The solution of $dy = \cos x\left( {2 - y{\rm{cosec}}x} \right)dx$, where $y = \sqrt 2 $, when $x = \frac{\pi }{4}$ is

$y = \sin x + \frac{1}{2}{\rm{cosec}}x$

$y = \tan \left( {\frac{x}{2}} \right) + \cot \left( {\frac{x}{2}} \right)$

$y = \left( {\frac{1}{{\sqrt 2 }}} \right)\sec \left( {\frac{x}{2}} \right) + \sqrt 2 \cos \left( {\frac{x}{2}} \right)$

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

The solution of

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

, where

y = \sqrt 2

, when

x = \frac{\pi }{4}

View answer and explanation

Correct option: A

y = \sin x + \frac{1}{2}\mathrm{cosec}x

Option A (

y = \sin x + \frac{1}{2}\mathrm{cosec}x

) is correct because it satisfies the given differential equation and the initial condition

y = \sqrt 2

when

x = \frac{\pi }{4}

Practice All Chapter MCQs →

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x

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y=y_{2}(x)

y=y(t)

\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}

\alpha>0, \beta>0

\gamma>0

\lim \limits_{{x \rightarrow \infty}} y(t)