On solving by options, we get
(a) $x^2+y^2+2 g x+4 y+2=0$
$$
\Rightarrow 2 x+2 y \frac{d y}{d x}+2 g+4 \frac{d y}{d x}=0
$$
[differentiating both sides w.r.t. ' $x$ ']
$$
\Rightarrow \quad-\left(x+y \frac{d y}{d x}+2 \frac{d y}{d x}\right)=g
$$
On substituting value of $g$ in Eq. (i), we get
$$
\begin{aligned}
& x^2+y^2-2\left(x+y \frac{d y}{d x}+2 \frac{d y}{d x}\right) x+4 y+2=0 \\
\Rightarrow & x^2+y^2-2 x^2-2 x y \frac{d y}{d x}-4 \frac{d y}{d x} x+4 y+2=0 \\
\Rightarrow \quad & y^2-x^2-2 x y \frac{d y}{d x}-4 x \frac{d y}{d x}+4 y+2=0
\end{aligned}
$$
Its order and degree both are 1
(b)
$$
\begin{aligned}
& x^2=a^2\left(1+y^2\right) \\
& 2 x=a^2\left(2 y \frac{d y}{d x}\right) \\
& a^2=\frac{x}{y \frac{d y}{d x}}
\end{aligned}
$$
[differentiating]


On substituting the value of $a^2$ in Eq. (ii), we get
$$
x^2=\frac{x}{y \frac{d y}{d x}}\left(1+y^2\right) \Rightarrow x^2 y \frac{d y}{d x}=x\left(1+y^2\right)
$$
Its order and degree both are 1.
(c)
$$
y^2=2 c(x+\sqrt{c})
$$
$$
\Rightarrow \quad 2 y \frac{d y}{d x}=2 c(1)
$$
[differentiating]
$$
\Rightarrow \quad c=y \frac{d y}{d x}
$$


On substituting the value of $C$ in Eq. (iii), we get
$$
\begin{aligned}
y^2 & =2 y \frac{d y}{d x}\left(x+\sqrt{y \frac{d y}{d x}}\right) \\
\Rightarrow y^2-2 x y \frac{d y}{d x} & =2 y \frac{d y}{d x} \sqrt{y \frac{d y}{d x}}
\end{aligned}
$$
On squaring both sides, we get
$$
\begin{aligned}
& \left(y^2-2 x y \frac{d y}{d x}\right)^2=4 y^2\left(\frac{d y}{d x}\right)^2 y \frac{d y}{d x} \\
\Rightarrow \quad & x\left(y^2-2 x y \frac{d y}{d x}\right)^2=4 y^3\left(\frac{d y}{d x}\right)^3
\end{aligned}
$$
It involves only first-order derivatives. Its order is 1 but its degree is 3 as $\left(\frac{d y}{d x}\right)^3$ is there
(d)
$$
\begin{aligned}
y^2 & =4 a x \\
2 y \frac{d y}{d x} & =4 a
\end{aligned}
$$
[differentiating]
On substituting $a$ in Eq. (iv), we get
$$
y^2=4 x\left(\frac{y \frac{d y}{d x}}{2}\right) \Rightarrow y^2=2 x y \frac{d y}{d x}
$$
Its order and degree both are 1 .