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Find the general solution $y\;\log\: y\;dx-x\;dy=0$

1 Answer

  • $\int \log x = \large\frac{1}{x}$
Step 1:
Given: $y\log ydx -xdy = 0$
$xdy = y\log ydx$
Seperating the variables we get,
$\large\frac{dy}{y\log y} =\frac{ dx}{x}$
Step 2:
Integrating on both sides we get,
$\int\large\frac{ y}{y\log y} = \int\large\frac{ dx}{x}$
$\log(\log y) = \log x + \log C$
$\log(\log y) = \log xC$
$\log_e y= xC$
$e^{Cx} = y$
This is the required general equation.
answered Aug 15, 2013 by sreemathi.v