MCQ
Solution of the equation $(x + \log y)dy + y\,dx = 0$ is
- A$xy + y\log y = c$
- ✓$xy + y\log y - y = c$
- C$xy + \log y - x = c$
- DNone of these
$y\frac{{dx}}{{dy}} + x = - \log y$ ==> $\frac{{dx}}{{dy}} + \frac{x}{y} = - \frac{{\log y}}{y}$
$I.F.$ =${e^{\int {\frac{1}{y}dy} }} = y$
Hence solution is $x.y = - \int {y.\frac{{\log ydy}}{y} + c} $
==> $xy = - (y\log y - y) + c$ ==> $xy + (y\log y - y) = c$.
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