Solution of the differential equation [3 x y2+x sin (x y)] d y+[y3+y sin (x y)] d x=0 is

Question

Solution of the differential equation [3 x y2+x sin (x y)] d y+[y3+y sin (x y)] d x=0 is (A) x y3- cos x y=c (B) x y3+ cos x y=c (C) x y2- cos x y=c (D) x y2+ sin x y=c

Tardigrade Admin · Accepted Answer

3 x y2 d y+y3 d x+ sin (x y)(x d y+y d x)=0 or d(x y3)+ sin (x y) d(x y)=0 On integrating, we get x y3- cos x y=c

Q. Solution of the differential equation
$[3 x y^{2} + x sin (x y)] d y + [y^{3} + y sin (x y)] d x = 0$ is

$x y_{0}^{3} - cos x y = c$

$x y^{3} + cos x y = c$

$x y^{2} - cos x y = c$

$x y^{2} + sin x y = c$

$3 x y^{2} d y + y^{3} d x + sin (x y) (x d y + y d x) = 0$
or $d (x y^{3}) + sin (x y) d (x y) = 0$
On integrating, we get
$x y^{3} - cos x y = c$

Q. Solution of the differential equation [3xy2+xsin(xy)]dy+[y3+ysin(xy)]dx=0 is

xy03​−cosxy=c

xy3+cosxy=c

xy2−cosxy=c

xy2+sinxy=c