Question

Find one-parameter families of solution curves of the following differential equation:-

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

Solve the following differential equation:-

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

Answer 1

We have,
\[\left( x + y \right)\frac{dy}{dx} = 1\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1}{x + y}\]
\[ \Rightarrow \frac{dx}{dy} = x + y\]
\[ \Rightarrow \frac{dx}{dy} - x = y . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form
\[\frac{dx}{dy} + Px = Q\]
where
\[P = - 1\]
\[Q = y\]
\[ \therefore I . F . = e^{\int P\ dy} \]
\[ = e^{\int - 1 dy} \]
\[ = e^{- y} \]
\[\text{ Multiplying both sides of }(1)\text{ by }e^{- y} ,\text{ we get }\]
\[ e^{- y} \left( \frac{dx}{dy} - x \right) = e^{- y} y\]
\[ \Rightarrow e^{- y} \frac{dx}{dy} - e^{- y} x = e^{- y} y\]
Integrating both sides with respect to y, we get

\[ \Rightarrow e^{- y} x = y\int e^{- y} dy - \int\left[ \frac{d}{dy}\left( y \right)\int e^{- y} dy \right]dy + C\]
\[ \Rightarrow e^{- y} x = - y e^{- y} - e^{- y} + C\]
\[ \Rightarrow e^{- y} x + y e^{- y} + e^{- y} = C\]
\[ \Rightarrow \left( x + y + 1 \right) e^{- y} = C\]
\[ \Rightarrow \left( x + y + 1 \right) = C e^y \]
\[\text{ Hence, }\left( x + y + 1 \right) = C e^y\text{ is the required solution.}\]