JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

y=y(x)

is a solution of the differential equation

(x+1)^4 d y+(2 x+2) \sin x d x=\left(x^2+1+2 x\right) \cos x d x

such that

y(0)=0

, then

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

is equal to-

View answer and explanation

Correct option: D

\frac{4}{(\pi+2)^2}

Option D (

\frac{4}{(\pi+2)^2}

) is correct because solving the differential equation and applying the initial condition yields

y(x) = \frac{\sin x}{(x+1)^2}

, and thus

y(\frac{\pi}{2}) = \frac{4}{(\pi+2)^2}

More Questions from DIFFERENTIATION

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x

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x y=1

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[x]

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y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x],

\frac{{dy}}{{dx}}

\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

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