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2 D 2 Y D X 2 + 3 √ 1 − ( D Y D X ) 2 − Y = 0 - Mathematics

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प्रश्न

\[2\frac{d^2 y}{d x^2} + 3\sqrt{1 - \left( \frac{dy}{dx} \right)^2 - y} = 0\]
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उत्तर

\[2\frac{d^2 y}{d x^2} + 3\sqrt{1 - \left( \frac{dy}{dx} \right)^2 - y} = 0\]
\[ \Rightarrow 2\frac{d^2 y}{d x^2} = - 3\sqrt{1 - \left( \frac{dy}{dx} \right)^2 - y}\]
Squaring both sides, we get
\[ \Rightarrow 4 \left( \frac{d^2 y}{d x^2} \right)^2 = 9\left[ 1 - \left( \frac{dy}{dx} \right)^2 - y \right]\]
\[ \Rightarrow 4 \left( \frac{d^2 y}{d x^2} \right)^2 + 9 \left( \frac{dy}{dx} \right)^2 + 9y - 9 = 0\]
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.
It is a non-linear differential equation, as it has degree 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 16 | पृष्ठ ५

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