The solution of differential equation `(1+x) y dx +(1-y) x dy = 0 ` is

Options

A. ` log_(e)(xy)+x-y=C`
B. ` log _(e)(x/y)+x+y = C`
C. ` log_(e) (x/y)-x+y = C`
D. `log_(e) (xy)-x+y = C`

Answer

a

Explanation

Given , `(1+x) y dx +(1-y) x dy = 0 `
` rArr ((1-y))/y dy +((1+x))/x dx = 0 `
On integrating both sides , we get
` rArr int (1/y -1) dy +int (1/x +1) dx = 0 `
` rArr log_(e)y - y + log_(e) x + x =C`
` rArr log_(e) (xy) +x-y = C`