Solve the following homogeneous differential equations \frac{dy}{dx}=\frac{x-y}{x+y}

Painevg 2021-12-31 Answered
Solve the following homogeneous differential equations
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{x}-{y}}}{{{x}+{y}}}}\)

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Expert Answer

Suhadolahbb
Answered 2022-01-01 Author has 821 answers
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{x}-{y}}}{{{x}+{y}}}}\) (1)
We have to solve the equation
Rearrange the equation
\(\displaystyle{\left({x}+{y}\right)}{\left.{d}{y}\right.}={\left({x}-{y}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle{x}{\left.{d}{y}\right.}+{y}{\left.{d}{y}\right.}={x}{\left.{d}{x}\right.}-{y}{\left.{d}{x}\right.}\)
\(\displaystyle{x}{\left.{d}{y}\right.}+{y}{\left.{d}{x}\right.}+{y}{\left.{d}{y}\right.}-{y}{\left.{d}{x}\right.}={0}\)
\(\displaystyle{d}{\left({x}{y}\right)}+{y}{\left.{d}{y}\right.}-{x}{\left.{d}{x}\right.}={0}\)
Integrating we get
\(\displaystyle{x}{y}+{\frac{{{y}^{{{2}}}}}{{{2}}}}-{\frac{{{x}^{{{2}}}+{c}}}{{{2}}}}={0}\)
\(\displaystyle{x}{y}+{\frac{{{y}^{{{2}}}}}{{{2}}}}-{\frac{{{x}^{{{2}}}}}{{{2}}}}+{c}={0}\)
It is solution of differential equation.
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Hector Roberts
Answered 2022-01-02 Author has 3403 answers
Put, \(\displaystyle{y}={v}{x}\)
\(\displaystyle\Rightarrow{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={v}+{x}{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}\)
\(\displaystyle\Rightarrow{v}+{x}{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}={\frac{{{1}-{v}}}{{{1}+{v}}}}\)
\(\displaystyle\Rightarrow{x}{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}={\frac{{{1}-{2}{v}-{v}^{{{2}}}}}{{{1}+{v}}}}\)
\(\displaystyle\Rightarrow\int{\frac{{{v}+{1}}}{{{\left({v}+{1}\right)}^{{{2}}}-{2}}}}{\left.{d}{x}\right.}=-\int{\frac{{{1}}}{{{x}}}}{\left.{d}{x}\right.}\)
\(\displaystyle\Rightarrow{\frac{{{1}}}{{{2}}}}{\ln}{\left|{\left({v}+{1}\right)}^{{{2}}}-{2}\right|}=-{\ln}{\left|{x}\right|}+{\ln}{\left|{c}_{{{1}}}\right|}\)
\(\displaystyle\Rightarrow{\ln}{\left|{\left[{\left({v}+{1}\right)}^{{{2}}}-{2}\right]}\right|}={2}{\ln}{\left|{\frac{{{c}_{{{1}}}}}{{{x}}}}\right|}\)
\(\displaystyle\Rightarrow{\left({v}+{1}\right)}^{{{2}}}-{2}={\frac{{{{c}_{{{1}}}^{{{2}}}}}}{{{x}^{{{2}}}}}}\)
\(\displaystyle\Rightarrow{x}^{{{2}}}{\left({v}^{{{2}}}+{2}{v}-{1}\right)}={C}\)
Where \(\displaystyle{C}={{c}_{{{1}}}^{{{2}}}}\)
Since, \(\displaystyle{v}={\frac{{{y}}}{{{x}}}}\)
\(\displaystyle\Rightarrow{y}^{{{2}}}+{2}{x}{y}-{x}^{{{2}}}={C}\)
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karton
Answered 2022-01-09 Author has 8454 answers

\(\begin{array}{}\text{Solution:} \\\frac{dy}{dx}=\frac{x-y}{x+y}=\frac{1-y/x}{1+y/x} (i) \\\text{Since, it is a homogeneous differential equation,} \\\text{Put}\ y=vx \\\Rightarrow \frac{dy}{dx}=v+x \frac{dv}{dx} \\\text{Hence, eq. (i) becomes} \\v+x \frac{dv}{dx}=\frac{1-v}{1+v} \\\Rightarrow x \frac{dv}{dx}=\frac{1-v}{1+v}-v \\\Rightarrow x \frac{dv}{dx}=\frac{1-2v-v^{2}}{1+v} \\\Rightarrow \frac{dx}{x}=\frac{-1}{2} (\frac{-2-2v}{1-2v-v^{2}})dv \\\text{On integration, we get} \\\int \frac{1}{x}dx=\frac{-1}{2} \int \frac{-2-2v}{1-2v-v^{2}}dv \\\Rightarrow \log c_{1}-2 \log x=\log (1-2v-v^{2}) \\\Rightarrow \log \frac{c_{1}}{x^{2}}=\log (\frac{x^{2}-2yx-y^{2}}{x^{2}}) \\\Rightarrow x^{2}-2yx-y^{2}=c_{1} \\\Rightarrow x^{2}-2yx-y^{2}+c=0 (c=-c_{1}) \end{array}\)

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