The general solution of the differential equation $(1+y^2) dx = (\tan^{-1} y - x) dy$ is

If $\cos x \frac{dy}{dx} - y \sin x = 6 x$,$0 < x < \frac{\pi}{2}$,then the general solution of the differential equation is

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

Let $f:(0, \infty) \rightarrow \mathbb{R}$ be a differentiable function such that $f^{\prime}(x)=2-\frac{f(x)}{x}$ for all $x \in(0, \infty)$ and $f(1) \neq 1$. Then

Let $f: R \to R$ be such that $f(xy) = f(x)f(y)$ for all $x, y \in R$ and $f(0) \ne 0$. Let $g: [1, \infty) \to R$ be a differentiable function such that $x^2 g(x) = \int_1^x (t^2 f(t) - t g(t)) dt$. Then $g(2)$ is equal to:

If $y=y(x)$ is the solution of the differential equation $x \frac{d y}{d x}+2 y=x e^{x}, y(1)=0$,then the local maximum value of the function $z(x)=x^{2} y(x)-e^{x}$,$x \in R$ is