Differential Equations

First Order Equations

1st Order Diff Equations Logo

Orthogonal Trajectories

Solved Problems

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 \(\left( { - \frac{1}{{y'}}} \right)\) 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 \(2C:\)

\[\frac{{4{y^2}}}{{2C}} + \frac{{{x^2}}}{{2C}} = \frac{{2C}}{{2C}},\;\; \Rightarrow \frac{{{y^2}}}{{\frac{C}{2}}} + \frac{{{x^2}}}{{2C}} = 1,\;\; \Rightarrow \frac{{{y^2}}}{{{{\left( {\sqrt {\frac{C}{2}} } \right)}^2}}} + \frac{{{x^2}}}{{{{\left( {\sqrt {2C} } \right)}^2}}} = 1.\]

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 \(4.\)

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

Example 4.

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

Solution.

\(1)\) Differentiating the given equation with respect to \(x\) gives:

\[y' = C\cos x.\]

By substituting \(C = \frac{y}{{\sin x}}\) we find the differential equation of the given sinusoidal curves:

\[y' = \frac{y}{{\sin x}}\cos x = y\cot x.\]

\(2)\) Replace \(y'\) with \(\left( { - \frac{1}{{y'}}} \right)\) 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 \({C_1} = \pm \frac{1}{C},\) we obtain the final implicit equation of the orthogonal trajectories:

\[\cos x = {C_1}\exp \left( {\frac{{{y^2}}}{2}} \right).\]
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