A homogeneous equation of the second degree in x, y, and z represents a cone with the vertex at the origin. The second statement is correct, as a cone can be generated by making a first-degree equation homogeneous with a guiding curve equation
LIST I | LIST II | ||
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A. | d²y/dx² + 13y = 0 | I. ex(c1 + c2x) | |
B. | d²y/dx² + 4dy/dx + 5y = cosh 5x | II. e2x(c1 cos 3x + c2 sin 3x) | |
C. | d²y/dx² + dy/dx + y = cos²x | III. c1ex + c2e3x | |
D. | d²y/dx² - 4dy/dx + 3y = sin 3x cos 2x | IV. e-2x(c1 cos x + c2 sin x) |