Question

The solution of the differential equation $(x^2{\sin }^{3}y-y^2\cos x)dx+(x^3\cos y{\sin }^{2}y-2y\sin x)dy=0$ is

• A. $x^3{\sin }^{3}y=3y^2\sin x+C$
• B. $x^3{\sin }^{3}y+3y^2\sin x=C$
• C. $x^2{\sin }^{3}y+y^3\sin x=C$
• D. $2x^2\sin y+y^2\sin x=C$

Answer 1

Answer: (A)

$x^3{\sin }^{3}y=3y^2\sin x+C$

${}^{\prime }\int (x^2{\sin }^{3}y-y^2\cos x)dx={\sin }^{3}y\frac{x^3}{3}-y^2\sin x+h(y\left)h^{\prime }\right(y)-2y\sin x+x^3{\sin }^{2}y\cos y=x^3{\sin }^{2}y\cos y-2y\sin x{h}^{\prime }(y)=0h(y)=Cf(x,y)={\sin }^{3}y\frac{x^3}{3}-y^2\sin x+C{x}^3{\sin }^{3}y-3y^2\sin x=C$