Equation of curve passing through `(1,1)` & satisfyng the differential equation `(xy^(2)+y^(2)+x^(2)y+2xy)dx+(2xy+x^(2))dy=0` is

Options

A. `xy(x+y)=2e^(1-x)`
B. `xy(x^(2)+y^(2))=2e^(x-1)`
C. `x^(2)y^(2)(x+y)=2`
D. `x^(2)y^(2)(x^(2)+y^(2))=2`

Answer

A

Explanation

`[(xy^(2)dx+(y^(2)dx+2xydy)]+x^(2)ydx+(2xydx+x^(2)dy)=0`
`e^(x) xy^(2)dx+e^(x)(y^(2)dx+xd(y^(2)))+e^(x)x^(2)ydx+e^(x)(2xydx)+x^(2)dy)=0`
`d(e^(x)xy^(2))+d(e^(x)x^(2)y)=0`
`e^(x)xy^(2)+e^(x)x^(2)y=C`
Passing through `(1,1)` so `C=2e` so, curve is `xy(x+y)=2e^(1-x) xy(x+y)=2e^(1-x)`