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√ D 2 Y D X 2 = √ D Y D X - Mathematics

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प्रश्न

\[\sqrt[3]{\frac{d^2 y}{d x^2}} = \sqrt{\frac{dy}{dx}}\]
योग
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उत्तर

\[\sqrt[3]{\frac{d^2 y}{d x^2}} = \sqrt{\frac{dy}{dx}}\]

\[ \Rightarrow \left( \frac{d^2 y}{d x^2} \right)^\frac{1}{3} = \left( \frac{dy}{dx} \right)^\frac{1}{2} \]

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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अध्याय 22: Differential Equations - Exercise 22.01 [पृष्ठ ५]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 22 Differential Equations
Exercise 22.01 | Q 6 | पृष्ठ ५

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