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The order and degree of the differential equation $\sqrt{1+\frac{1}{\left(\frac{d y}{d x}\right)^{2}}}=\left(\frac{d^{2} y}{d x^{2}}\right)^{\frac{3}{2}}$ are respectively
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Verified Answer
The correct answer is:
2,3
(A)
By raising the given equation to the power of 2 , we get
$\begin{aligned}
& 1+\frac{1}{\left(\frac{d y}{d x}\right)^{2}}=\left(\frac{d^{2} y}{d x^{2}}\right)^{3} \\
\therefore &\left(\frac{d y}{d x}\right)^{2}+1=\left(\frac{d^{2} y}{d x^{2}}\right)^{3}\left(\frac{d y}{d x}\right)^{2}
\end{aligned}$
It's order is 2 and degree is 3 .
By raising the given equation to the power of 2 , we get
$\begin{aligned}
& 1+\frac{1}{\left(\frac{d y}{d x}\right)^{2}}=\left(\frac{d^{2} y}{d x^{2}}\right)^{3} \\
\therefore &\left(\frac{d y}{d x}\right)^{2}+1=\left(\frac{d^{2} y}{d x^{2}}\right)^{3}\left(\frac{d y}{d x}\right)^{2}
\end{aligned}$
It's order is 2 and degree is 3 .
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