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If $m$ and $n$ are 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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Equation of family of parabola with focus at origin and the $X$-axis its axis is
$$
\begin{aligned}
& & y^2 & =-4 a(x-a) \\
\Rightarrow & & y^2 & =-4 a x+4 a^2 \\
\Rightarrow & & 2 y y^{\prime} & =-4 a \\
\Rightarrow & & a & =-\frac{y y^{\prime}}{2}
\end{aligned}
$$
Putting the value of $a$ in Eq. (i), we get
$$
\begin{aligned}
& y^2=2 y y^{\prime}\left(x+\frac{y y^{\prime}}{2}\right) \\
& y=2 y^{\prime} \frac{\left(2 x+y y^{\prime}\right)}{2} \\
\Rightarrow \quad & y=2 x y^{\prime}+y y^{\prime 2} \\
\Rightarrow \quad & y\left(\frac{d y}{d x}\right)^2+2 x \frac{d y}{d x}-y=0
\end{aligned}
$$
$$
\begin{gathered}
m=\text { order }=1 \\
n=\text { degree }=2 \\
\therefore m n-m+n=2-1+2=3
\end{gathered}
$$
$$
\begin{aligned}
& & y^2 & =-4 a(x-a) \\
\Rightarrow & & y^2 & =-4 a x+4 a^2 \\
\Rightarrow & & 2 y y^{\prime} & =-4 a \\
\Rightarrow & & a & =-\frac{y y^{\prime}}{2}
\end{aligned}
$$
Putting the value of $a$ in Eq. (i), we get
$$
\begin{aligned}
& y^2=2 y y^{\prime}\left(x+\frac{y y^{\prime}}{2}\right) \\
& y=2 y^{\prime} \frac{\left(2 x+y y^{\prime}\right)}{2} \\
\Rightarrow \quad & y=2 x y^{\prime}+y y^{\prime 2} \\
\Rightarrow \quad & y\left(\frac{d y}{d x}\right)^2+2 x \frac{d y}{d x}-y=0
\end{aligned}
$$
$$
\begin{gathered}
m=\text { order }=1 \\
n=\text { degree }=2 \\
\therefore m n-m+n=2-1+2=3
\end{gathered}
$$
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