The general solution of `e^(x) cos ydx-e^(x) sin dy=0` is

Options

A. `e^(x)cosy=k`
B. `e^(x)siny=k`
C. `e^(x)=k cos y`
D. `e^(x)=k sin y`

Explanation

Given that, `" "e^(x)cosydx-e^(x)sinydy=0`
`rArr" "e^(x)cosydx=e^(x)sinydy`
`rArr" "(dx)/(dy)=tany`
`rArr" "dx=tanydy`
On integrating both sides, we get
`" "x=logsecy+C`
`rArr" "x-C=logsecy`
`rArr" "secy=e^(x-C)`
`rArr" "secy=e^(x)e^(-C)`
`rArr" "(1)/(cosy)=(e^(x))/(e^(C))`
`rArr" "e^(x)cosy=e^(C)`
`rArr" "e^(x)cosy=K" "` [where, `K=e^(C)`]