Classify the following differential equations as separable, homogeneous, parallel line,

obrozenecy6 2021-12-31 Answered
Classify the following differential equations as separable, homogeneous, parallel line, or exact. Explain briefly your answers. Then, solve each equation according to their classification. \(\displaystyle{\left({2}{x}-{3}{y}\right)}{\left.{d}{x}\right.}+{\left({2}{y}-{3}{x}\right)}{\left.{d}{y}\right.}={0}\)

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Expert Answer

Archie Jones
Answered 2022-01-01 Author has 2734 answers
Classsify the following differential eq.
\(\displaystyle{\left({2}{x}-{3}{y}\right)}{\left.{d}{x}\right.}+{\left({2}{y}-{3}{x}\right)}{\left.{d}{y}\right.}={0}\)
Solution:
\(\displaystyle{\left.{d}{y}\right.}{\left({2}{y}-{3}{x}\right)}=-{\left({2}{x}-{3}{y}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}=-{\frac{{{\left({2}{x}-{3}{y}\right)}}}{{{\left({2}{y}-{3}{x}\right)}}}}\)
Finding homogeneous
\(\displaystyle{F}{\left(\lambda{x},\lambda{y}\right)}={\frac{{-{\left({2}\lambda{x}-{3}\lambda{y}\right)}}}{{{\left({2}\lambda{y}-{3}\lambda{x}\right)}}}}\)
\(\displaystyle={\frac{{-\lambda{\left({2}{x}-{3}{y}\right)}}}{{{\left({2}{y}-{3}{x}\right)}}}}\)
\(\displaystyle={\frac{{-{\left({2}{x}-{3}{y}\right)}}}{{{\left({2}{x}-{3}{x}\right)}}}}\)
\(\displaystyle=\lambda^{{\circ}}{F}{\left({r},{y}\right)}\)
The given equation is homogenous.
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Robert Pina
Answered 2022-01-02 Author has 1730 answers
Simplifying
\(\displaystyle{\left({2}{x}+-{3}{y}\right)}\cdot{\left.{d}{x}\right.}+{\left({2}{y}+-{3}{x}\right)}\cdot{\left.{d}{y}\right.}={0}\)
Reorder the terms for easier multiplication:
\(\displaystyle{\left.{d}{x}\right.}{\left({2}{x}+-{3}{y}\right)}+{\left({2}{y}+-{3}{x}\right)}\cdot{\left.{d}{y}\right.}={0}\)
\(\displaystyle{\left({2}{x}\cdot{\left.{d}{x}\right.}+-{3}{y}\cdot{\left.{d}{x}\right.}\right)}+{\left({2}{y}+-{3}{x}\right)}\cdot{\left.{d}{y}\right.}={0}\)
Reorder the terms:
\(\displaystyle{\left(-{3}{\left.{d}{x}\right.}{y}+{2}{\left.{d}{x}\right.}^{{{2}}}\right)}+{\left({2}{y}+-{3}{x}\right)}\cdot{\left.{d}{y}\right.}={0}\)
\(\displaystyle{\left(-{3}{\left.{d}{x}\right.}{y}+{2}{\left.{d}{x}\right.}^{{{2}}}\right)}+{\left({2}{y}+-{3}{x}\right)}\cdot{\left.{d}{y}\right.}={0}\)
Reorder the terms:
\(\displaystyle-{3}{\left.{d}{x}\right.}{y}+{2}{\left.{d}{x}\right.}^{{{2}}}+{\left(-{3}{x}+{2}{y}\right)}\cdot{\left.{d}{y}\right.}={0}\)
Reorder the terms for easier multiplication:
\(\displaystyle-{3}{\left.{d}{x}\right.}{y}+{2}{\left.{d}{x}\right.}^{{{2}}}+{\left.{d}{y}\right.}{\left(-{3}{x}+{2}{y}\right)}={0}\)
\(\displaystyle-{3}{\left.{d}{x}\right.}{y}+{2}{\left.{d}{x}\right.}^{{{2}}}+{\left(-{3}{x}\cdot{\left.{d}{y}\right.}+{2}{y}\cdot{\left.{d}{y}\right.}\right)}={0}\)
\(\displaystyle-{3}{\left.{d}{x}\right.}{y}+{2}{\left.{d}{x}\right.}^{{{2}}}+{\left(-{3}{\left.{d}{x}\right.}{y}+{2}{\left.{d}{y}\right.}^{{{2}}}\right)}={0}\)
Reorder the terms:
\(\displaystyle-{3}{\left.{d}{x}\right.}{y}+-{3}{\left.{d}{x}\right.}{y}+{2}{\left.{d}{x}\right.}^{{{2}}}+{2}{\left.{d}{y}\right.}^{{{2}}}={0}\)
Combine like terms: \(\displaystyle-{3}{\left.{d}{x}\right.}{y}+-{3}{\left.{d}{x}\right.}{y}=-{6}{\left.{d}{x}\right.}{y}\)
\(\displaystyle-{6}{\left.{d}{x}\right.}{y}+{2}{\left.{d}{x}\right.}^{{{2}}}+{2}{\left.{d}{y}\right.}^{{{2}}}={0}\)
Solving
\(\displaystyle-{6}{\left.{d}{x}\right.}{y}+{2}{\left.{d}{x}\right.}^{{{2}}}+{2}{\left.{d}{y}\right.}^{{{2}}}={0}\)
Solving for variable 'd'.
Move all terms containing d to the left, all other terms to the right.
Factor out the Greatest Common Factor (GCF), '2d'.
\(\displaystyle{2}{d}{\left(-{3}{x}{y}+{x}^{{{2}}}+{y}^{{{2}}}\right)}={0}\)
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karton
Answered 2022-01-09 Author has 8454 answers

\(2x-3y+(2y-3x) \frac{dy}{dx}=0\)
\(2x-3y+(2y-3x)y'=0\)
Verify that \(\frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x}\): True
Find \((x,y):(x,y)=y^{2}-3xy+x^{2}+c_{1}\)
\(y^{2}-3xy+x^{2}+c_{1}=c_{2}\)
\(y^{2}-3xy+x^{2}=c_{1}\)
Isolate y: \(y=\frac{3x+\sqrt{5x^{2}+4c_{1}}}{2}, y=\frac{3x-\sqrt{5x^{2}+4c_{1}}}{2}\)
\(y=\frac{3x+\sqrt{5x^{2}+c_{1}}}{2}, y=\frac{3x-\sqrt{5x^{2}+c_{1}}}{2}\)

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