Question

athuleee help........
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Answer 1

Answer:

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Answer 2

$Given \: \: Question \: \: Is \: \: \: \\ \\ x \frac{dy}{dx} + y = x \cos(x) + \sin(x) \: \: \: find \: \: y \: \: \: \\ when \: \: y(x) = 1 \: \: at \: \: x = \frac{\pi}{2} \\ Answer \: \\ \\ x \frac{dy}{dx} + y = x \cos(x) + \sin(x) \\ \\ xdy + ydx = (x \cos(x) + \sin(x) )dx \\ \\ d(xy) = (x \cos(x) + \sin(x) )dx \\ becoz \: \: d(xy) = xdy + ydx \\ \\ taking \: integral \: on \: both \: sides \\ \\ xy = int(x \cos(x) + \sin(x) )dx \\ \\ xy = xsin(x) - cos(x) - cos(x) + c \\ \\ xy = x \sin(x) - 2 \cos(x) + c \\ \\ xy(x) = x \sin(x) - 2 \cos(x) + c \: \: \: ... \: \: equation \: \: \: 01\\ \\ now \: impose \: conditions \: \: at \: \: x = \frac{\pi}{2} \: \: \: \: y(x) = 1 \\ \\ \frac{\pi}{2} \times (1) = \frac{\pi}{2} \times 1 + c \\ \\ c = 0 \\ from \: \: \:equation \: \: \: 01 \\ \\ \\ xy = x \sin(x) - 2\cos(x) \\ \\ therefore \: \: the \: \: general \: solution \: of \: given \: differential \: equation \: is \\ \\ \\ y = \sin(x ) - 2 \frac{ \cos(x) }{x}$