Given the solution,
$a x^2+b y^2=1$ $\ldots$ (i)
Now differentiating Eq. (i) w.r.t. $x$, we get
$2 a x+2 b y \frac{d y}{d x}=0 \text { or } \frac{y}{x}\left(\frac{d y}{d x}\right)=-\frac{a}{b}$
Again differentiating w.r.t. $x$, we get
$\frac{x\left\{\left(\frac{d y}{d x}\right)^2+y\left(\frac{d^2 y}{d x^2}\right\}-y\left(\frac{d y}{d x}\right)\right.}{x^2}=0$
or $\quad x y\left(\frac{d^2 y}{d x^2}\right)+x\left(\frac{d y}{d x}\right)^2-y\left(\frac{d y}{d x}\right)=0$
This is the differential equation of order 2 and degree 1 obtained from the solution (1).
$\therefore$ Comparing with $\alpha x^2+\beta y^2=1$
$\alpha=2, \beta=1 \Rightarrow 2 x^2+y^2=1$
or $\quad \frac{x^2}{\left(\frac{1}{2}\right)}+\frac{y^2}{1}=1 \Rightarrow a^2=\frac{1}{2}, b^2=1$
Eccentricity, $e=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{1 / 2}{1}}=\sqrt{1-\frac{1}{2}} \Rightarrow e=\frac{1}{\sqrt{2}}$