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EEJ MAIN Mathematics QUESTION #6006
Question 1

Given $(2 + \sin x)\dfrac{dy}{dx} + (y+1)\cos x = 0$ with $y(0) = 1$, find the value of $y\!\left(\dfrac{\pi}{2}\right)$.

  • $\dfrac{1}{3}$✔️
  • $-\dfrac{2}{3}$
  • $-\dfrac{1}{3}$
  • $\dfrac{4}{3}$
Correct Answer Explanation

Separate variables: $\frac{dy}{y+1} = -\frac{\cos x}{2+\sin x}dx$

Integrate: $\ln|y+1| = -\ln|2+\sin x| + C$

So $(y+1)(2+\sin x) = K$

At $x=0, y=1$: $(1+1)(2+0) = K \Rightarrow K = 4$

Thus $y+1 = \frac{4}{2+\sin x}$, so $y = \frac{4}{2+\sin x} - 1$

At $x = \frac{\pi}{2}$: $y = \frac{4}{2+1} - 1 = \frac{4}{3} - 1 = \mathbf{\frac{1}{3}}$