If $x=f(y)$ is the solution of the differential equation $(1+y^2)+(x-2 e^{\tan ^{-1} y}) \frac{d y}{d x}=0$,$y \in(-\frac{\pi}{2}, \frac{\pi}{2})$ with $f(0)=1$,then $f(\frac{1}{\sqrt{3}})$ is equal to :

Find the general solution of the differential equation $x \frac{dy}{dx} + 2y = x^2$ where $x \neq 0$.

By multiplying with $e^{\int P dx}$ on both sides of the equation $\frac{dy}{dx} + P(x)y = Q(x)$,the left side of the equation takes the form $\frac{d}{dx}(y f(x))$,then $f(x) =$

The solution of the differential equation $\frac{dy}{dx} - y \tan x = e^x \sec x$ is

Let $f$ be a real-valued differentiable function on $\mathbb{R}$ (the set of all real numbers) such that $f(1)=1$. If the $y$-intercept of the tangent at any point $P(x, y)$ on the curve $y=f(x)$ is equal to the cube of the abscissa of $P$,then the value of $f(-3)$ is equal to

Let $f:[1, \infty) \rightarrow \mathbb{R}$ be a differentiable function. If $6 \int_{1}^{x} f(t) dt = 3xf(x) + x^{3} - 4$ for all $x \ge 1$,then the value of $f(2) - f(3)$ is