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VITEEE 2012 Paper

Section: Mathematics

Question : 119 of 120

Marks: +1, -0

\frac{\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\frac{d^2y}{dx^2}}

are , respectively

Solution:

ρ =

\frac{\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\frac{d^2y}{dx^2}}

⇒ ρ

\left(\frac{d^2y}{dx^2}\right)

\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2}

On squaring both sides, we get

\rho^2

\left(\frac{d^2y}{dx^2}\right)^2

\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3}

Clearly, it is a second order differential equation of degree 2.

Note that the higher order derivative is in the transcendental, then we do not determined the degree of that equation

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