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If the relation y + d/dx(xy) = x(sin x + log x) holds, then which expression gives y?

1. y = cos x + (2)/(x)sin x + (2)/(x²)cos x + (x)/(3)log x - (x)/(9) + (c)/(x²)
2. y = -cos x - (2)/(x)sin x + (2)/(x²)cos x + (x)/(3)log x - (x)/(9) + (c)/(x²)
3. y = -cos x + (2)/(x)sin x + (2)/(x²)cos x - (x)/(3)log x - (x)/(9) + (c)/(x²)
4. None of these

Correct answer: None of these

Solution

The equation reduces to y' + (2/x)y = sinx + logx with integrating factor x^2, giving y = -cosx + (2/x)sinx + (2/x^2)cosx + (x/3)logx - x/9 + C/x^2. None of options (a)-(c) match this (they have a wrong sign on cosx, sinx, or the logx term), so the answer is 'None of these'.

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