Find the general solution of the differential equation $y dx - (x + 2y^2) dy = 0$.

Let $g$ be a differentiable function such that $\int_0^x g(t) dt = x - \int_0^x tg(t) dt$ for $x \geq 0$. Let $y = y(x)$ satisfy the differential equation $\frac{dy}{dx} - y \tan x = 2(x+1) \sec x g(x)$ for $x \in [0, \frac{\pi}{2})$. If $y(0) = 0$,then $y(\frac{\pi}{3})$ is equal to

The differential equation $y^2 dx + (2xy - 1) dy = 0$ 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 $f$ be a differentiable function in the interval $(0, \infty)$ such that $f(1)=1$ and $\lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$ for each $x>0$. Then $2 f(2)+3 f(3)$ is equal to ....................

The solution of $\frac{dy}{dx} + y = e^x$ is