The solution of the differential equation $\frac{dy}{dx} + \frac{y}{x \log_{e} x} = \frac{1}{x}$ under the condition $y = 1$ when $x = e$ is

Let the solution curve $x=x(y), 0 < y < \frac{\pi}{2}$,of the differential equation $(\log_e(\cos y))^2 \cos y dx - (1+3x \log_e(\cos y)) \sin y dy = 0$ satisfy $x(\frac{\pi}{3}) = \frac{1}{2 \log_e 2}$. If $x(\frac{\pi}{6}) = \frac{1}{\log_e m - \log_e n}$,where $m$ and $n$ are co-prime,then $mn$ is equal to $.....$.

The solution of $r dx + (x - r^2) dr = 0$ is

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

If $y=y(x)$ is the solution of the differential equation $e^{y}\left(\frac{dy}{dx}-1\right)=e^{x}$ such that $y(0)=0,$ then $y(1)$ is equal to

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