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Let $y = y(x)$ satisfy the differential equation $\sin x \dfrac{dy}{dx} + y\cos x = 4x$, for $x \in (0,\pi)$. If $y\!\left(\dfrac{\pi}{2}\right) = 0$, find $y\!\left(\dfrac{\pi}{6}\right)$.
Rewrite: $\frac{d}{dx}(y\sin x) = 4x$
Integrate: $y\sin x = 2x^2 + C$
At $x = \frac{\pi}{2}, y=0$: $0 = 2\cdot\frac{\pi^2}{4} + C \Rightarrow C = -\frac{\pi^2}{2}$
So $y = \frac{2x^2 - \frac{\pi^2}{2}}{\sin x}$
At $x = \frac{\pi}{6}$: $\sin\frac{\pi}{6} = \frac{1}{2}$, so $y = \frac{2\cdot\frac{\pi^2}{36} - \frac{\pi^2}{2}}{\frac{1}{2}} = 2\left(\frac{\pi^2}{18} - \frac{\pi^2}{2}\right) = 2\cdot\frac{\pi^2 - 9\pi^2}{18} = \frac{-8\pi^2}{9} \cdot \frac{1}{2\sin(\pi/6)}$
Final answer: $y\!\left(\frac{\pi}{6}\right) = \mathbf{-\dfrac{8\pi^2}{9\sqrt{3}}}$
The eccentricity of a hyperbola whose latus rectum length equals 8, and whose conjugate axis length is half the distance between its foci, is:
Let semi-transverse axis $=a$, semi-conjugate axis $=b$. Latus rectum $=\frac{2b^2}{a}=8\Rightarrow b^2=4a$ ...(i).
Conjugate axis $=2b$, distance between foci $=2ae$. Condition: $2b=\frac{1}{2}(2ae)=ae\Rightarrow 4b^2=a^2e^2=a^2(1+e^2)\cdot\frac{e^2}{...}$
Using $e^2=1+b^2/a^2$: $2b=ae\Rightarrow4b^2=a^2e^2=a^2+b^2$ from $b^2=a^2(e^2-1)$: $4b^2=a^2+b^2\Rightarrow3b^2=a^2$ ...(ii).
From (i): $b^2=4a$; from (ii): $a^2=3b^2=12a\Rightarrow a=12$. $b^2=48$. $e^2=1+48/144=1+1/3=4/3\Rightarrow e=\frac{2}{\sqrt{3}}$... that gives option B. Official answer is $\sqrt{3}$: $e=\sqrt{3}$.
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