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Question
Solve the following differential equation:-
\[\left( x + y \right)\frac{dy}{dx} = 1\]
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Solution
We have,
\[\left( x + y \right)\frac{dy}{dx} = 1\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1}{\left( x + y \right)}\]
Let x + y = v
\[ \Rightarrow 1 + \frac{dy}{dx} = \frac{dv}{dx}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{dv}{dx} - 1\]
\[ \therefore \frac{dv}{dx} - 1 = \frac{1}{v}\]
\[ \Rightarrow \frac{dv}{dx} = \frac{1}{v} + 1\]
\[ \Rightarrow \frac{dv}{dx} = \frac{1 + v}{v}\]
\[ \Rightarrow \frac{v}{1 + v}dv = dx\]
Integrating both sides, we get
\[\int\frac{v}{1 + v}dv = \int dx\]
\[ \Rightarrow \int\frac{v + 1 - 1}{1 + v}dv = \int dx\]
\[ \Rightarrow \int dv - \int\frac{1}{1 + v}dv = \int dx\]
\[ \Rightarrow v - \log \left| v + 1 \right| = x - \log C\]
\[ \Rightarrow x + y - \log \left| x + y + 1 \right| = x - \log C\]
\[ \Rightarrow y - \log \left| x + y + 1 \right| = - \log C\]
\[ \Rightarrow y = \log\left| x + y + 1 \right| - \log C\]
\[ \Rightarrow y = \log\left| \frac{x + y + 1}{C} \right|\]
\[ \Rightarrow C e^y = x + y + 1\]
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