Question

Let y = y(x) be the solution of the differential equation, `(2 + sinxdy)/(y + 1) (dy)/(dx)` = –cosx. If y > 0, y(0) = 1. If y(π) = a, and `(dy)/(dx)` at x = π is b, then the ordered pair (a, b) is equal to ______.

Options

• `(2, 3/2)`

• (1, 1)

• (2, 1)

• (1, –1)

Answer 1

Let y = y(x) be the solution of the differential equation, `(2 + sinxdy)/(y + 1) (dy)/(dx)` = –cosx. If y > 0, y(0) = 1. If y(π) = a, and `(dy)/(dx)` at x = π is b, then the ordered pair (a, b) is equal to (1, 1).

Explanation:

`(2 + sinxdy)/(y + 1) (dy)/(dx)` = –cosx     y(0) = 1,  y(π) = a

`(dy)/(y + 1) = (-cosx)/(2 + sinx)dx`

`int(dy)/(y + 1) = -intcosx/(2 + sinx)dx`

log(y + 1) = –log(2 + sinx) + C  ...(i)

y(0) = 1

log(1 + 1) = –log(2 + sin0) + C

C = 2log2

From equation (i),

log(y + 1) = –log(2 + sinx) + 2log2

y(π) = a

⇒ log(a + 1) = –log(2 + sinπ) + 2log2

⇒ log(a + 1) = –log2 + 2log2

⇒ log(a + 1) = log2

a + 1 = 2

⇒ a = 1

Again `1/(y + 1) (dy)/(dx) = (-1)/(2 + sinx)cosx`

`(dy)/(dx)` = b, at x = π  ...(ii)

`{{:(log(y + 1) = -log(2 + sinx) + 2log2),(x = π),(log(y + 1) = -log2 + 2log2),(y + 1 = 2 ⇒ y = 1):}}`

From equation (ii),

`1/2(dy)/(dx) = (-(-1))/(2 + 0)`

`(dy)/(dx)` = 1

b = 1

⇒ a = 1, b = 1