SATHEE: Differential Equations 1 Question 8

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 $y \frac{π}{2}$ equals

(2004, 1M)

(a) $1 / 3$

(b) $2 / 3$

(d) 1

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Answer:

Correct Answer: 8. (c)

Solution:

Given, $\frac{d y}{d x} = \frac{- \cos x (y + 1)}{2 + \sin x}$

$\Rightarrow \frac{d y}{y + 1} = \frac{- \cos x}{2 + \sin x} d x$

On integrating both sides

$\begin{aligned} \int \frac{d y}{y + 1} & = - \int \frac{\cos x}{2 + \sin x} d x \\ \Rightarrow \log (y + 1) & = - \log (2 + \sin x) + \log c \end{aligned}$

When $x = 0, y = 1 \Rightarrow c = 4$

$\begin{array}{ll} \Rightarrow & y + 1 = \frac{4}{2 + \sin x} \\ ∴ & y \frac{π}{2} = \frac{4}{3} - 1 \\ \Rightarrow y \frac{π}{2} = \frac{1}{3} \end{array}$

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 $y \frac{π}{2}$ equals

Answer:

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Differential Equations 1 Question 8

8. If y=y(x) and 2+sin⁡xy+1dydx=−cos⁡x,y(0)=1, then yπ2 equals

Answer:

Engineering Preparation

Medical Preparation

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Sample Mock Test

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NCERT Exercise Video Solution

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8. If $y = y (x)$ and $\frac{2 + \sin x}{y + 1} \frac{d y}{d x} = - \cos x, y (0) = 1$ , then $y \frac{π}{2}$ equals