The general solution of the differential equation $\left(\frac{1}{x^2}+x\right) \frac{d y}{d x}+3 y=1$ is

If $y'' - 3y' + 2y = 0$ where $y(0) = 1$ and $y'(0) = 0$,then the value of $y$ at $x = \log_{e} 2$ is

Let $f:[0, \infty) \rightarrow \mathbb{R}$ be a differentiable function such that $f(x) = 1 - 2x + \int_0^x e^{x-t} f(t) dt$ for all $x \in [0, \infty)$. Then the area of the region bounded by $y = f(x)$ and the coordinate axes is

Observe the following statements:
$A$. Integrating factor of $\frac{dy}{dx} + y = x^2$ is $e^x$.
$R$. Integrating factor of $\frac{dy}{dx} + P(x)y = Q(x)$ is $e^{\int P(x) dx}$.
Then,the true statement among the following is:

Let $f$ be a non-negative function defined on $\left[0, \frac{\pi}{2}\right]$. If $\int_0^x \left(f^{\prime}(t)-\sin 2t\right) dt = \int_x^0 f(t) \tan t dt$ and $f(0)=1$,then find $\int_0^{\frac{\pi}{2}} f(x) dx$.

If a curve $y=y(x)$ passes through the point $\left(1, \frac{\pi}{2}\right)$ and satisfies the differential equation $\left(7 x^4 \cot y-e^x \operatorname{cosec} y\right) \frac{d x}{d y}=x^5, x \geq 1$,then at $x=2$,the value of $\cos y$ is: