हिंदी

X + ( D Y D X ) = √ 1 + ( D Y D X ) 2

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प्रश्न

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

  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 21: Differential Equations - Exercise 22.01 [पृष्ठ ५]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 21 Differential Equations
Exercise 22.01 | Q 8 | पृष्ठ ५

वीडियो ट्यूटोरियलVIEW ALL [2]

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