From a Family of Curves to a Differential Equation
A first-order differential equation describes a relationship between \(x\), a function \(y(x)\), and its first derivative \(y'(x)\). Sometimes we begin not with the equation, but with a proposed family of curves containing an arbitrary constant. We can ask: what differential equation does every curve in that family satisfy?
The main idea is to differentiate a curve while treating its parameter as fixed, then eliminate the parameter from the original family relation and the differentiated relation. The result is an equation involving \(x\), \(y\), and \(y'\), but not the arbitrary constant.
The resulting equation may be explicit, such as \(y'=f(x,y)\), or implicit, such as \(G(x,y,y')=0\). An implicit equation is often the safer form: solving it for \(y'\) can require division, which may exclude points or branches.
The Differentiation and Elimination Principle
Suppose the family is given explicitly as \(y=\phi(x,C)\). Along any one curve, \(C\) is fixed, so differentiation with respect to \(x\) gives \(y'=\frac{\partial \phi}{\partial x}(x,C)\). The original equation and this derivative equation form a pair of relations involving \(x\), \(y\), \(y'\), and \(C\). Eliminate \(C\) from that pair.
For an implicit family \(F(x,y,C)=0\), differentiate along a differentiable curve \(y=y(x)\). The chain rule gives \(F_x+F_y y'=0\), where \(F_x\) and \(F_y\) are partial derivatives. This relation still contains \(C\), so it must be combined with the original family equation to eliminate it.
Why the Method Works
The justification is the chain rule. A particular member of the family is obtained by choosing one fixed value of \(C\). As \(x\) varies along that member, \(y\) changes, but the chosen \(C\) does not. Thus the derivative records the slope of that curve without introducing a derivative of the parameter.
This equation is necessary for every differentiable curve in the family. Combining it with \(F(x,y,C)=0\) removes the parameter and leaves a differential relation. The result is guaranteed to be satisfied by the original family wherever the calculations are valid. It need not describe only that family: the differential equation may also have other solutions. Therefore, after eliminating the parameter, check the domain and verify the original family directly.
Worked Examples
Worked Example: An Exponential Curve Family
Recover a first-order differential equation from \(y=C e^{2x}\), where \(C\) is any real constant.
Differentiate with \(C\) fixed:
The original family says \(Ce^{2x}=y\). Substituting this into the derivative relation eliminates \(C\):
Check: For any fixed \(C\), differentiating \(y=Ce^{2x}\) gives \(y'=2Ce^{2x}\), while \(2y=2Ce^{2x}\). Hence the family satisfies \(y'=2y\), including the member \(C=0\), for which both sides are zero.
Worked Example: An Implicit Family With Two Branches
Recover an equation from \(y^2-x=C\). The family is implicit, and its curves can have separate branches.
Differentiate both sides with respect to \(x\), holding \(C\) fixed:
This relation no longer contains \(C\). The recovered first-order differential equation can be kept in implicit form, \(2y y'-1=0\). At points where \(y\ne 0\), it can be written explicitly as
Solving explicitly for \(y'\) requires \(y\ne0\). The implicit family can be written \(y=\pm\sqrt{x+C}\) on intervals where the chosen branch is defined and \(y\ne0\). For instance, take \(C=3\) and the positive branch at \(x=1\). Then \(y=\sqrt{1+3}=2\), and differentiation of \(y=\sqrt{x+3}\) gives \(y'=1/(2\sqrt{x+3})=1/4\) there. The recovered equation gives \(1/(2y)=1/(2\cdot2)=1/4\), as required. Substitution into the implicit equation gives \(2(2)(1/4)-1=1-1=0\).
Worked Example: A Rational Family
Recover a differential equation from \(y=\frac{C}{1-Cx}\). Each curve is considered on an interval where \(1-Cx\ne0\).
Differentiate using the quotient rule, with \(C\) constant:
Squaring the original expression gives
Thus the parameter is eliminated and the family satisfies \(y'=y^2\).
Check: Choose \(C=2\) and \(x=0\), where the denominator is \(1-2(0)=1\). The family gives \(y=2/1=2\), and its derivative gives \(y'=2^2/1^2=4\). The differential equation gives \(y^2=2^2=4\), matching the derivative. In general, the derivative and \(y^2\) are both \(C^2/(1-Cx)^2\) throughout the family’s domain.
Keeping Track of What Elimination Establishes
Elimination produces a differential equation that the proposed curves satisfy; it does not automatically prove that the family lists every solution of that equation. The recovered equation can admit additional curves, and algebraic steps can also introduce or exclude points if their conditions are ignored.
When eliminating \(C\), it is useful to distinguish three claims:
- The family satisfies the equation: substitution or differentiation verifies this for every allowed family member.
- The equation describes exactly the family: this requires checking that its solutions correspond to the proposed parameterized curves, not merely that the family members solve it.
- An explicit form is valid: writing \(y'=f(x,y)\) requires that any division used to isolate \(y'\) is permitted on the region under consideration.
For example, the implicit equation \(2y y'-1=0\) can be divided by \(2y\) only where \(y\ne0\). That restriction is not cosmetic: at \(y=0\), the implicit equation would read \(-1=0\), so no differentiable solution of that equation passes through such a point. Stating the implicit equation first makes the scope of the division visible.
Common Pitfall: Dividing Away a Special Point
Consider the family \(y=Cx+1\). Differentiating with \(C\) fixed gives \(y'=C\). Since the original relation can be written \(y-1=Cx\), substituting \(C=y'\) yields the implicit equation
A tempting but incorrect global step is to divide by \(x\) and claim \(y'=(y-1)/x\) for every \(x\). That division is valid only when \(x\ne0\). At \(x=0\), every family member has \(y=1\), but its slope is \(y'=C\), which can be any real number. The implicit equation correctly gives \(0=1-1\) at that point without imposing a slope; the divided equation is undefined there.
Correction: Keep \(xy'=y-1\) as the recovered equation when describing the family across \(x=0\), or state explicitly that the explicit form applies only on intervals where \(x\ne0\). This is why checking restrictions after every division is part of parameter elimination.
A Reliable Procedure
For a new family, the following sequence keeps the reasoning organized.
Identify the arbitrary constant \(C\) and note any restrictions on the curve’s domain.
Use ordinary differentiation for an explicit family, or the chain rule for an implicit relation.
Combine the original and differentiated relations, preserving an implicit equation if division would impose restrictions.
Record where the manipulations apply and substitute the family into the final equation to check it.
Why Recovering the Equation Is Useful
A formula gives individual curves once its parameter is chosen. A differential equation expresses a shared rule for the slopes of those curves. Recovering that rule makes it possible to compare different members of the family through one relation, rather than treating each formula separately.
The procedure is also a useful check on a proposed solution family. If differentiation and elimination do not produce a consistent relation, there may be an algebraic or differentiation error. Conversely, agreement verifies that the family satisfies the recovered equation, while the domain and completeness checks clarify exactly what that verification does—and does not—show.
Check Your Understanding
For each question, pay attention to whether the parameter is held fixed and whether any division has a restriction.
- For \(y=Cx^2\), differentiate with \(C\) fixed and eliminate \(C\) to obtain a first-order differential equation.
- For \(y^2+x=C\), find an implicit differential equation and state the condition needed to solve it explicitly for \(y'\).
- Why must \(C\) remain constant while differentiating one member of a family?
- For \(y=Cx+1\), what information about slopes at \(x=0\) is lost by dividing \(xy'=y-1\) by \(x\)?
- Does showing that every member of a proposed family satisfies a differential equation prove that the family contains all solutions? Explain.
