Differential Equations

1st Order Equations

Orthogonal Trajectories

Page 1
Theory
Page 2
Problems 1-4

Example 1.

Find the orthogonal trajectories of the family of straight lines \(y = Cx,\) where \(C\) is a parameter.

Solution.

We apply the algorithm described above.

\(1)\) First, we construct the differential equation for the family of straight lines \(y = Cx.\) By differentiating the last equation with respect to \(x,\) we get:
\[y’ = C = \text{const}.\] Eiminate the constant \(C\) from the system of equations:
\[{\left\{ \begin{array}{l}
y = Cx\\
y’ = C
\end{array} \right.,\;\; }\Rightarrow {y’ = \frac{y}{x}.}\] We obtain the differential equation of the initial set of straight lines.

\(2)\) Replace \(y’\) with \(\left( { – \large\frac{1}{{y’}}\normalsize} \right).\) This gives the differential equation of the orthogonal trajectories:
\[{ – \frac{1}{{y’}} = \frac{y}{x},\;\; }\Rightarrow {y’ = – \frac{x}{y}.}\] \(3)\) Now we solve the last differential equation to find the algebraic equation of the family of orthogonal trajectories:
\[
{y’ = – \frac{x}{y},\;\;}\Rightarrow
{\frac{{dy}}{{dx}} = – \frac{x}{y},\;\;}\Rightarrow
{ydy = – xdx,\;\;}\Rightarrow
{\int {ydy} = – \int {xdx} ,\;\;}\Rightarrow
{\frac{{{y^2}}}{2} = – \frac{{{x^2}}}{2} + C,\;\;}\Rightarrow
{\frac{{{x^2}}}{2} + \frac{{{y^2}}}{2} = C,\;\;}\Rightarrow
{{x^2} + {y^2} = 2C.}
\] By replacing \(2C\) with \({R^2}\) we see that the orthogonal trajectories for the family of straight lines are concentric circles (Figure \(1\)):
\[{x^2} + {y^2} = {R^2}.\]

Example 2.

A family of hyperbolic curves is given by the equation \(y = {\large\frac{C}{x}\normalsize}.\) Find the orthogonal trajectories for these curves.

Solution.

\(1)\) Determine the differential equation for the given family of hyperbolas. Differentiating the equation with respect to \(x\) gives:
\[y’ = – \frac{C}{{{x^2}}}.\] Now we eliminate the parameter \(C\) from the system of two equations:
\[\left\{ \begin{array}{l}
y = \frac{C}{x}\\
y’ = – \frac{C}{{{x^2}}}
\end{array} \right..\] It follows from the first equation that \(C = xy.\) Substituting into the second equation yields:
\[y’ = – \frac{{xy}}{{{x^2}}} = – \frac{y}{x}.\] \(2)\) Replace \(y’\) with \(\left( { – \large\frac{1}{{y’}}\normalsize} \right):\)
\[{ – \frac{1}{{y’}} = – \frac{y}{x},\;\; }\Rightarrow {y’ = \frac{x}{y}.}\] \(3)\) Now we integrate the differential equation of the orthogonal trajectories:
\[
{y’ = \frac{x}{y},\;\;}\Rightarrow
{\frac{{dy}}{{dx}} = \frac{x}{y},\;\;}\Rightarrow
{ydy = xdx,\;\;}\Rightarrow
{\int {ydy} = \int {xdx} ,\;\;}\Rightarrow
{\frac{{{y^2}}}{2} = \frac{{{x^2}}}{2} + C,\;\;}\Rightarrow
{{x^2} – {y^2} = C.}
\]

In the last equation we replaced \(2C\) with just a constant \(C.\) Thus, we have obtained the equation of the family of orthogonal trajectories. As it can be seen, these orthogonal trajectories are also hyperbolas. Both the families of hyperbolas are shown schematically in Figure \(3.\)

Orthogonal trajectories for a family of hyperbolic curves

Figure 3.

Example 3.

Find the orthogonal trajectories of the family of curves given by the power function \(y = C{x^4}.\)

Solution.

\(1)\) Determine the differential equation for the given family of power curves:
\[{y = C{x^4},\;\; }\Rightarrow {y’ = 4C{x^3}.}\] By solving the system of two equations and eliminating \(C,\) we get:
\[
{C = \frac{y}{{{x^4}}},\;\;}\Rightarrow
{y’ = 4 \cdot \frac{y}{{{x^4}}} \cdot {x^3} = \frac{{4y}}{x}.}
\] \(2)\) Replacing \(y’\) with \(\left( { – \large\frac{1}{{y’}}\normalsize} \right)\) gives:
\[{ – \frac{1}{{y’}} = \frac{{4y}}{x},\;\; }\Rightarrow {y’ = – \frac{x}{{4y}}.}\] The last expression is the differential equation of the orthogonal trajectories.

\(3)\) By integrating we can find the algebraic equation of the orthogonal trajectories:
\[
{y’ = – \frac{x}{{4y}},\;\;}\Rightarrow
{\frac{{dy}}{{dx}} = – \frac{x}{{4y}},\;\;}\Rightarrow
{4ydy = – xdx,\;\;}\Rightarrow
{4\int {ydy} = – \int {xdx} ,\;\;}\Rightarrow
{4 \cdot \frac{{{y^2}}}{2} = – \frac{{{x^2}}}{2} + C,\;\;}\Rightarrow
{4{y^2} + {x^2} = 2C.}
\] Divide both sides by \(2C:\)
\[
{\frac{{4{y^2}}}{{2C}} + \frac{{{x^2}}}{{2C}} = \frac{{2C}}{{2C}},\;\;}\Rightarrow
{\frac{{{y^2}}}{{\frac{C}{2}}} + \frac{{{x^2}}}{{2C}} = 1,\;\;}\Rightarrow
{\frac{{{y^2}}}{{{{\left( {\sqrt {\frac{C}{2}} } \right)}^2}}} + \frac{{{x^2}}}{{{{\left( {\sqrt {2C} } \right)}^2}}} }={ 1.}
\]

Orthogonal trajectories of the family of curves given by the power function y=Cx^4

Figure 4.

We obtain the equation of the family of ellipses, which are the orthogonal trajectories for the given family of power curves \(y = C{x^4}.\) The ratio of the lengths of semiaxes for these ellipses is
\[{\frac{{\sqrt {2C} }}{{\sqrt {\frac{C}{2}} }} = \frac{{\sqrt 2 }}{{\sqrt {\frac{1}{2}} }} }={ {\left( {\sqrt 2 } \right)^2} }={ 2.}\] Schematically the graphs of the families of curves are shown in Figure \(4.\)

Example 4.

Determine the orthogonal trajectories of the family of sinusoids \(y = C\sin x.\)

\(1)\) Differentiating the given equation with respect to \(x\) gives:
\[y’ = C\cos x.\] By substituting \(C = {\large\frac{y}{{\sin x}}\normalsize}\) we find the differential equation of the given sinusoidal curves:
\[{y’ = \frac{y}{{\sin x}}\cos x }={ y\cot x.}\] \(2)\) Replace \(y’\) with \(\left( { – \large\frac{1}{{y’}}\normalsize} \right)\) to write the differential equation of the orthogonal curves:
\[
{- \frac{1}{{y’}} = y\cot x,\;\;}\Rightarrow
{y’ = – \frac{1}{{y\cot x}} }={ – \frac{{\tan x}}{y}.}
\] \(3)\) Now we can integrate this differential equation:
\[
{y’ = – \frac{{\tan x}}{y},\;\;}\Rightarrow
{\frac{{dy}}{{dx}} = – \frac{{\tan x}}{y},\;\;}\Rightarrow
{ydy = – \tan xdx,\;\;}\Rightarrow
{\int {ydy} = – \int {\tan xdx} ,\;\;}\Rightarrow
{\frac{{{y^2}}}{2} = \ln \left| {\cos x} \right| + \ln C,\;\;}\Rightarrow
{\frac{{{y^2}}}{2} = \ln \left( {C\left| {\cos x} \right|} \right).}
\] It follows from here that
\[
{C\left| {\cos x} \right| = \exp \left( {\frac{{{y^2}}}{2}} \right),\;\;}\Rightarrow
{\cos x = \pm \frac{1}{C}\exp \left( {\frac{{{y^2}}}{2}} \right).}
\] By denoting \({C_1} = \pm {\large\frac{1}{C}\normalsize},\) we obtain the final implicit equation of the orthogonal trajectories:
\[\cos x = {C_1}\exp \left( {\frac{{{y^2}}}{2}} \right).\]

Page 1
Theory
Page 2
Problems 1-4