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If $k$ and $l$ respectively are the order and degree of the differential equation whose general solution represents the family of circles of constant radius, then $k^2+l^2=$
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Verified Answer
The correct answer is:
8
Equation of circle with constant radius $r$ is
$$
\begin{gathered}
(x-a)^2+(y-b)^2=r^2 \\
x=a+r \cos \theta, y=b+r \sin \theta \\
\frac{d x}{d \theta}=r \sin \theta, \frac{d y}{d \theta}=r \cos \theta \\
\frac{d y}{d x}=-\cot \theta \Rightarrow \frac{d^2 y}{d x^2}=\operatorname{cosec}^2 \theta \frac{d \theta}{d x} \\
\frac{d^2 y}{d x^2}=\frac{-1}{r} \operatorname{cosec}^3 \theta \Rightarrow r^2\left(\frac{d^2 y}{d x^2}\right)^2=\left(\operatorname{cosec}^2 \theta\right)^3 \\
r^2\left(\frac{d^2 y}{d x^2}\right)^2=\left(1+\cot ^2 \theta\right)^3 \\
r^2\left(\frac{d^2 y}{d x^2}\right)^2=\left(1+\left(\frac{d y}{d x}\right)^2\right)^3
\end{gathered}
$$
$$
\begin{aligned}
& \text { Order }=2 \text { degree }=2 \\
& \therefore \quad k=2 \text { and } l=2 \\
& k^2+l^2=4+4=8
\end{aligned}
$$
$$
\begin{gathered}
(x-a)^2+(y-b)^2=r^2 \\
x=a+r \cos \theta, y=b+r \sin \theta \\
\frac{d x}{d \theta}=r \sin \theta, \frac{d y}{d \theta}=r \cos \theta \\
\frac{d y}{d x}=-\cot \theta \Rightarrow \frac{d^2 y}{d x^2}=\operatorname{cosec}^2 \theta \frac{d \theta}{d x} \\
\frac{d^2 y}{d x^2}=\frac{-1}{r} \operatorname{cosec}^3 \theta \Rightarrow r^2\left(\frac{d^2 y}{d x^2}\right)^2=\left(\operatorname{cosec}^2 \theta\right)^3 \\
r^2\left(\frac{d^2 y}{d x^2}\right)^2=\left(1+\cot ^2 \theta\right)^3 \\
r^2\left(\frac{d^2 y}{d x^2}\right)^2=\left(1+\left(\frac{d y}{d x}\right)^2\right)^3
\end{gathered}
$$
$$
\begin{aligned}
& \text { Order }=2 \text { degree }=2 \\
& \therefore \quad k=2 \text { and } l=2 \\
& k^2+l^2=4+4=8
\end{aligned}
$$
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