Question

(x + y − 1) dy = (x + y) dx

Answer 1

We have,

\[\left( x + y - 1 \right)dy = \left( x + y \right)dx\]

\[\frac{dy}{dx} = \frac{\left( x + y \right)}{\left( x + y - 1 \right)}\]

Putting `x + y = v,` we get

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

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

\[ \therefore \frac{dv}{dx} - 1 = \frac{v}{\left( v - 1 \right)}\]

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

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

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

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

Integrating both sides, we get

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

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

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

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

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

\[ \Rightarrow \frac{1}{2}v - \frac{1}{4}\log \left| 2v - 1 \right| = x + C\]

\[ \Rightarrow \frac{1}{2}\left( x + y \right) - \frac{1}{4}\log \left| 2x + 2y - 1 \right| = x + C\]

\[ \Rightarrow 2\left( x + y \right) - \log \left| 2x + 2y - 1 \right| = 4x + 4C\]

\[ \Rightarrow 2\left( x + y \right) - 4x - \log \left| 2x + 2y - 1 \right| = 4C \]

\[ \Rightarrow 2\left( y - x \right) - \log \left| 2x + 2y - 1 \right| = k,\text{ where }k = 4C\]