मराठी

√ 1 + ( D Y D X ) 2 = ( C D 2 Y D X 2 ) 1 / 3

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प्रश्न

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

\[\sqrt{1 + \left( \frac{dy}{dx} \right)^2} = \left( c\frac{d^2 y}{d x^2} \right)^\frac{1}{3} \]
Squaring both sides, we get
\[ \Rightarrow 1 + \left( \frac{dy}{dx} \right)^2 = \left( c\frac{d^2 y}{d x^2} \right)^\frac{2}{3} \]
Taking cubes of both sides, we get
\[ \Rightarrow \left( c\frac{d^2 y}{d x^2} \right)^2 = \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^3 \]
\[ \Rightarrow c^2 \left( \frac{d^2 y}{d x^2} \right)^2 = 1 + 3 \left( \frac{dy}{dx} \right)^2 + 3 \left( \frac{dy}{dx}\right)^4 + \left( \frac{dy}{dx} \right)^6\]
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 its degree is more than 1.
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पाठ 21: Differential Equations - Exercise 22.01 [पृष्ठ ५]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 21 Differential Equations
Exercise 22.01 | Q 4 | पृष्ठ ५

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

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