A curve is such that $\dfrac{\mathrm{d}y}{\mathrm{d}x}=2x+\dfrac{k}{(x+1)^2}$, where \(k

Question

A curve is such that $\dfrac{\mathrm{d}y}{\mathrm{d}x}=2x+\dfrac{k}{(x+1)^2}$, where $k$ is a constant. Given that the curve passes through the points $(0,2)$ and $(1,5)$, find the value of $k$ and the equation of the curve.[7]

Accepted Answer

$k=4,\quad y=x^2-\dfrac4{x+1}+6$

Reciprocal-square differential equation

Reciprocal-square differential equation

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