Solve the differential equation : (2x+3+2y)dy/dx = (x+1+y)

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$$\begin{gathered} Given differential equation: \left(2x + 3 + 2y\right)\frac{dy}{dx} = \left(x + 1 + y\right) \\ \Rightarrow \frac{dy}{dx} = \frac{x + y + 1}{2x + 2y + 3} = \frac{x + y + 1}{2\left(x + y + 1\right) + 1} ...\left(i\right) \\ Let x + y + 1 = t \\ \therefore 1 + \frac{dy}{dx} = \frac{dt}{dx} \Rightarrow \frac{dy}{dx} = \frac{dt}{dx} - 1 \\ Thus, equation \left(i\right) becomes \\ \frac{dt}{dx} - 1 = \frac{t}{2t + 1} \\ \Rightarrow \frac{dt}{dx} = \frac{t}{2t + 1} + 1 = \frac{3t + 1}{2t + 1} \\ \Rightarrow \int \frac{2t + 1}{3t + 1}dt = \int dx \\ \Rightarrow \frac{2}{3}\int \frac{t + {\displaystyle \frac{1}{2}}}{t + {\displaystyle \frac{1}{3}}}dt = \log \left|x\right| + C \\ \Rightarrow \frac{2}{3}\int \frac{t + {\displaystyle \frac{1}{2}} - {\displaystyle \frac{1}{6}} + {\displaystyle \frac{1}{6}}}{t + {\displaystyle \frac{1}{3}}}dt = \log \left|x\right| + C \\ \Rightarrow \frac{2}{3}\left[\int \frac{t + {\displaystyle \frac{1}{3}}}{t + {\displaystyle \frac{1}{3}}}dt + \int \frac{{\displaystyle \frac{1}{6}}}{t + {\displaystyle \frac{1}{3}}}dt\right] = \log \left|x\right| + C \\ \Rightarrow \frac{2}{3}\left[\int dt + \frac{1}{6}\int \frac{1}{t + {\displaystyle \frac{1}{3}}}dt\right] = \log \left|x\right| + C \\ \Rightarrow \frac{2t}{3} + \frac{1}{6}\log \left|t + \frac{1}{3}\right| = \log \left|x\right| + C \\ \Rightarrow \frac{2\left(x + y + 1\right)}{3} + \frac{1}{6}\log \left|x + y + \frac{4}{3}\right| = \log \left|x\right| + C \left[\because t = x + y + 1\right] \end{gathered}$$

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