Differential Equations of Plane Curves
Solved Problems
Example 3.
Write the corresponding differential equation for the family of plane curves defined by the equation \[y = \cot \left( {x - C} \right).\]
Solution.
By differentiating the given equation with respect to \(x,\) we obtain:
\[y' = - \frac{1}{{{{\sin }^2}\left( {x - C} \right)}}.\]
Notice that
\[1 + {y^2} = 1 + {\cot ^2}\left( {x - C} \right) = 1 + \frac{{{{\cos }^2}\left( {x - C} \right)}}{{{\sin^2}\left( {x - C} \right)}} = \frac{{{\sin^2}\left( {x - C} \right) + {{\cos }^2}\left( {x - C} \right)}}{{{\sin^2}\left( {x - C} \right)}} = \frac{1}{{{\sin^2}\left( {x - C} \right)}}.\]
Therefore we can write:
\[y' = - \left( {1 + {y^2}} \right).\]
Hence, we get the following differential equation describing the given family of curves:
\[y' = - 1 - {y^2},\;\; \Rightarrow y' + {y^2} = - 1.\]
Now we substitute this into the differential equation:
Example 4.
A family of curves is given by the expression \[y = \frac{1}{C} \cos \left( {Cx + \alpha } \right),\] where \(C\) is a parameter, \(\alpha\) is an arbitrary angle. Determine the differential equation for this family of plane curves.
Solution.
First, we differentiate the equation with respect to the variable \(x\) assuming that \(y\) is a function of \(x:\)
\[y' = \frac{1}{C}\left[ { - \sin\left( {Cx + \alpha } \right)} \right] \cdot C = - \sin\left( {Cx + \alpha } \right).\]
Eliminate \(C\) from the system of two equations:
\[\left\{ \begin{array}{l}
y' = - \sin\left( {Cx + \alpha } \right)\\
y = \frac{1}{C}\cos\left( {Cx + \alpha } \right)
\end{array} \right..\]
To make this we square both sides of the equation and then add them together:
\[
\left\{ \begin{array}{l}
{\left( {y'} \right)^2} = {\sin^2}\left( {Cx + \alpha } \right)\\
{y^2} = \frac{1}{{{C^2}}}{\cos^2}\left( {Cx + \alpha } \right)
\end{array} \right.,\;\; \Rightarrow
{\left( {y'} \right)^2} + {C^2}{y^2} = 1,\;\; \Rightarrow
{C^2}{y^2} = 1 - {\left( {y'} \right)^2},\;\; \Rightarrow
{C^2} = \frac{{1 - {{\left( {y'} \right)}^2}}}{{{y^2}}},\;\; \Rightarrow
C = \frac{{\sqrt {1 - {{\left( {y'} \right)}^2}} }}{y}.\]
Inserting the found expression for \(C\) into the differential equation, we have:
\[y' = - \sin \left( {Cx + \alpha } \right) = - \sin \left( {\frac{{x\sqrt {1 - {{\left( {y'} \right)}^2}} }}{y} + \alpha } \right).\]
Thus, the family of plane curves is described by the following implicit differential equation:
\[y' = - \sin \left( {\frac{{x\sqrt {1 - {{\left( {y'} \right)}^2}} }}{y} + \alpha } \right).\]
Example 5.
Derive the differential equation for the family of two-parameter plane curves \[y = {C_1}{x^2} + {C_2}x.\]
Solution.
We differentiate the given equation twice with respect to \(x\) and write the following system of three equations:
\[\left\{ \begin{array}{l}
y = {C_1}{x^2} + {C_2}x\\
y' = 2{C_1}x + {C_2}\\
y^{\prime\prime} = 2{C_1}
\end{array} \right..\]
Express \({C_1}\) from the last equation and substitute it into the first two equations:
\[
{C_1} = \frac{{y^{\prime\prime}}}{2},\;\; \Rightarrow
\left\{ {\begin{array}{*{20}{l}}
{y = \frac{{y^{\prime\prime}}}{2}{x^2} + {C_2}x}\\
{y' = y^{\prime\prime}x + {C_2}}
\end{array}} \right..\]
Now we can express \({C_2}\) in terms of the derivatives of \(y\) and put this into the first equation to get the final differential equation:
\[{C_2} = y' - y^{\prime\prime}x,\;\; \Rightarrow y = \frac{{y^{\prime\prime}}}{2}{x^2} + \left( {y' - y^{\prime\prime}x} \right)x,\;\; \Rightarrow y = \frac{{y^{\prime\prime}}}{2}{x^2} + y'x - y^{\prime\prime}{x^2},\;\; \Rightarrow y = y'x - \frac{{y^{\prime\prime}}}{2}{x^2},\;\; \Rightarrow 2y = 2y'x - y^{\prime\prime}{x^2},\;\; \Rightarrow y^{\prime\prime}{x^2} - 2y'x + 2y = 0.\]
