अवकल समीकरण `3e^(x)tany dx + (1+e^(x))sec^(2)y dy = 0` का हल क्या है?

Options

A. `(1+e^(x))tan y = C`
B. `(1+e^(x))^(3)tan y = C`
C. `(1+e^(x))^(2)tany=C`
D. `(1+e^(x))sec^(2)y=C`

Answer

B

Explanation

`3e^(x)tanydx+(1+e^(x))sec^(2)ydy=0`
`implies int(3e^(x))/(1+e^(x))dx+int(sec^(2)y)/(tany)dy=0`
`implies3log(1+e^(x))+logtany=logC`
`implies log(1+e^(x))^(3)tany=logC`
`therefore (1+e^(x))^(3)tany=C`