Differential Equations 1 Question 8
8. If $y=y(x)$ and $\frac{2+\sin x}{y+1} \frac{d y}{d x}=-\cos x, y(0)=1$, then $y \frac{\pi}{2}$ equals
(2004, 1M)
(a) $1 / 3$
(b) $2 / 3$
(c) $-1 / 3$
(d) 1
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Answer:
Correct Answer: 8. (c)
Solution:
- Given, $\frac{d y}{d x}=\frac{-\cos x(y+1)}{2+\sin x}$
$$ \Rightarrow \quad \frac{d y}{y+1}=\frac{-\cos x}{2+\sin x} d x $$
On integrating both sides
$$ \begin{aligned} \int \frac{d y}{y+1} & =-\int \frac{\cos x}{2+\sin x} d x \\ \Rightarrow \quad \log (y+1) & =-\log (2+\sin x)+\log c \end{aligned} $$
When $x=0, y=1 \Rightarrow c=4$
$$ \begin{array}{ll} \Rightarrow & y+1=\frac{4}{2+\sin x} \\ \therefore & y \frac{\pi}{2}=\frac{4}{3}-1 \\ \Rightarrow \quad y \frac{\pi}{2}=\frac{1}{3} \end{array} $$
