Question

Let \(y = y(x)\) be the solution of the differential equation,
\(\frac{2 + \sin x}{y + 1}\cdot\frac{dy}{dx} = -\cos x\),
\(y \gt 0\), \(y(0) = 1\).
If \(y(\pi) = a\) and \(\frac{dy}{dx}\) at \(x = \pi\) is b, then the ordered pair (a, b) is equal to :

• A. \((2, 1)\)
• B. \(\left(2, \frac32\right)\)
• C. \((1, -1)\)
• D. \((1, 1)\)

Hint:

Separate variables: \(\frac{dy}{y+1} = -\frac{\cos x}{2+\sin x}dx\) and integrate.

Answer 1

The Correct Option is D

\(\frac{dy}{y+1} = -\frac{\cos x}{2+\sin x}\,dx\).
Integrating: \(\ln(y+1) = -\ln(2+\sin x) + \ln C\), so \((y+1)(2+\sin x) = C\).
At \(x=0,\ y=1\): \(C = 2\cdot 2 = 4\).
At \(x=\pi\): \((a+1)\cdot 2 = 4 \Rightarrow a = 1\).
\(\frac{dy}{dx} = -\frac{(y+1)\cos x}{2+\sin x}\); at \(x=\pi\), \(y=1\): \(b = -\frac{2\cdot(-1)}{2} = 1\).
\((a,b) = (1,1)\).
Answer: (4) \((1, 1)\)