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प्रश्न
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उत्तर
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
\[\int\frac{2v + 5}{v + 2}dv = \int dx\]
\[ \Rightarrow \int\frac{2v + 4 + 1}{v + 2}dv = \int dx\]
\[ \Rightarrow \int\left( \frac{2v + 4}{v + 2} + \frac{1}{v + 2} \right)dv = \int dx\]
\[ \Rightarrow 2\int dv + \int\frac{1}{v + 2}dv = \int dx\]
\[ \Rightarrow 2v + \log \left| v + 2 \right| = x + C\]
\[ \Rightarrow 2\left( x - y \right) + \log\left| x - y + 2 \right| = x + C\]
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