Question

\[x\frac{dy}{dx} + y = y^2\]

Answer 1

We have,
\[x\frac{dy}{dx} + y = y^2 \]
\[ \Rightarrow x\frac{dy}{dx} = y^2 - y\]
\[ \Rightarrow \frac{1}{y^2 - y}dy = \frac{1}{x}dx\]
Integrating both sides, we get 
\[\int\frac{1}{y^2 - y}dy = \int\frac{1}{x}dx\]
\[ \Rightarrow \int\frac{1}{y\left( y - 1 \right)}dy = \int\frac{1}{x}dx . . . . . \left( 1 \right)\]
\[\text{ Let }\frac{1}{y\left( y - 1 \right)} = \frac{A}{y} + \frac{B}{y - 1}\]
\[ \Rightarrow 1 = A\left( y - 1 \right) + B\left( y \right)\]
\[\text{ Putting }y = 0,\text{ we get }\]
\[1 = - A\]
\[ \Rightarrow A = - 1\]
\[\text{ Putting }y = 1, \text{ we get }\]
\[1 = B\]
\[ \therefore \frac{1}{y\left( y - 1 \right)} = \frac{- 1}{y} + \frac{1}{y - 1}\]
\[ \Rightarrow \int\frac{1}{y\left( y - 1 \right)}dy = \int\frac{- 1}{y} dy + \int\frac{1}{y - 1}dy . . . . . \left( 2 \right) \]
From (1) & (2), we get 
\[\int\frac{- 1}{y} dy + \int\frac{1}{y - 1}dy = \int\frac{1}{x}dx \]
\[ \Rightarrow - \log \left| y \right| + \log \left| y - 1 \right| = \log \left| x \right| + \log C\]
\[ \Rightarrow \log \left| \frac{y - 1}{y} \right| - \log \left| x \right| = \log C\]
\[ \Rightarrow \log\left| \frac{y - 1}{xy} \right| = \log C\]
\[ \Rightarrow \frac{y - 1}{xy} = C\]
\[ \Rightarrow y - 1 = Cxy\]
\[\text{ Hence, }y - 1 = Cxy\text{  is the required solution .}\]