Let the solution curve $y=y(x)$ of the differential equation $(4+x^{2}) dy - 2x(x^{2}+3y+4) dx = 0$ pass through the origin. Then $y(2)$ is equal to

If a curve $y=y(x)$ passes through the point $\left(1, \frac{\pi}{2}\right)$ and satisfies the differential equation $\left(7 x^4 \cot y-e^x \operatorname{cosec} y\right) \frac{d x}{d y}=x^5, x \geq 1$,then at $x=2$,the value of $\cos y$ 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

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

If the solution curve $f(x, y)=0$ of the differential equation $(1+\log_e x) \frac{dx}{dy} - x \log_e x = e^y, x > 0$,passes through the points $(1,0)$ and $(\alpha, 2)$,then $\alpha^\alpha$ 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: