Question

The equation of the curve satisfying the differential equation $xe^{x}sin ydy-\left(x + 1\right)e^{x}cos ⁡ ydx=ydy$ and passing through the origin is

• A. $xe^{x}=y^{2}cosy$
• B. $2xe^{x}=ycos y$
• C. $2xe^{x}cos y+y^{2}=0$
• D. $2xe^{x}cos y=y^{2}$

Answer 1

Answer: (C)

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