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Worked Examples · Example 2

Q.Find the order and degree of 1+(dydx)2=d2ydx2\displaystyle\sqrt{1 + \left(\frac{dy}{dx}\right)^{2}} = \frac{d^2y}{dx^2}.

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✓ Free question

Clear the radical first. Degree is defined only when the equation is a polynomial in the derivatives, so square both sides:

1+(dydx)2=(d2ydx2)2.1 + \left(\frac{dy}{dx}\right)^{2} = \left(\frac{d^2y}{dx^2}\right)^{2}.

Order. The highest derivative is d2ydx2\dfrac{d^2y}{dx^2}, so the order is 22.

Degree. In this polynomial form the highest-order derivative d2ydx2\dfrac{d^2y}{dx^2} appears to the power 22, so the degree is 22.

Verify: before squaring, the second derivative had power 11 but sat opposite a radical, so degree was not yet defined; after squaring, the exponent of the highest-order derivative is unambiguously 22. Order (unchanged by squaring) is 22.

✓Final answer

Order =2= 2, Degree =2= 2.

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