The differential equation $y^2 dx + (2xy - 1) dy = 0$ is

If a curve $y = f(x)$ passes through the point $(1, 2)$ and satisfies $x \frac{dy}{dx} + y = bx^4$,then for what value of $b$ is $\int_{1}^{2} f(x) dx = \frac{62}{5}$?

Let $y = y(x)$ be the solution of the differential equation $\cos x \frac{dy}{dx} + 2y \sin x = \sin 2x$ for $x \in (0, \frac{\pi}{2})$. If $y(\frac{\pi}{3}) = 0$,then $y(\frac{\pi}{4})$ is equal to:

Let $y=y(x)$ be a differentiable function in the interval $(0, \infty)$ such that $y(1)=2$ and $\lim_{t \rightarrow x} \left( \frac{t^{2}y(x)-x^{2}y(t)}{x-t} \right) = 3$ for each $x>0$. Then $2y(2)$ is equal to

The general solution of the differential equation $\frac{dy}{dx} + \frac{y \ln y}{x} = \frac{y(\ln y)^2}{x^2}$ is (where $C$ is an arbitrary constant):

The solution of the differential equation $x \log x \frac{dy}{dx} + y = 2 \log x$ is