Explanation

(a)
Given, $\frac{d^{2}y}{dx^{2}}\left( x^{2} + 1 \right) = 2x\ \frac{dy}{dx}$
$\Rightarrow$ $\frac{\frac{d^{2}y}{dx^{2}}}{\frac{dy}{dx}} = \frac{2x}{x^{2} + 1}$
On integrating both sides, we get
$\log\frac{dy}{dx} = \log{(x^{2} + 1)} + \log c$
$\Rightarrow$ $\frac{dy}{dx} = c(x^{2} + 1)$ …(i)
As at $x = 0,\ \ \frac{dy}{dx} = 3$
$\therefore$ $3 = c(0 + 1)$
$\Rightarrow$ $c = 3$
$\therefore$ From Eq. (i),
$\frac{dy}{dx} = 3(x^{2} + 1)$
$\Rightarrow$ $dy = 3\left( x^{2} + 1 \right)dx$
Again, integrating both sides, we get
$y = 3\left( \frac{x^{3}}{3} + x \right) + c_{1}$
At point (0,1)
$1 = 3(0 + 0) + c_{1} \Rightarrow c_{1} = 1$
$\therefore$ $y = 3\left( \frac{x^{3}}{3} + x \right) + 1$
$\Rightarrow$ $y = x^{3} + 3x + 1$