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If the order and degree of the differential equation corresponding to the family of curves $y^2=4 a(x+a)$ ( $a$ is parameter) are $m$ and $n$ respectively, then $m+n^2=$
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$5$
$y^2=4 a(x+a)$
$\begin{aligned} & 2 y \frac{d y}{d x}=4 a \\ & \therefore y^2=2 y \frac{d y}{d x}(x+a)=2 y \frac{d y}{d x}\left(x+\frac{y}{2} \frac{d y}{d x}\right) \\ & y^2=2 x y \frac{d y}{d x}+y^2\left(\frac{d y}{d x}\right)^2 \\ & \therefore \text { order }=1=m \\ & \text { degree }=2=n \\ & \therefore m+n^2=1+2^2=5\end{aligned}$
$\begin{aligned} & 2 y \frac{d y}{d x}=4 a \\ & \therefore y^2=2 y \frac{d y}{d x}(x+a)=2 y \frac{d y}{d x}\left(x+\frac{y}{2} \frac{d y}{d x}\right) \\ & y^2=2 x y \frac{d y}{d x}+y^2\left(\frac{d y}{d x}\right)^2 \\ & \therefore \text { order }=1=m \\ & \text { degree }=2=n \\ & \therefore m+n^2=1+2^2=5\end{aligned}$
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