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

sibuzwaW 2021-09-26 Answered
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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Expert Answer

tabuordy
Answered 2021-09-27 Author has 13858 answers
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 \(\displaystyle{t}={1}+{\log{{v}}}\)
\(\displaystyle{t}-{\log{{t}}}={x}+{C}\)
\(\displaystyle{\left({1}+{\log{{v}}}\right)}-{\log{{\left[{1}+{\log{{v}}}\right]}}}={x}+{C}\)
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