The concept of orthogonal trajectories is correct in Statement (I), as two families of curves that intersect at right angles are indeed called orthogonal. For Statement (II), the orthogonal trajectory of the hyperbola \( xy = c \) is \( y = \frac{1}{x} \), which is correct.
LIST I | LIST II | ||
---|---|---|---|
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) |
Europium (Eu) resembles Calcium (Ca) in the following ways:
(A). Both are diamagnetic
(B). Insolubility of their sulphates and carbonates in water
(C). Solubility of these metals in liquid NH3
(D). Insolubility of their dichlorides in strong HCI
Choose the correct answer from the options given below: