Question

$if \: \frac{1 +sin \: x \: dy }{1 + y \: dx } = - cos \: x \: such \: that \: y(0) = 1 \: y(x) = a \: and \: \frac{dy}{dx} = b \: then \: (a \: b) =$
(1, -2)
(1, 2)
(-1, -2)
(-1, 2) ​

Answer 1

given : if (1 + sinx)(1 + y) dy/dx = -cosx such that y(0) = 1, y(π) = a and (dy/dx)_(x = π) = b

To find : the points (a , b)

solution : here, (1 + sinx)/(1 + y) dy/dx = -cosx

⇒dy/(1 + y) = - cosx/(1 + sinx) dx

⇒ln(1 + y) = - ln(1 + sinx) + C

at x = 0,

⇒ln(1 + y(0)) = - ln(1 + sin0) + C

⇒ln2 = - 0 + C

⇒C = ln2

so, ln(1 + y) = - ln(1 + sinx) + ln2

⇒1 + y = 2/(1 + sinx)

⇒y = (1 - sinx)/(1 + sinx)

now y(π) = 1 = a

now dy/dx = d[(1 - sinx)/(1 + sinx)]/dx = -2/(1 + sinx)²

at x = π, dy/dx = -2/(1 + 0)² = -2 = b

Therefore, (a , b ) = (1, -2)