Let the solution curve of the differential equation $x d y-y d x=\sqrt{x^2+y^2} d x, x>0$, $y(1)=0$, be $y=y(x)$. Then $y(3)$ is equal to

Let $y = y(x)$ be the solution of the differential equation $\sec x \dfrac{dy}{dx} - 2y = 2 + 3 \sin x$, $x \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$,

$y(0) = -\dfrac{7}{4}$. Then $y\left(\dfrac{\pi}{6}\right)$ is equal to :

Let $y=y(x)$ be the solution curve of the differential equation $\left(1+x^2\right) \mathrm{d} y+\left(y-\tan ^{-1} x\right) d x=0, y(0)=1$. Then the value of $y(1)$ is :

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