If $x \cdot \log _e\left(\log _e x\right)-x^2+y^2=4(y>0)$, then $\frac{d y}{d x}$ at $x=\mathrm{e}$ is

If p : switch $\mathrm{S}_1$ is closed, q : switch $\mathrm{S}_2$ is closed then correct interpretation from the following circuit is

The solution of $\log \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)=2 x-5 y, y(0)=0$ is

The integrating factor of the differential equation $x \frac{\mathrm{~d} y}{\mathrm{~d} x}+y \log x=x \cdot \mathrm{e}^x x^{-\frac{1}{2}} \log x(x>0)$ is