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Question

The equation of the curve satisfying the differential equation y′′(x2+1)=4xy, passing through the point (0,4) and having slope of the tangent at x=0 as 4 is

A
y=4(x55+x+2x331)
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B
y=4(x55x+2x331)
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C
y=4(x55+x+2x33+1)
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D
None of these
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Solution

The correct option is A y=4(x55+x+2x331)
We have, y2y1=2.2xx2+1.
Integrating with respect to. x, we get
logy1=2log(x2+1)+logcy1=c(x2+1)2
Since y1(0)=4c=4
y1=4(x2+1)dy=4(x4+1+2x2)dx
Integrating again, we get y=5(x55+2x33+x1)+k
Putting x=0,y=4, we get k=4.
The required equation of curve is
y=4(x55+2x33+x1).

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