The integrating factor of the differential equation $\frac{dy}{dx}(x \log x) + y = 2 \log x$ is given by

The general solution of the differential equation $(x + y) \frac{dy}{dx} = 1$ is

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

If for the solution curve $y=f(x)$ of the differential equation $\frac{dy}{dx}+(\tan x)y=\frac{2+\sec x}{(1+2\sec x)^2}$,$x \in \left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$,$f\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{10}$,then $f\left(\frac{\pi}{4}\right)$ is equal to:

For all $x > 0$,let $y_1(x), y_2(x)$,and $y_3(x)$ be the functions satisfying $\frac{dy_1}{dx} - (\sin x)^2 y_1 = 0, y_1(1) = 5$; $\frac{dy_2}{dx} - (\cos x)^2 y_2 = 0, y_2(1) = \frac{1}{3}$; and $\frac{dy_3}{dx} - \left(\frac{2-x^3}{x^3}\right) y_3 = 0, y_3(1) = \frac{3}{5e}$ respectively. Then $\lim_{x \rightarrow 0^{+}} \frac{y_1(x) y_2(x) y_3(x) + 2x}{e^{3x} \sin x}$ is equal to:

If $x \log x \frac{dy}{dx} + y = \log x^2$ and $y(e) = 0$,then $y(e^2) = $