Family of Curves and Their Differential Equations

A general solution containing $n$ arbitrary constants describes not one curve but an entire family of curves. This section runs the relationship in reverse: starting from a family such as $y = c x^{2}$ or $x^{2} + y^{2} = r^{2}$ , we eliminate the arbitrary constants by differentiation and recover the differential equation that the whole family satisfies.

Quick Reference

Idea	Statement
Family of curves	A collection of curves described by one equation containing one or more arbitrary constants
Parameter count	A family with $n$ arbitrary constants is called an $n$ -parameter family
Main result	An $n$ -parameter family generally satisfies an ODE of order $n$
Elimination method	Differentiate the family equation $n$ times, then eliminate the $n$ constants algebraically
Reverse direction	Solving an $n$ th order ODE produces an $n$ -parameter family, its general solution
Orthogonal trajectories	Replace y' by -1/y' in the family's differential equation and solve

What Is a Family of Curves?

A family of curves is a collection of curves described by a single equation containing one or more arbitrary constants, called parameters. Assigning a value to each parameter selects one particular curve from the family.

The equation

y = c x^{2}

is a one-parameter family. Taking $c = 1$ gives the parabola $y = x^{2}$ , taking $c = - 2$ gives the downward parabola $y = - 2 x^{2}$ , and taking $c = 0$ gives the $x$ -axis. Together, all choices of $c$ sweep out a whole page of parabolas sharing a vertex at the origin.

Likewise

x^{2} + y^{2} = r^{2}

is a one-parameter family of concentric circles centered at the origin, and

y = c_{1} e^{x} + c_{2} e^{- x}

is a two-parameter family.

The connection with differential equations is immediate. The general solution of an $n$ th order equation is exactly an $n$ -parameter family of curves, and each individual solution curve is one member of that family.

Finding the Differential Equation of a Family

Elimination of Arbitrary Constants. Given an $n$ -parameter family $Φ (x, y, c_{1}, \dots, c_{n}) = 0$ , differentiate the equation $n$ times with respect to $x$ . This produces $n + 1$ equations in total. Use them to eliminate the $n$ constants algebraically. What remains is a differential equation of order $n$ satisfied by every member of the family.

In words: each differentiation gives you one more equation, and you need exactly as many extra equations as you have constants to get rid of. This is why an $n$ -parameter family yields an equation of order $n$ , no more and no less.

Find the differential equation of the family $y = c x^{2}$ .

Solution

There is one arbitrary constant, so differentiate once:

y' = 2cx .

Now eliminate $c$ . From the original equation, $c = \frac{y}{x^{2}}$ for $x \neq 0$ . Substituting,

y' = 2x \cdot \frac{y}{x^2} = \frac{2y}{x} .

The differential equation of the family is

x y' = 2y \qquad \text{or} \qquad y' = \frac{2y}{x} .

As expected, one constant produced a first-order equation. Notice that the constant $c$ has vanished completely: the final equation involves only $x$ , $y$ , and y', which is what makes it a differential equation rather than a family.

Find the differential equation of the family of circles centered at the origin, $x^{2} + y^{2} = r^{2}$ .

Solution

One parameter, so differentiate once implicitly:

2x + 2y y' = 0 .

The constant $r$ has already disappeared, since it was on the right-hand side as a constant. Dividing by $2$ ,

x + y y' = 0 \qquad \text{or} \qquad y' = -\frac{x}{y} .

The geometric reading is pleasant: at any point of a circle centered at the origin, the tangent slope y' is the negative reciprocal of the radius slope $y / x$ , which is exactly the statement that the tangent is perpendicular to the radius.

Find the differential equation of the two-parameter family $y = c_{1} e^{x} + c_{2} e^{- x}$ .

Solution

Two constants, so differentiate twice:

\begin{aligned} y &= c_1 e^{x} + c_2 e^{-x} \\ y' &= c_1 e^{x} - c_2 e^{-x} \\ y'' &= c_1 e^{x} + c_2 e^{-x} \end{aligned}

The third line is identical to the first, so we can eliminate both constants at once:

y'' = y \qquad \text{or} \qquad y'' - y = 0 .

Two constants gave a second-order equation, as promised. This example also runs the argument backwards: the general solution of y'' - y = 0 is $y = c_{1} e^{x} + c_{2} e^{- x}$ .

Find the differential equation of the family of all straight lines $y = m x + b$ .

Solution

There are two parameters, $m$ and $b$ , so differentiate twice:

y' = m, \qquad y'' = 0 .

The differential equation is simply

y'' = 0 .

This says exactly what it should: a curve has zero curvature in this sense precisely when it is a straight line. Integrating twice recovers $y = m x + b$ with two constants of integration, closing the loop.

Find the differential equation of the family $y = c_{1} x + c_{2} x^{2}$ .

Solution

Two parameters, so differentiate twice:

\begin{aligned} y &= c_1 x + c_2 x^2 \\ y' &= c_1 + 2c_2 x \\ y'' &= 2c_2 \end{aligned}

From the last equation, c_2 = \dfrac{y''}{2}. Substituting into the middle equation,

c_1 = y' - 2x \cdot \frac{y''}{2} = y' - x y'' .

Now substitute both into the original equation:

\begin{aligned} y &= \left(y' - x y''\right)x + \frac{y''}{2}\, x^2 \\ &= x y' - x^2 y'' + \frac{x^2 y''}{2} \\ &= x y' - \frac{x^2 y''}{2} \end{aligned}

Multiplying through by $2$ and rearranging gives

x^2 y'' - 2x y' + 2y = 0 ,

a Cauchy-Euler equation, treated in detail in the higher order chapter.

Counting Parameters Correctly

The rule "an $n$ -parameter family gives an $n$ th order equation" assumes the constants are genuinely independent. Constants that can be merged do not count separately.

How many essential parameters does the family $y = A e^{x + B}$ contain?

Solution

Write $A e^{x + B} = A e^{B} e^{x}$ . Since $A$ and $B$ appear only through the single product $A e^{B}$ , setting $c = A e^{B}$ collapses the family to

y = c e^{x} .

There is only one essential parameter, so the differential equation has order $1$ , not $2$ . Differentiating once gives y' = ce^{x} = y, so the equation is

y' = y .

The lesson is to simplify the family algebraically before counting constants.

Orthogonal Trajectories

Once you can write down the differential equation of a family, a classical geometric problem becomes routine.

Given a family of curves, its orthogonal trajectories are the curves that intersect every member of the family at right angles.

Orthogonal trajectories appear throughout physics, where field lines and equipotential curves always meet at right angles: electric field lines cross equipotentials, streamlines cross lines of constant velocity potential, and heat flow lines cross isotherms.

Method. Find the differential equation y' = f(x,y) of the given family. Replace y' by -\dfrac{1}{y'}, giving -\dfrac{1}{y'} = f(x,y), that is, y' = -\dfrac{1}{f(x,y)}. Solve this new equation. Its solutions are the orthogonal trajectories.

The replacement encodes the perpendicularity condition from analytic geometry: two lines are perpendicular exactly when their slopes are negative reciprocals.

Find the orthogonal trajectories of the family of circles $x^{2} + y^{2} = r^{2}$ .

Solution

We found above that this family satisfies y' = -\dfrac{x}{y}. Replacing y' by -\dfrac{1}{y'}:

-\frac{1}{y'} = -\frac{x}{y} \quad \Longrightarrow \quad y' = \frac{y}{x} .

This is separable. Writing $\frac{d y}{y} = \frac{d x}{x}$ and integrating,

\ln | y | = \ln | x | + k ⟹ y = m x .

The orthogonal trajectories are the straight lines through the origin, which is geometrically obvious: a radius meets its circle at a right angle at every point.

Exercises

Find the differential equation of the family $y = c e^{- 3 x}$ .

Solution

One parameter, so differentiate once:

y' = -3c e^{-3x} = -3y .

The differential equation is y' + 3y = 0.

Find the differential equation of the family of parabolas $y^{2} = 4 a x$ with vertex at the origin and axis along the $x$ -axis.

Solution

Differentiate implicitly:

2y y' = 4a \quad \Longrightarrow \quad a = \frac{y y'}{2} .

Substitute into the original equation:

y^2 = 4x \cdot \frac{y y'}{2} = 2x y y' .

Dividing by $y$ , assuming $y \neq 0$ ,

y = 2x y' \qquad \text{or} \qquad 2x y' - y = 0 .

Find the differential equation of the family of circles of radius $1$ with centers on the $x$ -axis, $(x - h)^{2} + y^{2} = 1$ .

Solution

Differentiate implicitly:

2(x-h) + 2y y' = 0 \quad \Longrightarrow \quad x - h = -y y' .

Substitute back into the original equation:

\left(-y y'\right)^2 + y^2 = 1 \quad \Longrightarrow \quad y^2 \left(y'\right)^2 + y^2 = 1 .

This is a first-order equation, as expected from the single parameter $h$ , but its degree is $2$ . Nonlinear equations like this one are where singular solutions live: here $y = 1$ and $y = - 1$ both satisfy the equation and are the envelopes of the family.

Find the orthogonal trajectories of the family $y = c x$ .

Solution

Differentiating gives y' = c, and substituting $c = y / x$ gives the family's equation y' = \dfrac{y}{x}.

Replace y' by -\dfrac{1}{y'}:

-\frac{1}{y'} = \frac{y}{x} \quad \Longrightarrow \quad y' = -\frac{x}{y} .

Separating and integrating, $y d y = - x d x$ gives $\frac{y^{2}}{2} = - \frac{x^{2}}{2} + k$ , that is,

x^{2} + y^{2} = r^{2} .

The orthogonal trajectories are circles centered at the origin, the mirror image of the earlier example.

Frequently Asked Questions

Why does an

n

-parameter family give an

n

th order differential equation?

Each differentiation supplies one new equation while introducing one new derivative. Starting from the family equation and differentiating $n$ times gives $n + 1$ equations containing the $n$ unknown constants. That is exactly enough information to eliminate all $n$ constants, and what survives contains derivatives up to order $n$ .

Is the differential equation of a family unique?

Up to algebraic rearrangement, yes. You may obtain x y' = 2y or y' = 2y/x or x y' - 2y = 0, but these are the same equation written differently. What is not unique is the reverse direction in the nonlinear case: a differential equation may admit singular solutions that belong to no member of the original family.

What if two constants can be combined into one?

Then they were never two independent parameters. Simplify first. In $y = A e^{x + B}$ the constants merge into $c = A e^{B}$ , leaving a one-parameter family and a first-order equation. Counting the symbols on the page rather than the essential parameters is the most common mistake in these problems.

How is this related to the general solution of an ODE?

They are two directions of the same relationship. Solving an $n$ th order equation produces an $n$ -parameter family, its general solution. Eliminating the constants from an $n$ -parameter family produces the $n$ th order equation. Working the elimination direction is a good way to check that a claimed general solution really belongs to the equation you started with.

Why do orthogonal trajectories use -1/y'?

Because perpendicular lines have slopes that are negative reciprocals. If a curve of the original family passes through a point with slope $m$ , a curve crossing it at right angles must have slope $- 1 / m$ there. Substituting -1/y' for y' in the family's differential equation imposes that condition at every point simultaneously.

Family of Curves

Quick Reference

What Is a Family of Curves?

Finding the Differential Equation of a Family

Counting Parameters Correctly

Orthogonal Trajectories

Exercises

Frequently Asked Questions

Quick Links

Social Media