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EEJ MAIN Mathematics QUESTION #6007
Question 1

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 $y\!\left(\dfrac{\pi}{6}\right)$.

  • $-\dfrac{8}{9}\pi^2$
  • $-\dfrac{4}{9}\pi^2$
  • $\dfrac{4}{9\sqrt{3}}\pi^2$
  • $\dfrac{-8}{9\sqrt{3}}\pi^2$✔️
Correct Answer Explanation

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: $y\!\left(\frac{\pi}{6}\right) = \mathbf{-\dfrac{8\pi^2}{9\sqrt{3}}}$