If $y(x)$ satisfies the differential equation $y' - y\tan x = 2x\sec x$ and $y(0) = 0$, then

$y' - y\tan x = 2x\sec x, \qquad y(0) = 0$ $\text{Solve for } y(x), \text{ then test } y,\, y' \text{ at } \tfrac{\pi}{4}, \tfrac{\pi}{3}$ $y' + P y = Q, \quad P = -\tan x, \quad Q = 2x\sec x$ $\mu = e^{\int P\,dx} = e^{-\int \tan x\,dx} = e^{\ln\cos x} = \cos x$ $\cos x\,y' - \sin x\,y = 2x \Rightarrow (y\cos x)' = 2x$ $y\cos x = x^{2} + C \Rightarrow y = (x^{2} + C)\sec x$ $y(0) = C\cdot 1 = 0 \Rightarrow C = 0$ $\boxed{y(x) = x^{2}\sec x}$ $y\!\left(\tfrac{\pi}{4}\right) = \dfrac{\pi^{2}}{16}\cdot\sqrt{2} = \dfrac{\pi^{2}\sqrt{2}}{16} = \dfrac{\pi^{2}}{8\sqrt{2}} \Rightarrow (1) \text{ TRUE}$ $y' = 2x\sec x + x^{2}\sec x\tan x$ $y'\!\left(\tfrac{\pi}{4}\right) = \dfrac{\pi\sqrt{2}}{2} + \dfrac{\pi^{2}\sqrt{2}}{16} \neq \dfrac{\pi^{2}}{18} \Rightarrow (2) \text{ FALSE}$ $y\!\left(\tfrac{\pi}{3}\right) = \dfrac{\pi^{2}}{9}\cdot 2 = \dfrac{2\pi^{2}}{9} \neq \dfrac{\pi^{2}}{9} \Rightarrow (3) \text{ FALSE}$ $y'\!\left(\tfrac{\pi}{3}\right) = \dfrac{4\pi}{3} + \dfrac{2\pi^{2}\sqrt{3}}{9} = \dfrac{4\pi}{3} + \dfrac{2\pi^{2}}{3\sqrt{3}} \Rightarrow (4) \text{ TRUE}$ $\boxed{\text{TRUE statements: } (1) \text{ and } (4)}$ $y = x^{2}\sec x; \quad (1), (4) \text{ TRUE}$