Question

solve the differential equation xdy/dx+y=xcosx+sinx given that y=1 when x=π/2

Answer 1

Step-by-step explanation:

Given, x dy/dx + y = x cosx + sinx

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

or, d (xy) = d (x sinx)

Integrating we get

∫ d (xy) = ∫ d (x sinx) + C, where C = int. const.

or, xy = x sinx + C ..... (1)

Given y = 1 & x = π/2. From (1),

(π/2) * 1 = π/2 * sin(π/2) + C

or, π/2 = π/2 + C

or, C = 0

From (1), we write

xy = x sinx

This is the required integral. (Ans.)

Note 1.

d/dx (x sinx)

= x d/dx (sinx) + sinx d/dx (x)

= x cosx + sinx

or, d (x sinx) = (x cosx + sinx) dx

Note 2.

d/dx (xy)

= x d/dx (y) + y d/dx (x)

= x dy/dx + y

= (x dy + y dx)/dx

or, d (xy) = x dy + y dx