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Question
For the following differential equation, find the general solution:- `y log y dx − x dy = 0`
Solution
We have,
\[y \log y\ dx - x\ dy = 0\]
\[ \Rightarrow y \log y dx = x dy\]
\[ \Rightarrow \frac{1}{x}dx = \frac{1}{y \log y}dy\]
\[ \Rightarrow \frac{1}{y \log y}dy = \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\frac{1}{y \log y}dy = \int\frac{1}{x}dx . . . . . \left( 1 \right)\]
Putting log y = t
\[ \Rightarrow \frac{1}{y}dy = dt\]
Therefore (1) becomes
\[\int\frac{1}{t}dt = \int\frac{1}{x}dx\]
\[ \Rightarrow \log \left( t \right) = \log x + \log C\]
\[ \Rightarrow \log \left( \log y \right) = \log x + \log C\]
\[ \Rightarrow \log \left( \log y \right) = \log Cx\]
\[ \Rightarrow \log y = Cx\]
\[ \Rightarrow y = e^{Cx}\]
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