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प्रश्न
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)`.
उत्तर
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`.
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