Question

For the following differential equation, find the general solution:- `y log y dx − x dy = 0`

Answer 1

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

\[\int\frac{1}{t}dt = \int\frac{1}{x}dx\]

\[ \Rightarrow \log \left( t \right) = \log x + \log C\]

\[ \Rightarrow \log \left( \log y \right) = \log x + \log C\]

\[ \Rightarrow \log \left( \log y \right) = \log Cx\]

\[ \Rightarrow \log y = Cx\]

\[ \Rightarrow y = e^{Cx}\]