A body cools according to Newton's law of cooling from $100^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ in 15 minutes. If the temperature of the surrounding is $20^{\circ} \mathrm{C}$, then the temperature of the body after cooling down for one hour is

If $y=y(x)$ is the solution of the differential equation $\left(\frac{5+\mathrm{e}^x}{2+y}\right) \frac{\mathrm{d} y}{\mathrm{~d} x}+\mathrm{e}^x=0$ satisfying $y(0)=1$, then a value of $y(\log 13)$ is

If $(2+\sin x) \frac{\mathrm{d} y}{\mathrm{~d} x}+(y+1) \cos x=0$ and $y(0)=1$ then $y\left(\frac{\pi}{2}\right)$ is equal to

The solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=(x-y)^2$ when $y(1)=1$ is