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Question: The solution of the differential equation \(3 e ^ { x } \tan y d x + \left( 1 - e ^ { x } \right) \...

The solution of the differential equation

3extan⁡ydx+(1−ex)sec⁡2ydy=03 e ^ { x } \tan y d x + \left( 1 - e ^ { x } \right) \sec ^ { 2 } y d y = 0 is

A

tan⁡y=c(1−ex)3\tan y = c \left( 1 - e ^ { x } \right) ^ { 3 }

B

(1−ex)3tan⁡y=c\left( 1 - e ^ { x } \right) ^ { 3 } \tan y = c

C

tan⁡y=c(1−ex)\tan y = c \left( 1 - e ^ { x } \right)

D

(1−ex)tan⁡y=c\left( 1 - e ^ { x } \right) \tan y = c

Answer

tan⁡y=c(1−ex)3\tan y = c \left( 1 - e ^ { x } \right) ^ { 3 }

Explanation

Solution

It can be written in the form of

sec⁡2ytan⁡ydy=−3ex1−exdx\frac { \sec ^ { 2 } y } { \tan y } d y = - 3 \frac { e ^ { x } } { 1 - e ^ { x } } d x

∫sec⁡2ytan⁡ydy=−3∫ex1−exdx\int \frac { \sec ^ { 2 } y } { \tan y } d y = - 3 \int \frac { e ^ { x } } { 1 - e ^ { x } } d x

⇒ log⁡(tan⁡y)=3log⁡(1−ex)+log⁡c\log ( \tan y ) = 3 \log \left( 1 - e ^ { x } \right) + \log c ⇒ tan⁡y=c(1−ex)3\tan y = c \left( 1 - e ^ { x } \right) ^ { 3 }.