Question

For the following differential equation, find the general solution:- \[\frac{dy}{dx} + y = 1\]

Answer 1

We have,

\[\frac{dy}{dx} + y = 1\]

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

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

Integrating both sides, we get

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

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

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

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

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

\[ \Rightarrow \frac{y - 1}{C} = e^{- x} \]

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