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Question
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Solution
\[\sqrt[3]{\frac{d^2 y}{d x^2}} = \sqrt{\frac{dy}{dx}}\]
\[ \Rightarrow \left( \frac{d^2 y}{d x^2} \right)^\frac{1}{3} = \left( \frac{dy}{dx} \right)^\frac{1}{2} \]
Taking cubes of both the sides, we get
\[ \Rightarrow \frac{d^2 y}{d x^2} = \left( \frac{dy}{dx} \right)^\frac{3}{2} \]
Squaring both the sides, we get
\[ \Rightarrow \left( \frac{d^2 y}{d x^2} \right)^2 = \left( \frac{dy}{dx} \right)^3 \]
\[ \Rightarrow \left( \frac{d^2 y}{d x^2} \right)^2 - \left( \frac{dy}{dx} \right)^3 = 0\]
In this differential equation, the order of the highest order derivative is 2 and its power is 2. So, it is a differential equation of order 2 and degree 2.
Thus, it is a non-linear differential equation, as its degree is 2, which is greater than 1.
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