Let $f(x) = x - 1$ and $g(x) = e^x$ for $x \in \mathbb{R}$. If $\frac{dy}{dx} = \left( e^{-2\sqrt{x}} g\left(f(f(x))\right) - \frac{y}{\sqrt{x}} \right)$, $y(0) = 0$, then $y(1)$ is

Let y = y(x) be the solution of the differential equation $(x^2 + 1)y' - 2xy = (x^4 + 2x^2 + 1)\cos x$,

$y(0) = 1$. Then $ \int\limits_{-3}^{3} y(x) \, dx $ is :

Let $y=y(x)$ be the solution curve of the differential equation

$x\left(x^2+e^x\right) d y+\left(\mathrm{e}^x(x-2) y-x^3\right) \mathrm{d} x=0, x>0$, passing through the point $(1,0)$. 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 $\left(7 x^4 \cot y-\mathrm{e}^x \operatorname{cosec} y\right) \frac{\mathrm{d} x}{\mathrm{~d} y}=x^5, x \geq 1$, then at $x=2$, the value of $\cos y$ is :