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Question

# The equation of the curve satisfying the differential equation y(x+y3)dx=x(y3−x)dy and passing through the point (1,1) is

A
y32x+3x2y=0
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B
y3+2x+3x2y=0
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C
y3+2x3x2y=0
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D
None of these
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Solution

## The correct option is C y3+2x−3x2y=0y(x+y3)dx=x(y3−x)dyHere, (xy+y4)dx=(xy3−x2)dy⇒xydx+y4dx−xy3dy+x2dy=0⇒x(ydx+xdy)+y3(ydx−xdy)=0⇒xd(xy)+[−x2y3(ydx−xdy)x2]=0⇒xd(xy)+[−x2y3d(yx)]=0Dividing the whole equation by x3y2, we get,⇒d(xy)x2y2−yxd(yx)=0⇒d(xy)x2y2=yxd(yx)Integrating both sides, we get⇒∫d(xy)x2y2=∫yxd(yx)⇒−1xy=(y/x)22−c⇒y3+2x+2cx2y=0at (1,1)⇒1+2+2c=0⇒c=−32then y3+2x−2⋅32x2y=0⇒y3+2x−3x2y=0.

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