1. **Problem (A):** Given $x=c_1\cos t + c_2\sin t$ solves $x''+x=0$, find $c_1, c_2$ satisfying initial conditions $x(\pi/2)=0$ and $x'(\pi/2)=1$. 2. Differentiate $x(t)$ to get $x'(t)=-c_1\sin t + c_2\cos t$. 3. Substitute $t=\pi/2$ into $x(t)$: $$x\left(\frac{\pi}{2}\right) = c_1 \cos \frac{\pi}{2} + c_2 \sin \frac{\pi}{2} = 0 + c_2 \cdot 1 = c_2 = 0.$$ 4. Substitute $t=\pi/2$ into $x'(t)$: $$x'\left(\frac{\pi}{2}\right) = -c_1 \sin \frac{\pi}{2} + c_2 \cos \frac{\pi}{2} = -c_1 \cdot 1 + 0 = -c_1 = 1 \implies c_1 = -1.$$ 5. Solution (A) is: $$x(t) = -\cos t.$$ 6. **Problem (B):** $x=c_1\cos t + c_2\sin t$ with initial conditions $x(\pi/4)=\sqrt{2}$ and $x'(\pi/4)=2\sqrt{2}$. 7. Recall: $x'(t)=-c_1 \sin t + c_2 \cos t$. 8. Substitute $t=\pi/4$ into $x(t)$: $$x\left(\frac{\pi}{4}\right) = c_1 \cos \frac{\pi}{4} + c_2 \sin \frac{\pi}{4} = c_1 \frac{\sqrt{2}}{2} + c_2 \frac{\sqrt{2}}{2} = \sqrt{2}.$$ Multiply both sides by $\sqrt{2}$: $$c_1 + c_2 = 2.$$ 9. Substitute $t=\pi/4$ into $x'(t)$: $$x'\left(\frac{\pi}{4}\right) = -c_1 \sin \frac{\pi}{4} + c_2 \cos \frac{\pi}{4} = -c_1 \frac{\sqrt{2}}{2} + c_2 \frac{\sqrt{2}}{2} = 2\sqrt{2}.$$ Multiply both sides by $\sqrt{2}$: $$-c_1 + c_2 =4.$$ 10. Solve linear system: $$\begin{cases} c_1 + c_2 = 2 \\ -c_1 + c_2 = 4 \end{cases}$$ Add equations: $$2 c_2 =6 \implies c_2=3.$$ From $c_1 + 3 = 2 \implies c_1 = -1.$ 11. Solution (B) is: $$x(t) = -\cos t + 3 \sin t.$$ 12. **Exercise 10:** Water draining from cubical tank with friction factor $c$. Hole radius $r=2$ in = $\frac{2}{12} = \frac{1}{6}$ ft. 13. Area of hole: $$A_h = \pi r^2 = \pi \left(\frac{1}{6}\right)^2 = \frac{\pi}{36}.$$ 14. Volume $V(t) = A_w h$, where $A_w$ is area of water surface, cubical tank side 10 ft, so $$A_w = 10 \times 10 = 100.$$ 15. Flow rate using friction factor $c$ (0 < c < 1): $$\frac{dV}{dt} = - c A_h \sqrt{2 g h}.$$ 16. Substitute $V = A_w h$ and $dV/dt = A_w dh/dt$: $$A_w \frac{dh}{dt} = - c A_h \sqrt{2 g h} \implies \frac{dh}{dt} = - \frac{c A_h}{A_w} \sqrt{2 g h}.$$ 17. Substitute values: $$\frac{dh}{dt} = - \frac{c (\pi/36)}{100} \sqrt{2 \cdot 32 \cdot h} = - \frac{c \pi}{3600} \sqrt{64 h} = - \frac{c \pi}{3600} \times 8 \sqrt{h} = - \frac{8 c \pi}{3600} \sqrt{h} = - \frac{2 c \pi}{900} \sqrt{h}.$$ 18. Final differential equation for Exercise 10 is: $$\boxed{\frac{dh}{dt} = - \frac{2 c \pi}{900} \sqrt{h}}.$$ 19. **Exercise 11:** Water draining from right circular conical tank. Hole radius $r=2$ in $= \frac{1}{6}$ ft, $g=32$, $c=0.6$. 20. Area of hole: $$A_h = \pi \left(\frac{1}{6}\right)^2 = \frac{\pi}{36}.$$ 21. For the conical tank with height 20 ft and radius 8 ft, water surface radius depends on $h$ by similar triangles: $$\frac{r(h)}{h} = \frac{8}{20} = 0.4 \implies r(h) = 0.4 h.$$ 22. Surface area of water: $$A_w = \pi [r(h)]^2 = \pi (0.4 h)^2 = \pi (0.16 h^2) = 0.16 \pi h^2.$$ 23. Differential equation for height with friction: $$A_w \frac{dh}{dt} = - c A_h \sqrt{2 g h}.$$ 24. Substitute $A_w$, $A_h$, $c$, and $g$: $$0.16 \pi h^2 \frac{dh}{dt} = -0.6 \times \frac{\pi}{36} \sqrt{2 \times 32 \times h}.$$ 25. Simplify: $$0.16 \pi h^2 \frac{dh}{dt} = -0.6 \frac{\pi}{36} \sqrt{64 h} = -0.6 \frac{\pi}{36} 8 \sqrt{h} = -0.6 \times \frac{8 \pi}{36} \sqrt{h} = - \frac{4.8 \pi}{36} \sqrt{h} = - \frac{4.8 \pi}{36} \sqrt{h}.$$ 26. Divide both sides by $0.16 \pi h^2$: $$\frac{dh}{dt} = - \frac{4.8 \pi}{36} \sqrt{h} \times \frac{1}{0.16 \pi h^2} = - \frac{4.8}{36} \times \frac{1}{0.16} \times \frac{\sqrt{h}}{h^2} = - \frac{4.8}{36 \times 0.16} h^{-3/2}.$$ 27. Compute coefficients: $$36 \times 0.16 = 5.76, \quad \frac{4.8}{5.76} = \frac{20}{24} = \frac{5}{6}.$$ 28. Final differential equation: $$\boxed{\frac{dh}{dt} = - \frac{5}{6} h^{-3/2}}.$$