The degree of the differential equation (1 + (dy/dx)²) 3/2 = k·(d²y/dx²) is:

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The degree of the differential equation (1 + (dy/dx)²)<sup>3/2</sup> = k·(d²y/dx²) is:

Answer: 2.

A 1
B 2
C 3
D not defined

Correct answer: B. 2

Explanation: Squaring to clear the fractional power gives (1 + (y')²)³ = k²(y'')², so the degree in the highest derivative is 2.

The general solution y=Ax² represents an entire FAMILY of curves, one for each value of the arbitrary constant A; a particular solution (fixed by an initial condition) selects exactly one curve from this family.

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The degree of the differential equation (1 + (dy/dx)²)<sup>3/2</sup> = k·(d²y/dx²) is:

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