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MHTCET 22 Apr 2025 Shift 1 Paper

Question : 101 of 150

Marks: +1, -0

y=y(x)

\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}=-\cos x, y(0)=1

y\left(\frac{\pi}{2}\right)=

Solution:

Concept:

This is a first-order differential equation solvable by separating the variables.

Explanation:

Given:

\left(\frac{2+\sin x}{y+1}\right) \frac{dy}{dx} = -\cos x

Separating the variables:

\frac{1}{y+1} \, dy = \frac{-\cos x}{2+\sin x} \, dx

Integrating both sides:

\ln(y+1) = -\ln(2+\sin x) + C

Using the initial condition

y(0)=1

Since

\sin 0 = 0

, we get:

\ln(2) = -\ln(2) + C \Rightarrow C = 2\ln 2 = \ln 4

Substitute

C = \ln 4

\ln(y+1) = -\ln(2+\sin x) + \ln 4

Using logarithm properties:

\ln(y+1) = \ln\left(\frac{4}{2+\sin x}\right)

Therefore:

y+1 = \frac{4}{2+\sin x} \Rightarrow y = \frac{4}{2+\sin x} - 1

Now find

y\left(\frac{\pi}{2}\right)

\sin\frac{\pi}{2} = 1

, so:

y\left(\frac{\pi}{2}\right) = \frac{4}{2+1} - 1 = \frac{4}{3} - 1 = \frac{1}{3}

Answer:

y\left(\frac{\pi}{2}\right) = \frac{1}{3}

Hence, the correct option is A. $\frac{1}{3}$ .

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