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Step-by-Step Solution
Step 1: Rewrite the Differential Equation
From the given information, we can rearrange the differential equation to the form:
$$
\frac{dy}{dx} + 2\sin^2 x \;=\; 1 + y\cos 2x.
$$
Bringing like terms together, we write it as:
$$
\frac{dy}{dx} - \bigl(\cos 2x\bigr)\,y \;=\; \cos 2x.
$$
This is now in the standard linear form
$$
\frac{dy}{dx} + P(x) \, y \;=\; Q(x),
$$
where $P(x) = -\cos 2x$ and $Q(x)=\cos 2x.$
Step 2: Find the Integrating Factor (I.F.)
The integrating factor I.F. is given by:
$$
\text{I.F.} \;=\; e^{\int P(x)\,dx} \;=\; e^{\int -\cos 2x \,dx}.
$$
We evaluate the integral:
$$
\int -\cos 2x \,dx \;=\; -\frac{\sin 2x}{2}.
$$
Hence,
$$
\text{I.F.} \;=\; e^{-\frac{\sin 2x}{2}}.
$$
Step 3: Multiply the Differential Equation by the I.F. and Integrate
After multiplying both sides by $e^{-\frac{\sin 2x}{2}}$, we get:
$$
e^{-\frac{\sin 2x}{2}} \frac{dy}{dx} \;-\; \cos 2x \, y \, e^{-\frac{\sin 2x}{2}}
\;=\; \cos 2x \, e^{-\frac{\sin 2x}{2}}.
$$
The left-hand side simplifies to the derivative of
$
y\;e^{-\frac{\sin 2x}{2}},
$
i.e.,
$$
\frac{d}{dx}\Bigl(y \, e^{-\frac{\sin 2x}{2}}\Bigr).
$$
Thus we have:
$$
\frac{d}{dx}\Bigl(y \, e^{-\frac{\sin 2x}{2}}\Bigr)
\;=\; \cos 2x \; e^{-\frac{\sin 2x}{2}}.
$$
Integrating both sides with respect to $x$:
$$
y \; e^{-\frac{\sin 2x}{2}}
\;=\; \int \cos 2x \; e^{-\frac{\sin 2x}{2}}\; dx \;+\; C.
$$
Step 4: Evaluate the Integral and Form the General Solution
One finds (by recognizing the derivative of $-\sin 2x/2$ inside the integrand) that:
$$
\int \cos 2x \; e^{-\frac{\sin 2x}{2}}\; dx
\;=\; -\,e^{-\frac{\sin 2x}{2}} \;+\; \text{constant}.
$$
Therefore, we can write:
$$
y \; e^{-\frac{\sin 2x}{2}}
\;=\; -\,e^{-\frac{\sin 2x}{2}} \;+\; C.
$$
Simplifying,
$$
y \;=\; -1 \;+\; C\,e^{\frac{\sin 2x}{2}}.
$$
(It is often left in the form
$
y \; e^{-\frac{\sin 2x}{2}} = -\,e^{-\frac{\sin 2x}{2}} + C,
$
but rewriting is also acceptable.)
Step 5: Apply the Initial Condition $y\bigl(\frac{\pi}{4}\bigr)=0$
We use $x = \tfrac{\pi}{4}$ and $y\bigl(\tfrac{\pi}{4}\bigr) = 0$. Substitute into
$
y \; e^{-\frac{\sin 2x}{2}}
= -\,e^{-\frac{\sin 2x}{2}} + C.
$
At $x=\tfrac{\pi}{4}$, $\sin \left(2 \times \frac{\pi}{4}\right) = \sin \frac{\pi}{2} = 1.$ Hence:
$$
0 \cdot e^{-\frac{1}{2}}
= -\, e^{-\frac{1}{2}} + C
\quad\Longrightarrow\quad
C = e^{-\frac{1}{2}}.
$$
So the particular solution becomes
$$
y \; e^{-\frac{\sin 2x}{2}}
\;=\; -\,e^{-\frac{\sin 2x}{2}} \;+\; e^{-\frac{1}{2}}.
$$
Step 6: Find $y(0)$ and Compute $(y(0)+1)^2$
At $x=0$, $\sin(2 \cdot 0)=0,$ so
$
e^{-\frac{\sin(2\cdot 0)}{2}} = e^0 = 1.
$
Substituting $x=0$ into our particular solution:
$$
y(0) \cdot 1 = -1 + e^{-\frac{1}{2}}.
$$
Therefore,
$$
y(0) = -1 + e^{-\frac{1}{2}}.
$$
Now calculate
$$
(y(0) + 1)^2
= \Bigl(-1 + e^{-\frac{1}{2}} + 1\Bigr)^2
= \Bigl(e^{-\frac{1}{2}}\Bigr)^2
= e^{-1}.
$$
Final Answer
$$
(y(0) + 1)^2 = e^{-1}.
$$
