Order And Degree

Question

Examine the equation \sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \frac{d^2y}{dx^2} and determine the degree after rationalizing it into proper polynomial form.

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Degree is defined only when a differential equation can be expressed as a polynomial in its derivatives, free of radicals and fractional powers. The given equation contains a square root over a derivative expression, so it must be rationalized before the degree can be read off. The highest-order derivative present is \frac{d^2y}{dx^2}, so the order is 2. To clear the radical, square both sides: 1 + \left(\frac{dy}{dx}\right)^2 = \left(\frac{d^2y}{dx^2}\right)^2. Now the equation is a polynomial in the derivatives, and the highest-order derivative \frac{d^2y}{dx^2} appears raised to the power 2. The degree is the power of the highest-order derivative in this rationalized polynomial form, so the degree is 2. Option Degree 1 reads the power off the original radical form before rationalizing, which is not permitted. Option Degree not defined is incorrect because squaring successfully removes the radical and yields a polynomial in the derivatives. Option Degree 3 has no basis. As a final consistency check, after rationalization the highest-order derivative \frac{d^2y}{dx^2} is squared, so the degree is well defined and equals 2.

Order And Degree

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