# Solve differential equation. \frac{dy}{dx}=(x+y-1)+\frac{x+y}{\log(x+y)}Z

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

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