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Find the orthogonal trajectories of the family y = C cos x

by RoRi

Find the orthogonal trajectories of the family of curves given by

    \[ y = C \cos x. \]

First, we find that the family of curves satisfies the differential equation

    \[ y = C \cos x \quad \implies \quad y' = -C \sin x. \]

Since C = \frac{y}{\cos x} we then have

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

Therefore, the orthogonal trajectories satisfy the differential equation

    \begin{align*} y' = \frac{\cos x}{y \sin x} && \implies && yy' &= \cot x \\ && \implies && \int y \, dy &= \int \cot x \, dx \\ && \implies && \frac{1}{2} y^2 &= \log |\sin x | + C \\ && \implies && y^2 - 2 \log |\sin x| &= C \\ && \implies && y^2 - \log (\sin^2 x) &= C. \end{align*}

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