Question

The solution of the differential equation $(x^{2}si{n}^{3}y-y^{2}cosx)dx+(x^{3}cosysi{n}^{2}y-2ysinx)dy=0$ is

• A. $x^{3}si{n}^{3}y=3y^{2}sinx+C$
• B. $x^{3}si{n}^{3}y+3y^{2}sinx=C$
  
• C. $x^{2}si{n}^{3}y+y^{3}sinx=C$
  
• D. $2x^{2}siny+y^{2}sinx=C$

Answer 1

The correct option is A $x^{3}si{n}^{3}y=3y^{2}sinx+C$

$(x^{2}si{n}^{3}y-y^{2}cosx)dx+(x^{3}cosysi{n}^{2}y-2ysinx)dy=0\frac{dy}{dx}=\frac{y^{2}cosx-x^{2}si{n}^{3}y}{x^{3}cosysi{n}^2-2ysinx}(x^{3}cosysi{n}^{2}y-2ysinx)dy=(y^{2}cosx-x^{2}si{n}^{3}y)dx=0(\frac{x^3}{3}dsi{n}^{3}y-sind{y}^2)-si{n}^{3}yd\left(\frac{x^3}{3}\right)+y^{2}dsinx=0$
$\frac{x^3}{3}dsi{n}^{2}y+si{n}^{3}yd\left(\frac{x^3}{3}\right)-(sind{y}^2+y^{2}dsinx)$
$d\left(\frac{x^3}{3}si{n}^{3}y\right)-d\left(y^{2}sinx\right)=0\frac{x^3}{3}si{n}^{3}y-y^{2}sinx=c$