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If $m$ and $n$ are respectively the order and degree of the differential equation of the family of parabolas with focus at the origin and $X$-axis as its axis, then $m n-m+n=$
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The equation of the family of parabolas with focus at the origin and $X$-axis as its axis is given by
$$
\begin{array}{rlrl}
y^2 & =4 a(x+a)=4 a x+4 a^2 \\
\therefore & & 2 y \frac{d y}{d x} & =4 a \\
\Rightarrow & & a & =\frac{1}{2} y \frac{d y}{d x}
\end{array}
$$
$$
\begin{aligned}
\therefore & 2 y \frac{d y}{d x}=4 a \\
\Rightarrow & a=\frac{1}{2} y \frac{d y}{d x}
\end{aligned}
$$
From Eqs. (i) and (ii), we have
$$
y^2=2 x y \frac{d y}{d x}+y^2\left(\frac{d y}{d x}\right)^2
$$
$\therefore$ order $=m=1$ and degree $=n=2$
$$
\therefore \quad m n-m+n=1 \times 2-1+2=3 .
$$
$$
\begin{array}{rlrl}
y^2 & =4 a(x+a)=4 a x+4 a^2 \\
\therefore & & 2 y \frac{d y}{d x} & =4 a \\
\Rightarrow & & a & =\frac{1}{2} y \frac{d y}{d x}
\end{array}
$$
$$
\begin{aligned}
\therefore & 2 y \frac{d y}{d x}=4 a \\
\Rightarrow & a=\frac{1}{2} y \frac{d y}{d x}
\end{aligned}
$$
From Eqs. (i) and (ii), we have
$$
y^2=2 x y \frac{d y}{d x}+y^2\left(\frac{d y}{d x}\right)^2
$$
$\therefore$ order $=m=1$ and degree $=n=2$
$$
\therefore \quad m n-m+n=1 \times 2-1+2=3 .
$$
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