Question

Find the particular solution of the differential equation
(1 – y²) (1 + log x) dx + 2xy dy = 0, given that y = 0 when x = 1.

Answer 1

Given:
\[\left( 1 - y^2 \right)\left( 1 + \log x \right)dx + 2xydy = 0\]

\[ \Rightarrow \left( 1 - y^2 \right)\left( 1 + \log x \right)dx = - 2xydy\]

\[ \Rightarrow \left( \frac{1 + \log x}{2x} \right)dx = - \left( \frac{y}{1 - y^2} \right)dy . . . . . \left( 1 \right)\]

Let: 

\[1 + \log x = t \]

and 

\[\left( 1 - y^2 \right) = p\]

\[ \Rightarrow \frac{1}{x}dx = dt\text{ and }- 2ydy = dp\]

\[\text{ Therefore, }\left( 1 \right)\text{ becomes }\]
\[\int\frac{t}{2}dt = \int\frac{1}{2p}dp\]
\[ \Rightarrow \frac{t^2}{4} = \frac{\log p}{2} + C . . . . . \left( 2 \right)\]
\[\text{ Substituting the values of t and p in }\left( 2 \right), \text{ we get }\]
\[\frac{\left( 1 + \log x \right)^2}{4} = \frac{\log\left( 1 - y^2 \right)}{2} + C . . . . . \left( 3 \right)\]
\[\text{ At }x = 1 \text{ and }y = 0, \left( 3 \right)\text{ becomes }\]
\[C = \frac{1}{4}\]
\[\text{ Substituting the value of C in }\left( 3 \right),\text{ we get }\]
\[\frac{\left( 1 + \log x \right)^2}{4} = \frac{\log\left( 1 - y^2 \right)}{2} + \frac{1}{4}\]
\[ \Rightarrow \left( 1 + \log x \right)^2 = 2\log\left( 1 - y^2 \right) + 1\]
Or 
\[ \left( \log x \right)^2 + \log x^2 = \log \left( 1 - y^2 \right)^2 \]
It is the required particular solution .