what is the solution of the differential equaiton
` 3e^(x) tan y dx + (1+e^(x)) sec^(2) y dy =0` ?

Options

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

Answer

B

Explanation

` 3e^(x) tan y dx + (1 +e^(x))sec^(2) y dy =0`
By separating the variables, we get
`3e^(x)dx=(-(1+e^(x))sec^(2)y)/tany""dy`
` Rightarrow (3e^(x))/(1+e^(x)) dx = - (sec ^(2)y)/(tany) dy `
Integrate on both sides,
` Rightarrow int (3e^(x))/(1+e^(x)) dx + int (sec^(2)y)/(tany ) dy =0`
` Rightarrow 3 log ( 1+ e^(x)) + log tan y = log c `
` Rightarrow log ( 1 + e^(x))^(3) tan y = log c " "( log m + log n - log m) `
` Rightarrow (1 + e^(x))^(3) tan y =c`