Step 1 Given

\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\left({x}+{y}-{1}\right)}+{\frac{{{x}+{y}}}{{{\log{{\left({x}+{y}\right)}}}}}}\)

Step 2 solving Differential Equations

Let x+y=v.

Then, \(\displaystyle{1}+{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}\)

The given differential equation reduces to

\(\displaystyle{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}-{1}={v}+{\frac{{{v}}}{{{\log{{v}}}}}}\)

\(\displaystyle{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}={v}+{\frac{{{v}}}{{{\log{{v}}}}}}\)

\(\displaystyle{\frac{{{\log{{v}}}}}{{{v}{\left({1}+{\log{{v}}}\right)}}}}{d}{v}={\left.{d}{x}\right.}\)

On integrating, we get

\(\displaystyle\int{\frac{{{\log{{v}}}}}{{{\left({1}+{\log{{v}}}\right)}}}}{\frac{{{1}}}{{{v}}}}{d}{v}=\int{1}.{\left.{d}{x}\right.}+{C}\)

\(\displaystyle\int{\frac{{{t}-{1}}}{{{t}}}}{\left.{d}{t}\right.}={x}+{C}\)

Where \(\displaystyle{t}={1}+{\log{{v}}}\)

\(\displaystyle{t}-{\log{{t}}}={x}+{C}\)

\(\displaystyle{\left({1}+{\log{{v}}}\right)}-{\log{{\left[{1}+{\log{{v}}}\right]}}}={x}+{C}\)

\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\left({x}+{y}-{1}\right)}+{\frac{{{x}+{y}}}{{{\log{{\left({x}+{y}\right)}}}}}}\)

Step 2 solving Differential Equations

Let x+y=v.

Then, \(\displaystyle{1}+{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}\)

The given differential equation reduces to

\(\displaystyle{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}-{1}={v}+{\frac{{{v}}}{{{\log{{v}}}}}}\)

\(\displaystyle{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}={v}+{\frac{{{v}}}{{{\log{{v}}}}}}\)

\(\displaystyle{\frac{{{\log{{v}}}}}{{{v}{\left({1}+{\log{{v}}}\right)}}}}{d}{v}={\left.{d}{x}\right.}\)

On integrating, we get

\(\displaystyle\int{\frac{{{\log{{v}}}}}{{{\left({1}+{\log{{v}}}\right)}}}}{\frac{{{1}}}{{{v}}}}{d}{v}=\int{1}.{\left.{d}{x}\right.}+{C}\)

\(\displaystyle\int{\frac{{{t}-{1}}}{{{t}}}}{\left.{d}{t}\right.}={x}+{C}\)

Where \(\displaystyle{t}={1}+{\log{{v}}}\)

\(\displaystyle{t}-{\log{{t}}}={x}+{C}\)

\(\displaystyle{\left({1}+{\log{{v}}}\right)}-{\log{{\left[{1}+{\log{{v}}}\right]}}}={x}+{C}\)