Let y = y(x) be the solution of the differential equation $xdy = (y + x^{3} cosx) dx$ with $y (π) = 0$ then $y (\frac{π}{2})$ is equal to:

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$\frac{π^{2}}{2} - \frac{π}{2}$

$\frac{π^{2}}{4} - \frac{π}{2}$

$\frac{π^{2}}{2} + \frac{π}{2}$

$\frac{π^{2}}{4} + \frac{π}{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{d y}{d x} = \frac{y}{x} + x^{2} \cos x \\ \Rightarrow \frac{d y}{d x} - \frac{1}{x} y = x^{2} \cos x \\ I . F = e^{- \int \frac{1}{x} d x} = e^{- \log x} = e^{\log \frac{1}{x}} = \frac{1}{x} \\ S o l u t i o n i s \\ y (I . F) = \int x^{2} \cos x (I . F) d x \\ = \int x^{2} \cos x (\frac{1}{x}) d x \\ = \int x \cos x d x \\ = x \sin x - \int \sin x d x \\ y (\frac{1}{x}) = x \sin x + \cos x + c \\ p u t x = π \end{array}$

$\begin{array}{l} \frac{y (π)}{π} = π \sin π + \cos π + c \\ \Rightarrow 0 = 0 - 1 + c \\ \Rightarrow c = 1 \\ \frac{y}{x} = x \sin x + \cos x + 1 \\ \Rightarrow y = x^{2} \sin x + x \cos x + x \\ p u t x = \frac{π}{2} \\ ∴ y (\frac{π}{2}) = \frac{π^{2}}{4} \sin \frac{π}{2} + \frac{π}{2} \cos \frac{π}{2} + \frac{π}{2} \\ = \frac{π^{2}}{4} + \frac{π}{2} \end{array}$