Question

If y(x) is a solution of `((2 + sinx)/(1 + y))"dy"/"dx"` = – cosx and y (0) = 1, then find the value of `y(pi/2)`.

Answer 1

Given equation is `((2 + sinx)/(1 + y))"dy"/"dx"` = – cosx

⇒ `((2 + sin y)/(cos x))"dy"/"dx"` = –(1 + y)

⇒ `"dy"/((1 + y)) = -((cosx)/(2 + sinx))"d"x`

Integrating both sides, we get

`int "dy"/(1 + y) = - int cosx/(2 + sinx) "d"x`

⇒ `log|1 + y| = - log|2 + sinx| + logc`

⇒ `log|1 + y| + log|2 + sinx|` = log c

⇒ `log(1 + y)(2 + sinx)` = log c

⇒ `(1 + y)(2 + sinx)` = c

Put x = 0 and y = 1, we get

(1 + 1)(2 + sin 0) = c

⇒ 4 = c

∴ Equation is (1 + y)(2 + sinx) = 4

Now put x = `pi/2`

∴ `(1 + y)(2 + sin  pi/2)` = 4

⇒ (1 + y)(2 + 1) = 4

⇒ 1 + y = `4/3`

⇒ y = `4/3 - 1`

⇒ `1/3`

So, `y(pi/2) = 1/3`

Hence, the required solution is `y(pi/2) = 1/3`.