Find a particular solution satisfying the given condition: $\frac{dy}{dx} + 2y \tan x = \sin x$; $y = 0$ when $x = \frac{\pi}{3}$.

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

If $y=y(x)$ is the solution of the differential equation $\frac{dy}{dx}+\frac{4x}{x^2-1}y=\frac{x+2}{(x^2-1)^{5/2}}$ for $x > 1$,such that $y(2)=\frac{2}{9}\log_e(2+\sqrt{3})$ and $y(\sqrt{2})=\alpha\log_e(\sqrt{\alpha}+\beta)+\beta-\sqrt{\gamma}$,where $\alpha, \beta, \gamma \in N$,then $\alpha\beta\gamma$ is equal to $........$.

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:

The solution of the equation $(x-4y^3) \frac{dy}{dx}-y=0, (y>0)$ 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