$A$ continuous function $f: R \rightarrow R$ satisfies the equation $f(x) = x + \int_0^x f(t) \, dt$. Which of the following options is true?

Let $x = x(y)$ be the solution of the differential equation $y^2 dx + (x - \frac{1}{y}) dy = 0$. If $x(1) = 1$,then $x(\frac{1}{2})$ is:

If the function $y = f(x)$ satisfies the differential equation $(x^3 + 1)dy = x(1 - 3xy)dx$ and $f(0) = 0$,then $\mathop {\lim }\limits_{x \to 0} \frac{x^2}{f(x)}$ is equal to

Let $y = y(x)$ be the solution of the differential equation $(x^2 + 1)^2 \frac{dy}{dx} + 2x(x^2 + 1)y = 1$ such that $y(0) = 0$. If $\sqrt{a} y(1) = \frac{\pi}{32}$,then the value of $a$ is

Let $y=y(x)$ be the solution curve of the differential equation $\frac{dy}{dx} = \frac{y}{x}(1 + xy^2(1 + \log_e x))$ for $x > 0$ and $y(1) = 3$. Then $\frac{y^2(x)}{9}$ is equal to:

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx} + \frac{\sqrt{2}y}{2\cos^4 x - \cos 2x} = x e^{\tan^{-1}(\sqrt{2} \cot 2x)}$,$0 < x < \pi/2$ with $y(\pi/4) = \pi^2/32$. If $y(\pi/3) = \frac{\pi^2}{18} e^{-\tan^{-1}(\alpha)}$,then the value of $3\alpha^2$ is equal to