`3e^(x) tan y dx + (1+e^(x)) sec^(2) dy =0 , " when" x = 0 and y = pi`

Explanation

`3e^(x) tan y dx+(1_e^(x))sec^(2)y dy=0`
On separating the variables, we get
`3e^(x)tany dx = -(1+e^(x))sec^(2)ydy`
`(3e^(x))/(1+e^(x))dx=-(sec^(2)y)/(tany)dy`
On integrating, we get
`3int(e^(x))/(1+e^(x))dx=-int(sec^(2)y)/(tany)dy`
Let `1+e^(x)=t rArr e^(x)dx=dt`
and `tan y = v rArr sec^(2)y dy = dv`
`rArr" "3int(1)/(t)dt=-int(1)/(v)dv`
`rArr" "3log|t|=-log|v|+log|c|`
`rArr" "log|t|^(3)+log|v|=log|c|`
`rArr" "logt^(3).v=logc`
`rArr" "(1+e^(x))^(3).tany=c`
Which is the required general equation Now, when x = 0 and `y=pi`, we get
`(1+e^(0))^(3).tanpi =c`
`(2)^(3).0=c rArr c=0`
Hence proved.