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Find an implicit formula satisfied by solutions of (tan x)(cos y) = -y′ tan y

by RoRi

Assume solutions of the equation

    \[ \tan x \cos y = -y' \tan y. \]

exist and find an implicit formula satisfied by these solutions.

This is a separable first order equation. We compute

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

One comment

  1. Anonymous says:

    Greeting! Very good solution indeed! But there seems to be a tiny error. Which stems from the fact that \frac{d}{dy}\frac{1}{\cos(y)} = \frac{d}{dy} \cos^{-1}(y) = -\cos^{-2}(y)(-\sin(y)) = \frac{\sin(y)}{\cos^{2}(y)} and not -\frac{\sin(y)}{\cos^{2}(y)} as the anti-derivative would require. Therefore the sign of \frac{1}{\cos(y)} should be inverted.

    Have a good day :)

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