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The degree of the differential equation $\Large\sqrt{1+\left(\frac{dy}{dx}\right)^{1/3}}=\frac{d^{2}y}{dx^{2}}$

\[\begin{array}{1 1}(1)1&(2)2\\(3)3&(4)6\end{array}\]

Can you answer this question?
 
 

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$\sqrt {1+(\large\frac{dy}{dx})^{1/3} } =\frac{d^2 y}{dx^2}$
squaring $1+ (\large\frac{dy}{dx} )^{1/3}=( \frac{d^2 y}{dx^2})^2$
$(\large\frac{dy}{dx} )^{1/3}=( \frac{d^2 y}{dx^2})^2$$-1$
Raising both sides of the power 3.
$\large\frac{dy}{dx} = \bigg[ \bigg( \large\frac{d^2y)^2}{dx^2} \bigg)-1\bigg]^3$
Order - 2
degree -6
Hence 4 is the correct answer.
answered May 21, 2014 by meena.p
 

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