MCQ
The solution to the differential equation $y\, \ln y + xy' = 0,$ where $y (1) = e,$ is
  • $x (\ln y) = 1$
  • B
    $xy (\ln y) = 1$
  • C
    $(\ln y)^2 = 2$
  • D
    $\ln y + \left( {\frac{{{x^2}}}{2}} \right)y  = 1$

Answer

Correct option: A.
$x (\ln y) = 1$
a
$x  \frac{{dy}}{{dx}}  + y (\ln y) = 0$
$\Rightarrow \int {\frac{{dx}}{x}} +  \int {\frac{{dy}}{{y\,(\ln \,y)}}}  = C ;$
$\ln (x \ln y) = C$
$\Rightarrow C = 0 $

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