If $y=y(x)$ is the solution of $\frac{dy}{dx}=\frac{x-y \cos x}{1+\sin x}$ with $y\left(\frac{\pi}{2}\right)=\frac{\pi^2}{8}$,then $y(\pi)=$

The solution of the equation $x\frac{dy}{dx} + 3y = x$ is

The solution of the differential equation $\frac{dx}{dy} = \frac{x}{1 + x e^y \cos(x^2)}$ is (where $c$ is the constant of integration):

The integrating factor of the linear differential equation $\frac{dy}{dx} = \frac{1}{4x + 3y}$ is

The solution of the differential equation $2 \frac{dy}{dx} - \frac{y}{x} = \frac{y^2}{x^2}$,given that $y = 2$ when $x = 1$,is

If $y(x)$ satisfies the differential equation $y^{\prime}-y \tan x=2 x \sec x$ and $y(0)=0$, then which of the following is true?
$(A)$ $y(\frac{\pi}{4}) = \frac{\pi^2}{8\sqrt{2}}$
$(B)$ $y(\frac{\pi}{3}) = \frac{\pi^2}{9}$
$(C)$ $y'(\frac{\pi}{4}) = \frac{\pi^2}{8\sqrt{2}} + \frac{\pi}{\sqrt{2}}$
$(D)$ $y'(\frac{\pi}{3}) = \frac{4\pi}{3} + \frac{2\pi^2\sqrt{3}}{9}$