Correct answer is (b)
We know that
\(\tau\) = r × F
∵ \(\tau\) \(= \frac{dL}{dt},\) \(F = \frac{dP}{dt}\)
\(\left(\frac{dL}{dt}\right) = r \times \left(\frac{dP}{dt}\right)\)
\(\left(\frac{dL}{dt}\right) - r \times \left(\frac{dP}{dt}\right) = 0\)
Given differential equation is,
\(\frac{d^2y}{dx^2}+\frac{dy}{dx}+x=\sqrt{1+\frac{d^2y}{dx^2}}\)
By Squaring both sides, we get
\((\frac{d^2y}{dx^2}+\frac{dy}{dx}+x)^2=1+\frac{d^2y}{dx^2}\)
\(\Rightarrow\) \((\frac{d^2y}{dx^2})^ 2-\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2+2\,\frac{d^2y}{dx^2}\frac{dy}{dx}+2x\frac{d^2y}{dx^2}+2x\frac{dy}{dx}+x^2=1\)
which is differential equation of order 2 and degree 2.
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