Question

Write the differential equation obtained by eliminating the arbitrary constant c in the equation representing the family of curves xy = c cos x.

Answer 1

$xy = c \cos(x) \\ xy \div \cos(x) = c \\ differentiate \\ (cosx(y + x(dy \div dx) + sinx(xy)) \div { \cos }^{2} x = 0 \\ cosx(y + x(dy \div dx) = - sinx \times (xy) \\ ycosx + xcosx(dy \div dx) = - xysinx \\ xcosx(dy \div dx) = - (ycosx + xysinx) \\ dy \div dx = - ((y \div x) + ytanx)$
here it is........

Answer 2

EXPLANATION 1.

The given family of curves is

xy = c cosx ..... (1)

Differentiating both sides of (1) with respect to x, we get

x dy/dx + y = - c sinx ..... (2)

Again, differentiating both sides of (2) with respect to x, we get

x d²y/dx² + dy/dx + dy/dx = - c cosx

or, x d²y/dx² + 2 dy/dx = - xy [ by (1) ]

or, x d²y/dx² + 2 dy/dx + xy = 0

This is the required differential equation.

EXPLANATION 2.

The given family of curves is

xy = c cosx

or, xy/cosx = c ..... (1)

Differentiating both sides of (1) with respect to x, we get

{cosx (x dy/dx + y) + xy sinx}/cos²x = 0

[ Use the formula for d/dx (u/v) where u, v are functions of x. ]

or, cosx (x dy/dx + y) + xy sinx = 0

or, x dy/dx + y + xy tanx = 0

This is the required differential equation.

Note:

In EXPLANATION 1, the differential equation is of order 2 and in EXPLANATION 2, the differential equation is of order 1. However both equations represent the curve in differentials.