Why Compare a Formula With a Direction Field?
A formula and a direction field describe a differential equation in different ways. The formula gives a particular curve; the field displays the slopes that solution curves must have at each point. Comparing them is a practical way to check whether a proposed solution is consistent with the equation and its initial condition.
The comparison is useful even when a formula looks plausible. A curve might pass through the correct initial point but have the wrong tangent slope there. It might match the field near the initial point but have a domain that stops too soon. Or its overall increasing and decreasing pattern might conflict with the slopes displayed across the field.
A direction field is a visual guide, not a replacement for calculation. A careful comparison combines what the field suggests with a direct check of the formula.
What the Direction Field Represents
Consider a first-order differential equation written in the form \(y'=f(x,y)\), where \(f\) is defined at the points under consideration. At a point \((x,y)\), the equation assigns the slope \(f(x,y)\). A direction field represents these slopes with short line segments: the segment drawn at \((x,y)\) has slope \(f(x,y)\).
To compare a formula with a direction field, check three things. First, does its graph pass through the specified initial point? Second, does its tangent slope agree with the field at that point and along the proposed curve? Third, do its domain and qualitative behavior remain consistent with the field where the solution is claimed to exist?
The slope comparison has an exact test. At each point \((x,\phi(x))\) on the graph, the field prescribes slope \(f(x,\phi(x))\). The graph’s actual slope is \(\phi'(x)\). They agree precisely when
Why the Slope Test Works
The graph of a differentiable function has tangent direction represented by the vector \((1,\phi'(x))\): as \(x\) changes by a small amount, the corresponding change in height is governed by \(\phi'(x)\). The differential equation assigns the direction \((1,f(x,\phi(x)))\) to the field at that same point.
Both direction vectors have first component \(1\), so they point in the same direction exactly when their second components agree. Thus the graph follows the field at every point precisely when \(\phi'(x)=f(x,\phi(x))\). This explains why visual alignment is the right idea, while differentiation provides the decisive check.
For an implicit formula, the graph may be given by a relation \(F(x,y)=0\) rather than an explicit expression for \(y\). On a differentiable branch, differentiate the relation to obtain its slope, then compare that slope with \(f(x,y)\). The check applies only where the branch is defined and the derivative calculation gives a valid finite slope.
A Practical Comparison
Substitute the initial \(x\)-value into the formula and check that the resulting point is the one specified.
Differentiate the formula and evaluate its slope at the initial point. Separately evaluate \(f(x,y)\) there.
Substitute the formula into the differential equation and verify the equality throughout the claimed interval.
Check whether the field’s slopes agree with the formula’s increasing or decreasing behavior and whether the domain is appropriate.
Matching the initial point and the slope at just one point is necessary for an initial-value solution, but it is not sufficient. The differential equation must hold along the whole interval on which the formula is claimed to be a solution.
Worked Examples
Worked Example: An Explicit Solution With a Turning Point
Consider the equation \(y'=x-y\) with initial condition \(y(0)=1\). Check the proposed formula \(\phi(x)=x-1+2e^{-x}\) against the equation and describe its behavior in relation to the field.
First check the initial point:
Thus the graph passes through \((0,1)\). Differentiate the formula:
Now evaluate the right-hand side of the differential equation along the proposed graph:
Therefore \(\phi'(x)=x-\phi(x)\) for every real \(x\), so the formula satisfies the differential equation on \(\mathbb{R}\) and the initial condition. At the initial point, the field slope is \(0-1=-1\), and the formula’s slope is \(\phi'(0)=1-2=-1\).
The field has horizontal segments where \(x-y=0\), or \(y=x\). The proposed curve has a horizontal tangent when \(1-2e^{-x}=0\). Solving gives \(e^{-x}=1/2\), hence \(x=\ln 2\). Its height there is \(\phi(\ln 2)=\ln 2-1+1=\ln 2\). For \(x<\ln 2\), \(e^{-x}>1/2\), so \(\phi'(x)<0\); for \(x>\ln 2\), \(\phi'(x)>0\). The curve decreases, reaches a minimum at \((\ln 2,\ln 2)\), and then increases, as the field’s slopes along it indicate.
As a final check, substituting \(x=\ln 2\) into the equation’s right side gives \(\ln 2-\ln 2=0\), matching the derivative there.
Worked Example: Checking an Implicit Solution Branch
Consider \(y'=x/y\) and the implicit relation \(y^2-x^2=1\). Check the branch through the initial point \((0,1)\), and compare its behavior with the direction field.
At \(x=0\), the relation gives \(y^2=1\). The specified initial value \(y(0)=1\) selects the upper branch. Solving for this branch gives \(y=\sqrt{1+x^2}\), which is defined for every real \(x\) and is positive.
Differentiate the implicit relation along this branch:
Since \(y=\sqrt{1+x^2}>0\), division by \(2y\) is valid, and
This is the differential equation, so the branch satisfies it wherever it is defined. It also meets the initial condition because \(\sqrt{1+0^2}=1\). At \((0,1)\), the field slope is \(0/1=0\), agreeing with the horizontal tangent obtained from the relation.
For \(x>0\), the field slope \(x/y\) is positive on the upper branch; for \(x<0\), it is negative. The formula reflects this: \(y'=x/\sqrt{1+x^2}\) is negative to the left of zero and positive to the right. The branch reaches its minimum value \(1\) at \(x=0\). The denominator in the equation is never zero on this branch, so the formula encounters no point on its domain where the equation is undefined.
As a pointwise check away from the initial point, at \(x=1\) the branch has \(y=\sqrt{2}\). Its slope is \(1/\sqrt{2}\), and the field prescribes \(x/y=1/\sqrt{2}\), as required.
Worked Example: A Field Whose Slopes Depend Only on \(x\)
Consider \(y'=x^2\) with initial condition \(y(0)=2\). Check the proposed formula \(\phi(x)=x^3/3+2\), then interpret what the direction field shows.
At the initial value,
so the graph passes through \((0,2)\). Its derivative is
The differential equation’s right-hand side is \(x^2\), independent of \(y\). Hence \(\phi'(x)=x^2=f(x,\phi(x))\) for every real \(x\). At the initial point, both the formula and the field give slope \(0^2=0\).
Every segment in a fixed vertical column of this direction field has the same slope, because changing \(y\) does not change \(x^2\). The slopes are nonnegative everywhere and are zero only at \(x=0\). Correspondingly, \(\phi'(x)\geq 0\) for every \(x\), and \(\phi'(x)>0\) when \(x\neq 0\). The curve has a horizontal tangent at the initial point and increases on either side of it.
Finally, substituting the formula into the right-hand side gives \(x^2\), exactly the derivative computed above. This checks the equation across its entire domain, not just at the initial point.
A Common Mistake: Checking Only the Initial Point
A candidate can pass through the correct initial point and still fail the differential equation. For example, take the candidate \(\phi(x)=x-1+2e^{-x}\) with initial condition \(\phi(0)=1\), but compare it with the different equation \(y'=x+y\). The initial point is correct, since \(\phi(0)=1\). However, its derivative is \(\phi'(x)=1-2e^{-x}\), while the equation’s right-hand side along the curve is
At \(x=0\), the formula’s slope is \(-1\), but the field prescribes \(0+1=1\). The slopes disagree, so the candidate is not a solution of this initial-value problem. The error is not repaired by the correct initial point: a solution must match both the initial condition and the differential equation.
How to Use Qualitative Behavior Carefully
A direction field helps reveal patterns that may be hard to see from a formula alone. If the segments along a curve tilt upward as \(x\) increases, the solution should be increasing there; if they tilt downward, it should be decreasing. A row or curve of horizontal segments signals locations where the differential equation prescribes zero slope. These observations can help identify a formula that is inconsistent with the field.
Qualitative agreement is a useful consistency check, but it is weaker than direct verification. A sketch has only finitely many drawn segments and may not show exact slopes. A curve can also appear to fit a coarse field while failing the differential equation algebraically. When a formula is available, calculate its derivative and compare it with the right-hand side.
Pay attention to the domain as well. The equation may be undefined at some points, and an implicit relation may describe multiple branches. A proposed solution must stay on a branch where its derivative and the differential equation are both defined. In the implicit example above, the upper branch remains away from \(y=0\), so division by \(y\) is valid throughout that branch.
Check Your Understanding
Use the definition of a solution and the direction-field comparison method to answer these questions.
- For \(y'=f(x,y)\), what slope does the direction field assign at the point \((x_0,y_0)\)?
- Why does matching the initial point and initial slope not, by itself, establish that a formula solves an initial-value problem?
- For \(y'=x-y\), what slope does the field prescribe at \((2,1)\)?
- When checking an implicit relation, why must you specify a branch and consider where division in implicit differentiation is valid?
- What is the difference between qualitative agreement with a sketch and direct verification of a proposed formula?
