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Question

Find the equation of the curve satisfying the differential equation y(x+y3)dx=x(y3−x)dy, and passing through the point (4,2)

A
1xy=12(yx)214
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B
1xy=12(yx)214
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C
1xy=12(yx)2+14
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D
1xy=12(yx)2+14
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Solution

The correct option is A 1xy=12(yx)214
y(x+y3)dx=x(y3x)dy,
x(ydx+xdy)=y3(xdyydx)
xd(xy)=y3x2d(yx)
d(xy)x2y2=yxd(yx)
Integrating we get,
1xy=12(yx)2+c
Given it passes through (4,2). 18=18+cc=14
Hence required solution is,
1xy=12(yx)214

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