मराठी

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

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प्रश्न

\[x + \left( \frac{dy}{dx} \right) = \sqrt{1 + \left( \frac{dy}{dx} \right)^2}\]
बेरीज
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उत्तर

\[x + \left( \frac{dy}{dx} \right) = \sqrt{1 + \left( \frac{dy}{dx} \right)^2}\]

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

Squaring both sides, we get

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

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

\[ \Rightarrow 2x\frac{dy}{dx} + x^2 = 1\]

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

Hence, it is a linear differential equation.

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Notes

The answer given in the book has some error. The solution here is created according to the question given in the book.

  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 22: Differential Equations - Exercise 22.01 [पृष्ठ ५]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 22 Differential Equations
Exercise 22.01 | Q 8 | पृष्ठ ५

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