Let $y=y(x)$ be the solution of the differential equation, $xy' - y = {x^2}(x\cos x + \sin x),x > 0.\,\,{\rm{ If }}\,\,\,y(\pi ) = \pi ,$ then $y''\left( {\frac{\pi }{2}} \right) + y\left( {\frac{\pi }{2}} \right)$ is equal to

$2 + \frac{\pi }{2} + \frac{{{\pi ^2}}}{4}$

$1 + \frac{\pi }{2} + \frac{{{\pi ^2}}}{4}$

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Application +4 | -1

Let

y=y(x)

be the solution of the differential equation,

xy' - y = {x^2}(x\cos x + \sin x),x > 0.\,\,\mathrm{ If }\,\,\,y(\pi ) = \pi ,

then

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

is equal to

View answer and explanation

Correct option: D

2 + \frac{\pi }{2}

Option D (

2 + \frac{\pi }{2}

) is correct because it is the result of solving the given differential equation and evaluating the expression at the specified point.

Practice All Chapter MCQs →

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