The solution of the differential equation (x2sin3y−y2cosx)dx+(x3cosysin2y−2ysinx)dy=0 is
Options:
A .  
x3sin3y=3y2sinx+C
B .  
x3sin3y+3y2sinx=C
C .  
x2sin3y+y3sinx=C
D .  
2x2siny+y2sinx=C
Answer: Option A : A (x2sin3y−y2cosx)dx+(x3cosysin2y−2ysinx)dy=0dydx=y2cosx−x2sin3yx3cosysin2−2ysinx(x3cosysin2y−2ysinx)dy=(y2cosx−x2sin3y)dx=0(x33dsin3y−sindy2)−sin3yd(x33)+y2dsinx=0 x33dsin2y+sin3yd(x33)−(sindy2+y2dsinx) d(x33sin3y)−d(y2sinx)=0x33sin3y−y2sinx=c
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