Question

\[\frac{dy}{dx} = \frac{\left( x - y \right) + 3}{2\left( x - y \right) + 5}\]

Answer 1

We have,
\[\frac{dy}{dx} = \frac{\left( x - y \right) + 3}{2\left( x - y \right) + 5}\]

Putting x - y = v

\[ \Rightarrow 1 - \frac{dy}{dx} = \frac{dv}{dx}\]

\[ \Rightarrow \frac{dy}{dx} = 1 - \frac{dv}{dx}\]

\[ \therefore 1 - \frac{dv}{dx} = \frac{v + 3}{2v + 5}\]

\[ \Rightarrow \frac{dv}{dx} = 1 - \frac{v + 3}{2v + 5}\]

\[ \Rightarrow \frac{dv}{dx} = \frac{2v + 5 - v - 3}{2v + 5}\]

\[ \Rightarrow \frac{dv}{dx} = \frac{v + 2}{2v + 5}\]

\[ \Rightarrow \frac{2v + 5}{v + 2}dv = dx\]

Integrating both sides, we get

\[\int\frac{2v + 5}{v + 2}dv = \int dx\]

\[ \Rightarrow \int\frac{2v + 4 + 1}{v + 2}dv = \int dx\]

\[ \Rightarrow \int\left( \frac{2v + 4}{v + 2} + \frac{1}{v + 2} \right)dv = \int dx\]

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

\[ \Rightarrow 2v + \log \left| v + 2 \right| = x + C\]

\[ \Rightarrow 2\left( x - y \right) + \log\left| x - y + 2 \right| = x + C\]