \(x\), and the second is \(y\). Along a solution graph, the point at input \(x\) has coordinates \((x,g(x))\). The equation says that the graph’s derivative there must equal the value of \(f\) at those coordinates.
Why the Right-Hand Side Gives the Tangent Slope
The derivative of a differentiable function at \(x_0\) is the slope of its tangent line at that input. In limit form, it is the limit of the slopes of secant lines between \((x_0,g(x_0))\) and nearby points on the graph:
provided this limit exists. If \(g\) is a solution of \(y'=f(x,y)\), then its derivative at \(x_0\) must satisfy the differential equation. Substituting the point on the graph, whose coordinates are \((x_0,g(x_0))\), gives the required slope.
This is the reason the differential equation acts as a slope rule: the derivative describes the tangent slope, and the right-hand side specifies what that derivative must be at the point on the graph.
Use the coordinates \((x_0,y_0)\) where the slope is requested.
Substitute \(x_0\) for \(x\) and \(y_0\) for \(y\) in \(f(x,y)\).
The resulting number is the derivative required at that point for a solution graph passing through it.
Worked Examples: Evaluating the Slope Rule
Worked Example: A Linear Expression in the Coordinates
Suppose \(y'=2x-y+1\). Find the slope assigned at each of the points \((0,0)\), \((1,3)\), and \((-1,2)\).
Here the rule is \(f(x,y)=2x-y+1\). At \((0,0)\), substitute \(x=0\) and \(y=0\):
The assigned slope at \((0,0)\) is \(1\). Check by putting the same coordinates into the right-hand side: it evaluates to \(1\).
At \((1,3)\):
The assigned slope at \((1,3)\) is \(0\), as confirmed by the substitution.
At \((-1,2)\):
The assigned slope at \((-1,2)\) is \(-3\). Re-evaluating \(2x-y+1\) at \(x=-1\), \(y=2\) gives \(-2-2+1=-3\), so the calculation checks.
Each answer describes the derivative required at its own point. None of the three slope values alone specifies an entire function.
Worked Example: A Quotient as the Slope Rule
Suppose \(y'=\dfrac{x+y}{1+x^2}\). Find the assigned slopes at \((2,1)\), \((-1,3)\), and \((0,-2)\).
The rule is \(f(x,y)=\dfrac{x+y}{1+x^2}\). Its denominator is positive for every real \(x\), since \(1+x^2\geq 1\), so these substitutions are defined.
At \((2,1)\), the numerator is \(2+1=3\) and the denominator is \(1+2^2=1+4=5\). Thus:
The slope assigned there is \(\frac{3}{5}\). Substitution into the original rule gives the same quotient, confirming the result.
At \((-1,3)\), the numerator is \(-1+3=2\) and the denominator is \(1+(-1)^2=2\):
The assigned slope is \(1\), and the original rule at \((-1,3)\) evaluates to \(1\).
At \((0,-2)\), the numerator is \(0+(-2)=-2\), while the denominator is \(1+0^2=1\):
The assigned slope is \(-2\). Substituting the coordinates again gives \(-2/1=-2\), as required.
Worked Example: Substituting Symbolic Coordinates
Suppose \(y'=y^2-x\). First express the slope assigned at an arbitrary point \((a,b)\), then evaluate it at \((2,3)\), \((2,-3)\), and \((1,0)\).
Replace \(x\) by \(a\) and \(y\) by \(b\). The general pointwise rule is:
At \((2,3)\), \(b=3\) and \(a=2\), so:
The assigned slope is \(7\), which checks because \(y^2-x=9-2=7\) at that point.
At \((2,-3)\):
This also gives slope \(7\); substitution into the original expression confirms it. The result illustrates that different points can receive the same slope.
At \((1,0)\):
The assigned slope is \(-1\), and direct substitution gives the same value. These evaluations apply the rule at particular points; they do not by themselves determine a solution function.
A Slope at a Point Is Not a Solution
A differential equation gives a condition that a solution function must satisfy throughout an interval. Evaluating its right-hand side at one point gives only one required derivative value. It does not provide all the function’s values or derivatives elsewhere.
For example, the equation \(y'=2x-y+1\) requires derivative \(1\) at \((0,0)\). That statement is local: it specifies what the derivative must be if a solution graph passes through \((0,0)\). To be a solution on an interval, a function must also satisfy the equation at every other input in that interval. A single pointwise calculation cannot establish that broader condition.
The converse requires care: prescribing the slope at one point does not prove that a differentiable function satisfies the equation over an interval. The equation is a continuing requirement, not just a list of isolated slope calculations.
Common Pitfall: Reversing the Coordinates
In \(y'=x+2y\), the first coordinate is substituted for \(x\), and the second is substituted for \(y\). At \((1,2)\), a common error is to reverse them and calculate \(2+2(1)=4\). That uses \(x=2\) and \(y=1\), which are the coordinates of a different point.
The correct substitution is \(x=1\), \(y=2\):
Thus the required slope at \((1,2)\) is \(5\). A quick check is to write the coordinate assignments before evaluating: first coordinate \(x=1\), second coordinate \(y=2\). Keeping the order explicit prevents the mistake.
Why the Slope-Rule View Matters
Reading a differential equation this way gives a precise interpretation before attempting any further analysis. The expression \(f(x,y)\) is evaluated at a point, and its value is the derivative required there. This viewpoint separates three ideas that are easy to confuse:
- The equation is a rule that specifies derivatives from coordinates.
- A pointwise evaluation gives the required slope at one point.
- A solution is a differentiable function satisfying the rule throughout an interval.
These distinctions make it easier to interpret equations accurately and to understand what a numerical slope calculation does—and does not—tell you. The essential habit is simple: identify the point, substitute its coordinates in the stated order, and interpret the output as a local derivative.
Check Your Understanding
For each question, interpret the right-hand side as a slope rule and show the coordinate substitutions.
- For \(y'=3x+y\), what slope is assigned at \((2,-1)\)?
- For \(y'=x^2-y\), what slope is assigned at \((-2,1)\)?
- In \(y'=f(x,y)\), what does \(f(x_0,y_0)\) represent at the point \((x_0,y_0)\)?
- Why does finding the assigned slope at one point not establish that a function is a solution on an interval?
- For \(y'=x+2y\), identify the correct values of \(x\) and \(y\) at \((3,1)\) before evaluating the slope rule.
