The equation of the curve satisfying the differential equation xexsin ydy-(x + 1)excos ⁡ ydx=ydy and passing through the origin is

Question

The equation of the curve satisfying the differential equation xexsin ydy-(x + 1)excos ⁡ ydx=ydy and passing through the origin is (A) xex=y2cosy (B) 2xex=ycos y (C) 2xexcos y+y2=0 (D) 2xexcos y=y2

Tardigrade Admin · Accepted Answer

The given equation is d(- x ex cos y)=ydy On integrating, we get -xexcos y=(y2/2)+c As it passes through the origin, c=0 ∴ the equation of the curve is 2xexcos y+y2=0

Q. The equation of the curve satisfying the differential equation $x e^{x} s in y d y - (x + 1) e^{x} cos y d x = y d y$ and passing through the origin is

$x e^{x} = y^{2} cosy$

$2 x e^{x} = ycosy$

$2 x e^{x} cosy + y^{2} = 0$

$2 x e^{x} cosy = y^{2}$

The given equation is $d (- x e^{x} cosy) = y d y$
On integrating, we get $- x e^{x} cosy = \frac{y ^{2}}{2} + c$
As it passes through the origin, $c = 0$
$∴$ the equation of the curve is $2 x e^{x} cosy + y^{2} = 0$

Q. The equation of the curve satisfying the differential equation xexsinydy−(x+1)excos⁡ydx=ydy and passing through the origin is

xex=y2cosy

2xex=ycosy

2xexcosy+y2=0

2xexcosy=y2