Question

Solve : 3e^x tanydx + (1 +e^x) sec² ydy = 0

Also, find the particular solution when x = 0 and y = π.

Answer 1

3e^x tanydx + (1 +e^x) sec² ydy = 0

Divide by tany(1+e^x)

`(3e^x)/(1+e^x)dx+(sec^2y)/(tany)dy=0`

`int(3e^x)/(1+e^x)dx+int(sec^2y)/(tany)dy=0`

Put tan y = t

dt = sec²y

`3 log|1+e^x|+int1/tdt=c`

`3log|1+e^x|+log|t|=c`

`3log|1+e^x|+log|tany|=c ................(1)`

put x=0 and y=pi in (1)

`3log|1+e^0|+log|0|=c`

`3log|2|=c................(2)`

`3log|1+e^x|+log|tany|=3log|2| `

`3log|1+e^x|+log|tany|-3log|2|=0`

`3(log|1+e^x|-log|2|)+log|tany|=0`

`3log((1+e^x)/2)+log|tany|=0`