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  • Find the general solution: e^x tan y dx + (1 - e^x)sec y^2 dy = 0

 10.Find the general solution:

    e^x\tan y dx + (1-e^x)\sec^2 y dy = 0

Answers (1)

best_answer

Given,

e^x\tan y dx + (1-e^x)\sec^2 y dy = 0

\\ \implies e^x\tan y dx = - (1-e^x)\sec^2 y dy \\ \implies \int \frac{\sec^2 y }{\tan y}dy = -\int \frac{e^x }{(1-e^x)}dx

\\ let\ tany = t \ and \ 1-e^x = u \\ \implies \sec^2 ydy = dt\ and \ -e^xdx = du

\\ \therefore \int \frac{dt }{t} = \int \frac{du }{u} \\ \implies \log t = \log u + \log k \\ \implies t = ku \\ \implies \tan y= k (1-e^x)

Posted by

HARSH KANKARIA

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