If the solution of the differential equation $x \frac{dy}{dx} + y = x e^x$ is $xy = e^x \phi(x) + C$, then $\phi(x)$ is equal to

If $x=f(y)$ is the solution of the differential equation $(1+y^2)+(x-2 e^{\tan ^{-1} y}) \frac{d y}{d x}=0$,$y \in(-\frac{\pi}{2}, \frac{\pi}{2})$ with $f(0)=1$,then $f(\frac{1}{\sqrt{3}})$ is equal to :

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}=(y+1)((y+1)e^{x^{2}/2}-x)$,with $y(2)=0$. Then $y'(1)$ is equal to . . . .

Let $y=y(x)$ be the solution of the differential equation $x\frac{dy}{dx}-y=x^{2}\cot x, x\in(0,\pi)$. If $y(\frac{\pi}{2})=\frac{\pi}{2}$, then $6y(\frac{\pi}{6})-8y(\frac{\pi}{4})$ is equal to :

The solution of the differential equation $(1+y^2) + (x - e^{\tan^{-1} y}) \frac{dx}{dy} = 0$ is

If $y=y(x)$ is the solution of $\frac{dy}{dx}=\frac{x-y \cos x}{1+\sin x}$ with $y\left(\frac{\pi}{2}\right)=\frac{\pi^2}{8}$,then $y(\pi)=$