Question

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

Answer 1

We have,

\[\left( x + y + 1 \right)\frac{dy}{dx} = 1\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1}{\left( x + y + 1 \right)}\]
\[\text{ Let }x + y + 1 = v\]
\[ \therefore 1 + \frac{dy}{dx} = \frac{dv}{dx}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{dv}{dx} - 1\]
\[ \therefore \frac{dv}{dx} - 1 = \frac{1}{v}\]
\[ \Rightarrow \frac{dv}{dx} = \frac{1}{v} + 1\]
\[ \Rightarrow \frac{v}{v + 1}dv = dx\]
Integrating both sides, we get
\[\int\frac{v}{v + 1}dv = \int dx\]
\[ \Rightarrow \int\frac{v + 1 - 1}{v + 1}dv = \int dx\]
\[ \Rightarrow \int\left( 1 - \frac{1}{v + 1} \right)dv = \int dx\]
\[ \Rightarrow v - \log\left| v + 1 \right| = x + K\]
\[ \Rightarrow x + y + 1 - \log\left| x + y + 1 + 1 \right| = x + K\]
\[ \Rightarrow y - \log\left| x + y + 2 \right| = K - 1\]
\[ \Rightarrow y - \log\left| x + y + 2 \right| = C_1 ...........\left( C_1 = K - 1 \right)\]
\[ \Rightarrow y - C_1 = \log\left| x + y + 2 \right|\]
\[ \Rightarrow e^{y - C_1} = x + y + 2\]
\[ \Rightarrow \frac{e^y}{e^{C_1}} = x + y + 2\]
\[ \Rightarrow e^{- C_1} e^y = x + y + 2\]
\[ \Rightarrow C e^y = x + y + 2 .............\left( C = e^{- C_1} \right)\]
\[ \Rightarrow x = C e^y - y - 2\]