1. Given differential equation

$\frac{2+\sin x}{y+1}\frac{dy}{dx}=-\cos x,$ with initial condition $y(0)=1, \qquad y>0.$

We need to find:

• $a=y(\pi)$
• $b=\left.\dfrac{dy}{dx}\right|_{x=\pi}$

2. Separate the variables

Rearrange:

$\frac{1}{y+1}\frac{dy}{dx}=-\frac{\cos x}{2+\sin x}.$

So,

$\frac{dy}{y+1}=-\frac{\cos x}{2+\sin x}\,dx.$

3. Integrate both sides

Integrate:

$\int \frac{dy}{y+1}= -\int \frac{\cos x}{2+\sin x}\,dx.$

Left side:

$\int \frac{dy}{y+1}=\ln(y+1).$

For the right side, let $u=2+\sin x \implies du=\cos x\,dx.$ Then

$-\int \frac{\cos x}{2+\sin x}\,dx = -\int \frac{du}{u}=-\ln(2+\sin x).$

Hence,

$\ln(y+1)=-\ln(2+\sin x)+C.$

So,

$\ln\big((y+1)(2+\sin x)\big)=C,$ which gives

$(y+1)(2+\sin x)=K,$ where $K$ is a constant.

4. Use the initial condition $y(0)=1$

At $x=0$, $\sin 0=0, \qquad y=1.$ Thus,

$(1+1)(2+0)=K \implies 2\cdot 2=4.$

So,

$(y+1)(2+\sin x)=4.$

Therefore,

$y+1=\frac{4}{2+\sin x}$

and

$y=\frac{4}{2+\sin x}-1.$

5. Find $a=y(\pi)$

Since $\sin \pi=0$,

$y(\pi)=\frac{4}{2+0}-1=2-1=1.$

Thus,

$a=1.$

6. Find $b=\left.\dfrac{dy}{dx}\right|_{x=\pi}$

Use the original differential equation:

$\frac{2+\sin x}{y+1}\frac{dy}{dx}=-\cos x.$

At $x=\pi$:

• $\sin \pi=0$
• $\cos \pi=-1$
• $y(\pi)=1 \Rightarrow y+1=2$

Substitute:

$\frac{2+0}{2}\frac{dy}{dx}=-(-1)=1.$

So,

$1\cdot \frac{dy}{dx}=1 \implies \left.\frac{dy}{dx}\right|_{x=\pi}=1.$

Hence,

$b=1.$

7. Ordered pair

$(a,b)=(1,1).$

So the correct option is D.

8. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.