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

प्रश्न

\sqrt{1 - y^{2}} d x + \sqrt{1 - x^{2}} d x = 0

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उत्तर

$\sqrt{1 - y^{2}} d x + \sqrt{1 - x^{2}} d y = 0$
$\Rightarrow \sqrt{1 - y^{2}} d x = - \sqrt{1 - x^{2}} d y$
$\Rightarrow - \frac{\sqrt{1 - y^{2}}}{\sqrt{1 - x^{2}}} = \frac{d y}{d x}$
$\Rightarrow \sqrt{1 - x^{2}} \frac{d y}{d x} + \sqrt{1 - y^{2}} = 0$

In this differential equation, the order of the highest order derivative is 1 and its power is 1. So, it is a differential equation of order 1 and degree 1.

It is a non-linear equation, as the exponent of dependent variable $(y)$ is more than 1 (on expanding $\sqrt{1 - y^{2}}$ binomially).

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Order and Degree of a Differential Equation

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Q 14 Q 13 Q 15

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

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पाठ 22 Differential Equations
Exercise 22.01 | Q 14 | पृष्ठ ५

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

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

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