Let $f$ be a differentiable function $f : R \rightarrow R$ satisfying the equation $f(x) = (1+x^2) \left[ 1 + \int_{0}^{x} \frac{f(t)}{1+t^2} dt \right]$ for all $x \in R$. Then $f(1)$ is:

Let $y=y(x)$ be the solution curve of the differential equation $\sec y \frac{dy}{dx} + 2x \sin y = x^3 \cos y$,with the initial condition $y(1) = 0$. Then the value of $y(\sqrt{3})$ is equal to:

The integrating factor of the differential equation $(x^2 + 1)\frac{dy}{dx} + 2xy = x^2 - 1$ is

The solution of the differential equation $\frac{dy}{dx} = \sec x(\sec x + \tan x)$ is

If the solution of the differential equation $\frac{dy}{dx} + e^x(x^2 - 2)y = (x^2 - 2x)(x^2 - 2)e^{2x}$ satisfies $y(0) = 0$,then the value of $y(2)$ is

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}+2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x$ such that $y(0)=\frac{5}{4}$. Then $12\left(y\left(\frac{\pi}{4}\right)-e^{-2}\right)$ is equal to . . . . . . .