Differential Equations of Plane Curves

As it is known, the solution of a differential equation is displayed graphically as a family of integral curves. It turns out that one can also solve the inverse problem: construct a differential equation of the family of plane curves defined by an algebraic equation!

Suppose that a family of plane curves is described by the implicit one-parameter equation:
\[F\left( {x,y,C} \right) = 0.\] We assume that the function \(F\) has continuous partial derivatives in \(x\) and \(y.\) To write the corresponding differential equation of first order, it’s necessary to perform the following steps:

1. Differentiate \(F\) with respect to \(x\) considering \(y\) as a function of \(x:\)
   \[\frac{{\partial F}}{{\partial x}} + \frac{{\partial F}}{{\partial y}} \cdot y’ = 0;\]
2. Solve the system of equations:
   \[\left\{ \begin{array}{l}
   \frac{{\partial F}}{{\partial x}} + \frac{{\partial F}}{{\partial y}} \cdot y’ = 0\\
   F\left( {x,y,C} \right) = 0
   \end{array} \right.\] by eliminating the parameter \(C\) from it.

If a family of plane curves is given by the two-parameter equation
\[F\left( {x,y,{C_1},{C_2}} \right) = 0,\] we should differentiate the last formula twice by considering \(y\) as a function of \(x\) and then eliminating the parameters \({{C_1}}\) and \({{C_2}}\) from the system of three equations.

The similar rule is applied to the case of \(n\)-parametric family of plane curves.

Solved Problems

Click on problem description to see solution.

Example 1.

Determine the differential equation for the family of curves defined by the equation \(y = {e^{x + C}}.\)

Solution.

Differentiating the given equation with respect to \(x\) gives:
\[y’ = {e^{x + C}}.\] We can easily eliminate the parameter \(C\) from the system of equations:
\[\left\{ \begin{array}{l}
y’ = {e^{x + C}}\\
y = {e^{x + C}}
\end{array} \right..\] As a result, we obtain the following simplest homogeneous equation:
\[{y’ = y,\;\; }\Rightarrow {y’ – y = 0.}\]

Example 2.

Derive the differential equation for the family of plane curves defined by the equation \(y = {x^2} – Cx.\)

Solution.

We differentiate the implicit equation with respect to \(x:\)
\[y’ = 2x – C.\] Write this equation jointly with the original algebraic equation and eliminate the parameter \(C:\)
\[
{\left\{ \begin{array}{l}
y’ = 2x – C\\
y = {x^2} – Cx
\end{array} \right.,\;\;}\Rightarrow
{C = y’ – 2x,\;\;}\Rightarrow
{y = {x^2} – \left( {y’ – 2x} \right)x,\;\;}\Rightarrow
{y = {x^2} – y’x + 2{x^2},\;\;}\Rightarrow
{y’x + y = 3{x^2}.}
\] As a result, we obtain the implicit differential equation corresponding to the given family of plane curves.