Find the general solution of the differential equation `(1+tany)(dx-dy)+2x dy=0`

Options

A. `x(sin y + cos y)=sin y + ce^(y)`
B. `x(sin y + cos y) = sin y + ce^(-y)`
C. `y(sin x + cos x)=sin x + ce^(x)`
D. None of these

Answer

B

Explanation

We have, `(1 + tan y)(dx - dy) + 2x dy = 0`
`rArr" "(1+tan y)dx =(1+tan y - 2x)dy`
`rArr" "(dx)/(dy)+(2)/(1+tan y)x=1`, which is linear differential equation.
I.F.`=e^(2int(dy)/(1+tan y))=e^(int(2 cos y)/(sin y + cos y)dy)`
`=e^(int(1+(cos y-sin y)/(sin y + cos y))dy)=e^(y+log(cos y + sin y))`
`=(cos y + sin y)e^(y)`
So, the solution is
`Xe^(y)(sin y + cos y) = int e^(y) (sin y + cos y) dy + c`
`i.e. xe^(y)(sin y + cos y)=e^(y) sin y + c`.
`i.e. x(sin y + cos y)=sin y + ce^(-y)`