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  • Find the general solution: sec to the power 2 x tan y dx plus sec to the power 2 y tan x dy equals 0

4.   Find the general solution: 

\sec^2 x \tan y dx + \sec^2 y \tan x dy = 0

Answers (1)

best_answer

Given,

\sec^2 x \tan y dx + \sec^2 y \tan x dy = 0

\\ \implies \frac{sec^2 y}{tan y}dy = -\frac{sec^2 x}{tan x}dx \\ \implies \int \frac{sec^2 y}{tan y}dy = - \int \frac{sec^2 x}{tan x}dx

Now, let tany = t and tanx = u

sec^2 y dy = dt\ and\ sec^2 x dx = du

\\ \implies \int \frac{dt}{t} = -\int \frac{du}{u} \\ \implies log t = -log u +logk \\ \implies t = \frac{1}{ku} \\ \implies tany = \frac{1}{ktanx}

Posted by

HARSH KANKARIA

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