हिंदी

Y = X D Y D X + a √ 1 + ( D Y D X ) 2

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प्रश्न

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

\[y = x\frac{dy}{dx} + a\sqrt{1 + \left( \frac{dy}{dx} \right)^2}\]
\[ \Rightarrow y - x\frac{dy}{dx} = a\sqrt{1 + \left( \frac{dy}{dx} \right)^2}\]
Squaring both sides, we get
\[ \Rightarrow \left( y - x\frac{dy}{dx} \right)^2 = a^2 \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]\]
\[ \Rightarrow y^2 - 2xy\frac{dy}{dx} + x^2 \left( \frac{dy}{dx} \right)^2 = a^2 + a^2 \left( \frac{dy}{dx} \right)^2 \]
\[ \Rightarrow \left( x^2 - a^2 \right) \left( \frac{dy}{dx} \right)^2 - 2xy\frac{dy}{dx} + \left( y^2 - a^2 \right) = 0\]
In this differential equation, the order of the highest order derivative is 1 and its highest power is 2. So, it is a differential equation of order 1 and degree 2.
It is a non-linear differential equation, as its degree is 2, which is greater than 1.

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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 18 | पृष्ठ ५

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