Question

Solve the following differential equation:-

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

Answer 1

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

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

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

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

\[ \Rightarrow v - \log \left| v + 1 \right| = x - \log C\]

\[ \Rightarrow x + y - \log \left| x + y + 1 \right| = x - \log C\]

\[ \Rightarrow y - \log \left| x + y + 1 \right| = - \log C\]

\[ \Rightarrow y = \log\left| x + y + 1 \right| - \log C\]

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

\[ \Rightarrow C e^y = x + y + 1\]