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If the order and degree of the differential equation corresponding to the family of curves $(x-2)^2+(y-a)^2=b^2$, (where $a$ and $b$ are parameters) are $m$ and $n$ respectively, then $m^2+n=$
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The correct answer is:
5
Given, $(x-2)^2+(y-a)^2=b^2$
On differentiating w.r.t. ‘x’
Again differentiating w.r.t ‘x’
From Eqs. (i) and (ii)
$\begin{aligned}
& \quad 1+\frac{-(x-2) y^{\prime \prime}}{y^{\prime}}+y^{\prime 2}=0 \Rightarrow \\
& (x-2) y^{\prime \prime}=y^{\prime 3}+y^{\prime} \\
& \because \text { Order }=2, \text { degree }=1 \\
& \therefore \quad m^2+n=(2)^2+1=5
\end{aligned}$
On differentiating w.r.t. ‘x’
Again differentiating w.r.t ‘x’
From Eqs. (i) and (ii)
$\begin{aligned}
& \quad 1+\frac{-(x-2) y^{\prime \prime}}{y^{\prime}}+y^{\prime 2}=0 \Rightarrow \\
& (x-2) y^{\prime \prime}=y^{\prime 3}+y^{\prime} \\
& \because \text { Order }=2, \text { degree }=1 \\
& \therefore \quad m^2+n=(2)^2+1=5
\end{aligned}$
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