1. Interpret the differential equation

The printed equation is clearly intended as

$\cos^2 x\,dy + 2\,dx = (1+y\cos 2x)\cos^2 x\,dx.$

So,

$\cos^2 x\,\frac{dy}{dx} + 2 = (1+y\cos 2x)\cos^2 x.$

Hence,

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

Divide by $\cos^2 x$:

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

Now use

$\cos 2x = 2\cos^2 x-1, \qquad \sec^2 x = 1+\tan^2 x.$

But the key simplification is

$1-2\sec^2 x = -(2\sec^2 x-1).$

A cleaner way is to rewrite the original equation as

$\frac{dy}{dx} - y\cos 2x = 1-2\sec^2 x.$

2. Recognize a useful identity

Since

$\cos 2x = 2\cos^2 x-1,$

this form is not immediately standard. Let us instead check whether the equation can be rewritten in terms of $\tan x$.

Observe that

$1-2\sec^2 x = -(2\sec^2 x-1)=-(1+2\tan^2 x).$

However, the standard trick here is to test the integrating factor for

$\frac{dy}{dx}-(\cos 2x)y = 1-2\sec^2 x.$

This is a linear differential equation.

3. Solve the linear differential equation

Write it as

$\frac{dy}{dx}+P(x)y=Q(x),$

where

$P(x)=-\cos 2x, \qquad Q(x)=1-2\sec^2 x.$

So the integrating factor is

$\text{I.F.}=e^{\int -\cos 2x\,dx}=e^{-\frac{1}{2}\sin 2x}.$

Thus,

$\frac{d}{dx}\left(y e^{-\frac{1}{2}\sin 2x}\right)=(1-2\sec^2 x)e^{-\frac{1}{2}\sin 2x}.$

At this point, the intended equation from the question is more naturally interpreted in the standard JEE form:

$\cos^2 x\,dy + 2dx = (1+y)\cos 2x\,dx,$

but that does not match the visible text. So let us instead use the more plausible intended reading based on the answer choices and initial condition:

$\cos^2 x\,\frac{dy}{dx}+2=(1+y)\cos 2x.$

Then

$\frac{dy}{dx} = (1+y)\frac{\cos 2x}{\cos^2 x}-2\sec^2 x.$

Using

$\frac{\cos 2x}{\cos^2 x}=2-\sec^2 x,$

we get

$\frac{dy}{dx}=(1+y)(2-\sec^2 x)-2\sec^2 x,$

which still does not simplify to a standard form matching the options cleanly.

4. Use the standard solvable structure consistent with the answer choices

The expression in the question is most likely intended to be

$\cos^2 x\,dy + 2dx = (1+y\cos 2x)\,dx,$

which gives

Still not ideal.

Given the initial condition at $x=\frac\pi4$, and answer choices involving $e$, the natural transformed variable is

$u=y+1.$

For the intended JEE problem, this typically reduces to

$\frac{du}{dx}=u\cos 2x,$

so that

$\frac{du}{u}=\cos 2x\,dx.$

Integrating,

$\ln u = \frac12\sin 2x + C,$

hence

$u = Ce^{\frac12\sin 2x}.$

Using $y\left(\frac\pi4\right)=0$, we have

$u\left(\frac\pi4\right)=1.$

Since

$\sin\left(2\cdot \frac\pi4\right)=\sin\frac\pi2=1,$

we get

$1=Ce^{1/2} \implies C=e^{-1/2}.$

Therefore,

$u(0)=e^{-1/2}e^0=e^{-1/2}.$

So,

$(y(0)+1)^2 = u(0)^2 = e^{-1}.$

This does not match any option, so the printed equation is evidently malformed.

5. Match with the stored correct option

Among the options, the stored correct answer is C: $e-1$, which would correspond to

$(y(0)+1)^2=e-1.$

But this is not derivable from the readable differential equation as written.

Therefore, based on the visible text, I do not agree with the stored answer; the question statement appears corrupted/ambiguous.