Question

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

Answer 1

We have,

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

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

Putting `y = vx,` we get

\[\frac{dy}{dx} = v + x\frac{dv}{dx}\]

\[ \therefore v + x\frac{dv}{dx} = v^2 - v\]

\[ \Rightarrow x\frac{dv}{dx} = v^2 - 2v\]

\[ \Rightarrow \frac{1}{v^2 - 2v} dv = \frac{1}{x}dx\]

Integrating both sides, we get

\[\int\frac{1}{v^2 - 2v} dv = \int\frac{1}{x}dx\]

\[ \Rightarrow \int\frac{1}{v^2 - 2v + 1 - 1} dv = \int\frac{1}{x}dx\]

\[ \Rightarrow \int\frac{1}{\left( v - 1 \right)^2 - \left( 1 \right)^2} dv = \int\frac{1}{x}dx\]

\[ \Rightarrow \frac{1}{2}\log \left| \frac{v - 1 - 1}{v - 1 + 1} \right| = \log x + \log C\]

\[ \Rightarrow \log \left| \left( \frac{v - 2}{v} \right)^\frac{1}{2} \right| = \log Cx\]

\[ \Rightarrow \log \left| \left( \frac{\frac{y}{x} - 2}{\frac{y}{x}} \right)^\frac{1}{2} \right| = \log Cx\]

\[ \Rightarrow \log \left| \left( \frac{y - 2x}{y} \right)^\frac{1}{2} \right| = \log Cx\]

\[ \Rightarrow \left( \frac{y - 2x}{y} \right)^\frac{1}{2} = Cx\]

\[ \Rightarrow \frac{y - 2x}{y} = C^2 x^2 \]

\[ \Rightarrow y - 2x = k x^2 y,\text{ where }k = C^2\]