MCQ
Solution of the equation $ydx - xdy + \log xdx = 0$ is
  • $y = cx - (1 + \log x)$
  • B
    $y = cx + (1 + \log x)$
  • C
    $y + cx + (1 + \log x) = 0$
  • D
    None of these

Answer

Correct option: A.
$y = cx - (1 + \log x)$
a
(a) The equation is $\frac{{dy}}{{dx}} - \frac{y}{x} = \frac{{\log x}}{x}$
$I.F.$ $ = {e^{\int_{}^{} { - \frac{1}{x}dx} }} = {e^{ - \log x}} = \frac{1}{x}$
Hence solution is $y.\frac{1}{x} = \int_{}^{} {\frac{{\log x}}{x} \times \frac{1}{x}dx} $
==> $\frac{y}{x} = - \frac{{\log x}}{x} - \frac{1}{x} + c$
==> $y = cx - (1 + \log x)$.

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