The integrating factor of the linear differential equation $\frac{dy}{dx} = \frac{1}{4x + 3y}$ is

Let the solution curve $y=f(x)$ of the differential equation $\frac{dy}{dx}+\frac{xy}{x^{2}-1}=\frac{x^{4}+2x}{\sqrt{1-x^{2}}}, x \in(-1,1)$ pass through the origin. Then $\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) dx$ is equal to

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

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx} + \frac{\sqrt{2}y}{2\cos^4 x - \cos 2x} = x e^{\tan^{-1}(\sqrt{2} \cot 2x)}$,$0 < x < \pi/2$ with $y(\pi/4) = \pi^2/32$. If $y(\pi/3) = \frac{\pi^2}{18} e^{-\tan^{-1}(\alpha)}$,then the value of $3\alpha^2$ is equal to

If $y = \left(\frac{2}{\pi} x - 1\right) \operatorname{cosec} x$ is the solution of the differential equation $\frac{dy}{dx} + p(x) y = \frac{2}{\pi} \operatorname{cosec} x$ for $0 < x < \frac{\pi}{2}$,then the function $p(x)$ is equal to

The solution of the differential equation $\frac{dy}{dx} - y \tan x = e^x \sec x$ is