Determine which of the following differential equations are homogeneous, then

Cheexorgeny 2021-12-31 Answered
Determine which of the following differential equations are homogeneous, then solve the general solution. If initial condition is given, find the particular solution:
\(\displaystyle{x}{y}{\left.{d}{x}\right.}-{\left({x}+{2}{y}\right)}^{{{2}}}{\left.{d}{y}\right.}={0}\)

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Expert Answer

Jim Hunt
Answered 2022-01-01 Author has 1929 answers

\(\displaystyle{x}{y}{\left.{d}{x}\right.}-{\left({x}+{2}{y}\right)}^{{{2}}}{\left.{d}{y}\right.}={0}\)
\(\displaystyle{x}{y}{\left.{d}{x}\right.}={\left({x}+{2}{y}\right)}^{{{2}}}{\left.{d}{y}\right.}\)
\(\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}={\frac{{{x}^{{{2}}}+{4}{y}^{{{2}}}+{4}{x}{y}}}{{{x}{y}}}}\)
\(\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}={\left({\frac{{{x}}}{{{y}}}}\right)}+{4}{\left({\frac{{{y}}}{{{x}}}}\right)}+{4}\)
It is a Homogeneous differential equation.
Let \(\displaystyle{\frac{{{x}}}{{{y}}}}={v}\Rightarrow{x}={v}{y}\)
So, \(\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}={v}+{y}{\frac{{{d}{v}}}{{{\left.{d}{y}\right.}}}}\)
\(\displaystyle\therefore v+{y}{\frac{{{d}{v}}}{{{\left.{d}{y}\right.}}}}=v+{\frac{{{4}}}{{{v}}}}+{4}\)
\(\displaystyle{y}{\frac{{{d}{v}}}{{{\left.{d}{y}\right.}}}}={4}{\left({\frac{{{v}+{1}}}{{{v}}}}\right)}\)
\(\displaystyle{\frac{{{v}{d}{v}}}{{{v}+{1}}}}={4}{\frac{{{\left.{d}{y}\right.}}}{{{y}}}}\)
It becomes variable separable equation. So, now integrate both sides.
\(\displaystyle\int{\frac{{{v}{d}{v}}}{{{v}+{1}}}}={4}\int{\frac{{{\left.{d}{y}\right.}}}{{{y}}}}\)
\(\displaystyle\int{\left({\frac{{{v}+{1}-{1}}}{{{v}+{1}}}}\right)}{d}{v}={4}\int{\frac{{{\left.{d}{y}\right.}}}{{{y}}}}\)
\(\displaystyle\int{\left({1}-{\frac{{{1}}}{{{v}+{1}}}}\right)}{d}{v}={4}{\ln}{\left|{y}\right|}+{\ln{{c}}}\)
\(\displaystyle{v}-{\ln}{\left|{v}+{1}\right|}={\ln{{\left({y}^{{{4}}}{c}\right)}}}\)
\(\displaystyle\therefore{\frac{{{x}}}{{{y}}}}+{\ln}{\left|{\frac{{{x}}}{{{y}}}}+{1}\right|}={\ln}{\left|{y}^{{{4}}}{c}\right|}\)
\(\displaystyle{\frac{{{x}}}{{{y}}}}={\ln}{\left|{y}^{{{4}}}{c}\right|}-{\ln}{\left|{\frac{{{x}+{y}}}{{{y}}}}\right|}\)
\(\displaystyle{\frac{{{x}}}{{{y}}}}={\ln}{\left|{\frac{{{\left({y}^{{{4}}}{c}\right)}}}{{{\left({\frac{{{x}+{y}}}{{{y}}}}\right)}}}}\right|}\)
\(\displaystyle{x}={y}{\ln}{\left|{\frac{{{c}{y}^{{{5}}}}}{{{x}+{y}}}}\right|}\)

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Serita Dewitt
Answered 2022-01-02 Author has 5298 answers
\(\displaystyle{x}{y}{\left.{d}{x}\right.}-{\left({x}+{2}{y}\right)}^{{{2}}}{\left.{d}{y}\right.}={0}\)
\(\displaystyle{x}{y}{\left.{d}{x}\right.}={\left({x}+{2}{y}\right)}^{{{2}}}{\left.{d}{y}\right.}\)
\(\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}={\frac{{{\left({x}+{2}{y}\right)}^{{{2}}}}}{{{x}{y}}}}\)
\(\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}={\frac{{{x}^{{{2}}}+{4}{y}^{{{2}}}+{4}{x}{y}}}{{{x}{y}}}}\)
\(\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}={\frac{{{x}}}{{{y}}}}+{4}{\frac{{{y}}}{{{x}}}}+{4}\).
Substituting; \(\displaystyle{x}={y}{v}\)
Diff. w.r.t. y
\(\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}={v}+{y}{\frac{{{d}{v}}}{{{\left.{d}{y}\right.}}}}\)
\(\displaystyle{v}+{y}{\frac{{{d}{v}}}{{{\left.{d}{y}\right.}}}}={v}+{4}{\frac{{{1}}}{{{v}}}}+{4}\)
\(\displaystyle\neg{\left\lbrace{v}\right\rbrace}+{y}{\frac{{{d}{v}}}{{{\left.{d}{y}\right.}}}}=\neg{\left\lbrace{v}\right\rbrace}+{4}{\left({\frac{{{1}}}{{{v}}}}+{1}\right)}\)
\(\displaystyle{\frac{{{y}}}{{{4}}}}{\frac{{{d}{v}}}{{{\left.{d}{y}\right.}}}}={\frac{{{1}+{v}}}{{{v}}}}\)
Integrating \(\displaystyle\int{\frac{{{v}}}{{{v}+{1}}}}{d}{v}={4}\int{\frac{{{1}}}{{{y}}}}{\left.{d}{y}\right.}\)
\(\displaystyle\int{\frac{{{v}+{1}-{1}}}{{{v}+{1}}}}{d}{v}={4}\int{\frac{{{1}}}{{{y}}}}{\left.{d}{y}\right.}\)
\(\displaystyle{v}-{\ln}{\left|{v}+{1}\right|}={4}{\log{{y}}}+{c}\)
applying \(\displaystyle{v}={\frac{{{x}}}{{{y}}}}\)
\(\displaystyle{\frac{{{x}}}{{{y}}}}-{\ln}{\left|{\frac{{{x}}}{{{y}}}}+{1}\right|}={4}{\log{{y}}}+{c}\)
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karton
Answered 2022-01-10 Author has 8454 answers

\(\begin{array}{} xydx-(x+2y)^{2}dy=0 \\xydx=(x+2y)^{2}dy \\\frac{dx}{dy}=\frac{x^{2}+4y^{2}+4xy}{xy} \\\frac{dx}{dy}=(\frac{x}{y})+4(\frac{y}{x})+4 \\\frac{x}{y}=v \Rightarrow x=vy \\\frac{dx}{dy}=v+y \frac{dv}{dy} \\\therefore \not{v}+y \frac{dv}{dy}=\not{v}+\frac{4}{v}+4 \\y \frac{dv}{dy}=4 (\frac{v+1}{v}) \\\frac{vdv}{v+1}=4 \frac{dy}{y} \\\int \frac{vdv}{v+1}=4 \int \frac{dy}{y} \\\int (\frac{v+1-1}{v+1})dv=4 \int \frac{dy}{y} \\\int (1-\frac{1}{v+1})dv=4 \ln |y|+\ln c \\v-\ln |v+1|=\ln (y^{4}c) \\\therefore \frac{x}{y}+\ln |\frac{x}{y}+1|=\ln |y^{4}c| \\\frac{x}{y}=\ln |y^{4}c|-\ln |\frac{x+y}{y}| \\\frac{x}{y}=\ln |\frac{(y^{4}c)}{(\frac{x+y}{y})}| \\x=y \ln |\frac{cy^{5}}{x+y}| \end{array}\)

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