The general solution of d y/d x+ ≤ft(x+y/2)= ≤ft(x-y/2) is

Correct Answer is : log⁡tan⁡(y4)=C−2sin⁡(x2) ≤ft(y/4)=C-2 ≤ft(x/2)logtan(4y)=C−2sin(2x)

Correct Answer is : log⁡tan⁡(y4)=C−2sin⁡(x2) ≤ft(y/4)=C-2 ≤ft(x/2)logtan(4y)=C−2sin(2x)

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MHTCET 11 May 2024 Shift 1 Paper

Question : 122 of 150

Marks: +1, -0

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

$\log \tan \left(\frac{y}{2}\right)=C-2 \sin x$
$\log \tan \left(\frac{y}{4}\right)=C-2 \sin \left(\frac{x}{2}\right)$
$\log \tan \left(\frac{y}{2}+\frac{\pi}{4}\right)=C-2 \sin x$
$\log \tan \left(\frac{y}{2}+\frac{\pi}{4}\right)=C-2 \sin \left(\frac{x}{2}\right)$

Solution:

Concept:

Use the identity

\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)

and then separate the variables.

Explanation:

Given:

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

Rewrite it as:

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

Taking

A = \frac{x-y}{2}

and

B = \frac{x+y}{2}

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

Separating the variables:

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

Integrating both sides:

\int \cosec\left(\frac{y}{2}\right) dy = -2 \int \cos\left(\frac{x}{2}\right) dx

This gives:

2 \ln\left|\tan\left(\frac{y}{4}\right)\right| = -4 \sin\left(\frac{x}{2}\right) + C

Dividing throughout by 2:

\log \tan\left(\frac{y}{4}\right) = C - 2 \sin\left(\frac{x}{2}\right)

Answer:

\log \tan\left(\frac{y}{4}\right) = C - 2 \sin\left(\frac{x}{2}\right)

, which matches option B.

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