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:

The solution of the differential equation $(x+2y^3) \frac{dy}{dx} = y$ is

Let $y=y(x)$ be the solution of the differential equation $2 \cos x \frac{d y}{d x}=\sin 2 x-4 y \sin x$,where $x \in \left(0, \frac{\pi}{2}\right)$. If $y\left(\frac{\pi}{3}\right)=0$,then $y^{\prime}\left(\frac{\pi}{4}\right)+y\left(\frac{\pi}{4}\right)$ is equal to:

Let $y(x)$ be a solution of $(1+x^{2}) \frac{dy}{dx} + 2xy - 4x^{2} = 0$ and $y(0) = -1$. Then $y(1)$ is equal to

Let $y = y(x)$ be the solution of the differential equation: $\frac{dy}{dx} + \left( \frac{6x^2 + (3x^2 + 2x^3 + 4)e^{-2x}}{(x^3 + 2)(2 + e^{-2x})} \right) y = 2 + e^{-2x}, x \in (-1, 2)$,satisfying $y(0) = \frac{3}{2}$. If $y(1) = \alpha(2 + e^{-2})$,then $\alpha$ is equal to:

If the solution of the differential equation $x y^{\prime}=y+x^2 \sin x$ subject to the condition $y(\pi)=0$ is $y=f(x)$ and $f(x)$ has an extreme value at $x=\alpha$,then