Question

Find the particular solution of the differential equation \[\frac{dy}{dx} = \frac{xy}{x^2 + y^2}\] given that y = 1 when x = 0.

 

Answer 1

\[\frac{dy}{dx} = \frac{xy}{x^2 + y^2}\]          .....(1)
\[\Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}\]
Substituting the value of y = xv and \[\frac{dy}{dx} = v + x\frac{dv}{dx}\] in (1), we get
\[\therefore v + x\frac{dv}{dx} = \frac{x^2 v}{x^2 + x^2 v^2}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v}{1 + v^2} - v\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{- v^3}{1 + v^2}\]
\[ \Rightarrow \frac{1 + v^2}{- v^3}dv = \frac{1}{x}dx\]
\[ \Rightarrow \int\frac{1 + v^2}{- v^3}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow \frac{1}{2 v^2} - \log v = \log x + C\]
\[\Rightarrow \frac{1}{2 \left( \frac{y}{x} \right)^2} - \log\frac{y}{x} = \log x + C\]
\[ \Rightarrow \frac{x^2}{2 y^2} - \log\frac{y}{x} = \log x + C . . . . . \left( 2 \right)\]
\[ \Rightarrow \frac{0}{2} - \log\frac{1}{0} = \log0 + C\]
\[ \Rightarrow C = 0\]
Substituting the value of C in (2), we get
\[\frac{x^2}{2 y^2} - \log\frac{y}{x} = \log x\]
\[ \Rightarrow \frac{x^2}{2 y^2} = \log x + \log\frac{y}{x}\]
\[ \Rightarrow \frac{x^2}{2 y^2} = \log y\]