This is a linear ODE with $P = 3\sec^2 x$ and $Q = \sec^2 x$.

Integrating factor: $e^{\int 3\sec^2 x\,dx} = e^{3\tan x}$

$\frac{d}{dx}(ye^{3\tan x}) = \sec^2 x \cdot e^{3\tan x}$

Integrate: $ye^{3\tan x} = \frac{1}{3}e^{3\tan x} + C$

At $x=\pi/4$, $\tan(\pi/4)=1$, $y=4/3$: $\frac{4}{3}e^3 = \frac{1}{3}e^3 + C \Rightarrow C = e^3$

So $y = \frac{1}{3} + e^3 \cdot e^{-3\tan x}$

At $x=-\pi/4$, $\tan(-\pi/4)=-1$: $y = \frac{1}{3} + e^3 \cdot e^{3} = \frac{1}{3} + e^3$... wait: $e^{-3(-1)}=e^3$, so $y = \frac{1}{3}+e^3 \cdot e^3 = \frac{1}{3}+e^6$.

Correct: $y(-\pi/4) = \mathbf{\frac{1}{3}+e^6}$. So correct option index is 0.

Ball 1 falls for 18 s: $h_1 = \dfrac{1}{2}(10)(18)^2 = 1620\text{ m}$

Ball 2 falls for $18 - 6 = 12\text{ s}$: $h_2 = 12v + \dfrac{1}{2}(10)(12)^2 = 12v + 720$

Setting $h_1 = h_2$: $12v + 720 = 1620 \Rightarrow 12v = 900 \Rightarrow v = \mathbf{75\text{ m/s}}$

Extension $\Delta L = \frac{FL}{AY} = \frac{FL}{\pi (d/2)^2 Y} \propto \frac{L}{d^2}$ [cite: 482, 484].

• New extension $\Delta L' \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{1}{2} \Delta L$
• $\Delta L' = \frac{1}{2} \times 0.04 = 0.02\text{ m}$[cite: 484].

Since each pair of circles has exactly one point in common, the circles are externally tangent to each other. If all circles have the same radius $r$, then the distance between any two centers equals $2r$.

Since triangle $PQR$ is equilateral with side length $2r$:

• Perimeter of $\triangle PQR = 3(2r) = 6r$
• Circumference of circle with center $Q = 2\pi r$

Comparing: $6r$ vs $2\pi r$

Dividing by $2r$: $3$ vs $\pi$

Since $\pi \approx 3.14159 > 3$, we have $2\pi r > 6r$.

Therefore, Column B (the circumference) is greater than Column A (the perimeter).