© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Write down the given differential equation
We have the differential equation:
$$
\frac{2 + \sin x}{y + 1} \cdot \frac{dy}{dx} = -\cos x.
$$
The initial condition given is $y(0) = 1$, and we want to find $y(\pi)$ and $\left.\frac{dy}{dx}\right|_{x=\pi}$.
Step 2: Separate the variables
Rewrite the equation to separate $y$ and $x$ terms:
$$
\frac{dy}{y + 1} = \frac{-\cos x}{2 + \sin x} \, dx.
$$
Step 3: Integrate both sides
Integrate the left-hand side with respect to $y$ and the right-hand side with respect to $x$:
$$
\int \frac{dy}{y + 1} = \int \frac{-\cos x}{2 + \sin x}\, dx.
$$
The integral on the left is:
$$
\int \frac{dy}{y + 1} = \ln|y + 1|.
$$
For the right-hand side, let us integrate:
$$
\int \frac{-\cos x}{2 + \sin x} \, dx.
$$
Notice that $2 + \sin x$ has derivative $\cos x$, so this suggests a natural substitution $u = 2 + \sin x$. Then $du = \cos x \, dx$. Incorporating the negative sign, we get
$$
- \int \frac{\cos x}{2 + \sin x} \, dx
= - \int \frac{1}{u} \, du
= -\ln|u|
= -\ln|2 + \sin x|.
$$
Hence,
$$
\ln|y + 1| = -\ln|2 + \sin x| + C,
$$
where $C$ is an integration constant.
Step 4: Simplify the solution form
Rewrite:
$$
\ln\bigl| (y + 1)(2 + \sin x)\bigr| = \ln|C'|
\quad\text{where}\quad C' = e^C.
$$
Thus,
$$
(y + 1)(2 + \sin x) = C'.
$$
Step 5: Apply the initial condition to find the constant
We are given $y(0) = 1$. Plug in $x=0$ and $y(0)=1$:
$$
(1 + 1)\bigl(2 + \sin 0\bigr) = C'.
$$
Since $\sin 0 = 0$, this becomes
$$
(2)(2) = 4,
$$
so $C' = 4$. Hence, the specific solution is
$$
(y + 1)(2 + \sin x) = 4,
$$
which gives
$$
y = \frac{4}{2 + \sin x} - 1
= \frac{4 - (2 + \sin x)}{2 + \sin x}
= \frac{2 - \sin x}{2 + \sin x}.
$$
Step 6: Compute y(π)
To find $y(\pi)$, substitute $x = \pi$ into $y(x)$:
$$
y(\pi) = \frac{2 - \sin\pi}{2 + \sin\pi}.
$$
Since $\sin \pi = 0$, we get
$$
y(\pi) = \frac{2 - 0}{2 + 0} = \frac{2}{2} = 1.
$$
Hence, $y(\pi) = 1$.
Step 7: Find $ \frac{dy}{dx} $ and evaluate at x = π
The function is
$$
y(x) = \frac{2 - \sin x}{2 + \sin x}.
$$
Differentiate with respect to $x$ using the quotient rule:
$$
\frac{dy}{dx}
= \frac{(2 + \sin x)(-\cos x) - (2 - \sin x)(\cos x)}{(2 + \sin x)^2}.
$$
Simplify the numerator:
$$
(2 + \sin x)(-\cos x) - (2 - \sin x)(\cos x)
= -2\cos x - \sin x \cos x - 2\cos x + \sin x \cos x
= -2\cos x - 2\cos x
= -4\cos x.
$$
So
$$
\frac{dy}{dx} = \frac{-4 \cos x}{(2 + \sin x)^2}.
$$
Now evaluate at $x = \pi$ (where $\sin \pi = 0$ and $\cos \pi = -1$):
$$
\left.\frac{dy}{dx}\right|_{x = \pi}
= \frac{-4 \cdot (-1)}{(2 + 0)^2}
= \frac{4}{4}
= 1.
$$
Therefore, $\left.\frac{dy}{dx}\right|_{x=\pi} = 1$.
Step 8: State the final answer
We have found:
$$
y(\pi) = 1 \quad\text{and}\quad \left.\frac{dy}{dx}\right|_{x=\pi} = 1.
$$
Hence, the ordered pair $(a, b)$ is $(1, 1)$.
